problem_id
stringlengths 32
32
| link
stringlengths 75
84
| problem
stringlengths 14
5.33k
| solution
stringlengths 15
6.63k
| letter
stringclasses 5
values | answer
stringclasses 957
values |
|---|---|---|---|---|---|
b2826c174d9431679aeb59f06d277da5 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_2 | Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$ | Let $b$ be the number of days Walter does his chores but doesn't do them well, and let $w$ be the number of days he does his chores exceptionally well.
$b + w = 10$ since there are $10$ days Walter does chores.
$3b + 5w = 36$ since $3b$ is the amount he earns from doing his chores not well, and $5w$ is the amount he earnes from doing his chores exceptionally well, and those two sum to $36$ dollars.
Multiply the first equation by $3$ to get:
$3b + 3w = 30$
$3b + 5w = 36$
Subtract the first equation from the second equation to get:
$5w - 3w = 36 - 30$
$2w = 6$
$w = 3$
Thus, he does his chores well $3$ days, and the answer is $\boxed{3}$ | A | 3 |
efd835768e55e5604e1bdc1a9196af75 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_3 | $\frac{(3!)!}{3!}=$
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 6\qquad\text{(D)}\ 40\qquad\text{(E)}\ 120$ | The numerator is $(3!)! = 6!$
The denominator is $3! = 6$
Using the property that $6! = 6 \cdot 5!$ in the numerator, the sixes cancel, leaving $5! = 120$ , which is answer $\boxed{120}$ | E | 120 |
e288592273e7dd06678f9b4ea49af7a5 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_4 | Six numbers from a list of nine integers are $7,8,3,5,9$ and $5$ . The largest possible value of the median of all nine numbers in this list is
$\text{(A)}\ 5\qquad\text{(B)}\ 6\qquad\text{(C)}\ 7\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ | First, put the six numbers we have in order, since we are concerned with the median: $3, 5, 5, 7, 8, 9$
We have three more numbers to insert into the list, and the median will be the $5^{th}$ highest (and $5^{th}$ lowest) number on the list. If we top-load the list by making all three of the numbers greater than $9$ , the median will be the highest it can possibly be. Thus, the maximum median is the fifth piece of data in the list, which is $8$ , giving an answer of $\boxed{8}$ | D | 8 |
d4406b8d60642c3213217a4fed931fe4 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_7 | A father takes his twins and a younger child out to dinner on the twins' birthday. The restaurant charges $4.95$ for the father and $0.45$ for each year of a child's age, where age is defined as the age at the most recent birthday. If the bill is $9.45$ , which of the following could be the age of the youngest child?
$\text{(A)}\ 1\qquad\text{(B)}\ 2\qquad\text{(C)}\ 3\qquad\text{(D)}\ 4\qquad\text{(E)}\ 5$ | The bill for the three children is $9.45 - 4.95 = 4.50$ . Since the charge is $0.45$ per year for the children, they must have $\frac{4.50}{0.45} = 10$ years among the three of them.
The twins must have an even number of years in total (presuming that they did not dine in the 17 minutes between the time when the first twin was born and the second twin was born). If we let the twins be $5$ years old, that leaves $10 - 2\cdot 5 = 0$ years leftover for the youngest child. But if the twins are each $4$ years old, then the youngest child could be $10 - 2\cdot 4 = 2$ years old, which is choice $\boxed{2}$ | B | 2 |
6001c3953fdccce55c19fc80abff4de6 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_10 | How many line segments have both their endpoints located at the vertices of a given cube
$\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$ | There are $8$ choices for the first endpoint of the line segment, and $7$ choices for the second endpoint, giving a total of $8\cdot 7 = 56$ segments. However, both $\overline{AB}$ and $\overline{BA}$ were counted, while they really are the same line segment. Every segment got double counted in a similar manner, so there are really $\frac{56}{2} = 28$ line segments, and the answer is $\boxed{28}$ | D | 28 |
6001c3953fdccce55c19fc80abff4de6 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_10 | How many line segments have both their endpoints located at the vertices of a given cube
$\text{(A)}\ 12\qquad\text{(B)}\ 15\qquad\text{(C)}\ 24\qquad\text{(D)}\ 28\qquad\text{(E)}\ 56$ | Each segment is either an edge, a facial diagonal, or a long/main/spacial diagonal.
A cube has $12$ edges: Four on the top face, four on the bottom face, and four that connect the top face to the bottom face.
A cube has $6$ square faces, each of which has $2$ facial diagonals, for a total of $6\cdot 2 = 12$
A cube has $4$ spacial diagonals: each diagonal goes from one of the bottom vertices to the "opposite" top vertex.
Thus, there are $12 + 12 + 4 = 28$ segments, and the answer is $\boxed{28}$ | D | 28 |
cfff55550e63a0096fa45d2a2fecc021 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_12 | A function $f$ from the integers to the integers is defined as follows:
\[f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ n/2 &\text{if n is even}\end{cases}\]
Suppose $k$ is odd and $f(f(f(k))) = 27$ . What is the sum of the digits of $k$
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 15$ | Going out the final step, if you have $f(f(f(k))) = 27$ , you would have to have $f(f(k))) = 51$ or $f(f(k)) = 108$
If you doubled either of these, $k$ would not be odd. So you must subtract $3$
If you subtract $3$ from $51$ , you would compute $f(48)$ , which would halve it, and not add the $3$ back.
If you subtract $3$ from $108$ , you would compute $f(105)$ , which would add the $3$ back.
Thus, $f(f(f(105))) = f(f(108)) = f(54) = 27$ , and $105$ is odd. The desired sum of the digits is $6$ , and the answer is $\boxed{6}$ | B | 6 |
ed7d0c8df6a6d8017d85157a2ba20847 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_14 | Let $E(n)$ denote the sum of the even digits of $n$ . For example, $E(5681) = 6+8 = 14$ . Find $E(1)+E(2)+E(3)+\cdots+E(100)$
$\text{(A)}\ 200\qquad\text{(B)}\ 360\qquad\text{(C)}\ 400\qquad\text{(D)}\ 900\qquad\text{(E)}\ 2250$ | The problem is asking for the sum of all the even digits in the numbers $1$ to $100$ . We can remove $100$ from the list, add $00$ to the list, and tack on some leading zeros to the single digit numbers without changing the sum of the even digits. This gives the list:
$00, 01, 02, 03, ..., 10, 11, ..., 98, 99$
There are $2\cdot 100 = 200$ digits on that list, and each digit appears $\frac{200}{10} = 20$ times.
Thus, each even digit appears $20$ times, and the sum of all the even digits is $0 \cdot 20 + 2\cdot 20 + 4\cdot 20 + 6\cdot 20 + 8\cdot 20 = (0 + 2 + 4 + 6 + 8)\cdot 20 = 400$ , and the correct answer is $\boxed{400}$ | C | 400 |
00d3433e73bb281384e53ad5a87cdc86 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_17 | In rectangle $ABCD$ , angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$ , where $E$ is on $\overline{AB}$ $F$ is on $\overline{AD}$ $BE=6$ and $AF=2$ . Which of the following is closest to the area of the rectangle $ABCD$ [asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--cycle, linewidth(0.8)); draw(E--C--F); dot(A^^B^^C^^D^^E^^F); label("$A$", A, dir((5, 3.5)--A)); label("$B$", B, dir((5, 3.5)--B)); label("$C$", C, dir((5, 3.5)--C)); label("$D$", D, dir((5, 3.5)--D)); label("$E$", E, dir((5, 3.5)--E)); label("$F$", F, dir((5, 3.5)--F)); label("$2$", (0,1), dir(0)); label("$6$", (7.5,0), N);[/asy] $\text{(A)}\ 110\qquad\text{(B)}\ 120\qquad\text{(C)}\ 130\qquad\text{(D)}\ 140\qquad\text{(E)}\ 150$ | Since $\angle C = 90^\circ$ , each of the three smaller angles is $30^\circ$ , and $\triangle BEC$ and $\triangle CDF$ are both $30-60-90$ triangles.
[asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--cycle, linewidth(0.8)); draw(E--C--F); dot(A^^B^^C^^D^^E^^F); label("$A$", A, dir((5, 3.5)--A)); label("$B$", B, dir((5, 3.5)--B)); label("$C$", C, dir((5, 3.5)--C)); label("$D$", D, dir((5, 3.5)--D)); label("$E$", E, dir((5, 3.5)--E)); label("$F$", F, dir((5, 3.5)--F)); label("$2$", (0,1), plain.E, fontsize(10)); label("$x$", (9,3.5), E, fontsize(10)); label("$x-2$", (0,5), plain.E, fontsize(10)); label("$y$", (5,7), N, fontsize(10)); label("$6$", (7.5,0), S, fontsize(10));[/asy]
Defining the variables as illustrated above, we have $x = 6\sqrt{3}$ from $\triangle BEC$
Then $x-2 = 6\sqrt{3} - 2$ , and $y = \sqrt{3} (6 \sqrt{3} - 2) = 18 - 2\sqrt{3}$
The area of the rectangle is thus $xy = 6\sqrt{3}(18 - 2\sqrt{3}) = 108\sqrt{3} - 36$
Using the approximation $\sqrt{3} \approx 1.7$ , we get an area of just under $147.6$ , which is closest to answer $\boxed{3} > 1.7$ ). | E | 3 |
00d3433e73bb281384e53ad5a87cdc86 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_17 | In rectangle $ABCD$ , angle $C$ is trisected by $\overline{CF}$ and $\overline{CE}$ , where $E$ is on $\overline{AB}$ $F$ is on $\overline{AD}$ $BE=6$ and $AF=2$ . Which of the following is closest to the area of the rectangle $ABCD$ [asy] pair A=origin, B=(10,0), C=(10,7), D=(0,7), E=(5,0), F=(0,2); draw(A--B--C--D--cycle, linewidth(0.8)); draw(E--C--F); dot(A^^B^^C^^D^^E^^F); label("$A$", A, dir((5, 3.5)--A)); label("$B$", B, dir((5, 3.5)--B)); label("$C$", C, dir((5, 3.5)--C)); label("$D$", D, dir((5, 3.5)--D)); label("$E$", E, dir((5, 3.5)--E)); label("$F$", F, dir((5, 3.5)--F)); label("$2$", (0,1), dir(0)); label("$6$", (7.5,0), N);[/asy] $\text{(A)}\ 110\qquad\text{(B)}\ 120\qquad\text{(C)}\ 130\qquad\text{(D)}\ 140\qquad\text{(E)}\ 150$ | Use the process above, but use $\sqrt{3} \approx 1.73$ . You should get $[ABCD]=150.84$ , which then you select $\boxed{150}$ . Notice that the actual area, when plugged into a calculator, yields about $151.0614872$ | E | 150 |
75add752db07c95c2c28c9d00666ff89 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_23 | The sum of the lengths of the twelve edges of a rectangular box is $140$ , and
the distance from one corner of the box to the farthest corner is $21$ . The total surface area of the box is
$\text{(A)}\ 776\qquad\text{(B)}\ 784\qquad\text{(C)}\ 798\qquad\text{(D)}\ 800\qquad\text{(E)}\ 812$ | Let $x, y$ , and $z$ be the unique lengths of the edges of the box. Each box has $4$ edges of each length, so: \[4x + 4y + 4z = 140 \ \Longrightarrow \ x + y + z = 35.\] The spacial diagonal (longest distance) is given by $\sqrt{x^2 + y^2 + z^2}$ . Thus, we have $\sqrt{x^2 + y^2 + z^2} = 21$ , so $x^2 + y^2 + z^2 = 21^2$
Our target expression is the surface area of the box:
\[S = 2xy + 2xz + 2yz.\]
Since $S$ is a symmetric polynomial of degree $2$ , we try squaring the first equation to get:
\[35^2 = (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy +2yz + 2xz = 35^2.\]
Substituting in our long diagonal and surface area expressions, we get: $21^2 + S = 35^2$ , so $S = (35 + 21)(35 - 21) = 56\cdot 14 = 784$ , which is option $\boxed{784}$ | B | 784 |
9687910f26e9baa522fa5e40c2548f90 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_24 | The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$ ’s separated by blocks of $2$ ’s with $n$ $2$ ’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is
$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449$ | The sum of the first $1$ numbers is $1$
The sum of the next $2$ numbers is $2 + 1$
The sum of the next $3$ numbers is $2 + 2 + 1$
In general, we can write "the sum of the next $n$ numbers is $1 + 2(n-1)$ ", where the word "next" follows the pattern established above.
Thus, we first want to find what triangular numbers $1234$ is between. By plugging in various values of $n$ into $f(n) = \frac{n(n+1)}{2}$ , we find:
$f(50) = 1275$
$f(49) = 1225$
Thus, we want to add up all those sums from "next $1$ number" to the "next $49$ numbers", which will give us all the numbers up to and including the $1225^{th}$ number. Then, we can manually tack on the remaining $2$ s to hit $1234$
We want to find:
$\sum_{n=1}^{49} 1 + 2(n-1)$
$\sum_{n=1}^{49} 2n - 1$
$\sum_{n=1}^{49} 2n - \sum_{n=1}^{49} 1$
$2 \sum_{n=1}^{49} n - 49$
$2\cdot \frac{49\cdot 50}{2} - 49$
$49^2$
$2401$
Thus, the sum of the first $1225$ terms is $2401$ . We have to add $9$ more $2$ s to get to the $1234^{th}$ term, which gives us $2419$ , or option $\boxed{2419}$ | B | 2419 |
9687910f26e9baa522fa5e40c2548f90 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_24 | The sequence $1, 2, 1, 2, 2, 1, 2, 2, 2, 1, 2, 2, 2, 2, 1, 2, 2, 2, 2, 2, 1, 2,\ldots$ consists of $1$ ’s separated by blocks of $2$ ’s with $n$ $2$ ’s in the $n^{th}$ block. The sum of the first $1234$ terms of this sequence is
$\text{(A)}\ 1996\qquad\text{(B)}\ 2419\qquad\text{(C)}\ 2429\qquad\text{(D)}\ 2439\qquad\text{(E)}\ 2449$ | The $k$ th appearance of 1 is at position $1 + 2 + \dots + k = \frac{k(k + 1)}{2}$ . Then there are $k$ 1's and $\frac{k(k + 1)}{2} - k = \frac{k(k - 1)}{2}$ 2's among the first $\frac{k(k + 1)}{2}$ numbers, so the sum of these $\frac{k(k + 1)}{2}$ terms is $k + k(k - 1) = k^2$
When $k = 49$ $\frac{k(k + 1)}{2} = 1225$ , and when $k = 50$ $\frac{k(k + 1)}{2} = 1275$
The sum of the first 1225 terms is $49^2 = 2401$ . The numbers in positions 1226 through 1234 are all 2's, so their sum is $(1234 - 1226 + 1) \cdot 2 = 18$ . Therefore, the sum of the first 1234 terms is $2401 + 18 = \boxed{2419}$ | null | 2419 |
48cb3b38af65ce8b3febd28bb25df7de | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | Complete the square to get \[(x-7)^2 + (y-3)^2 = 64.\] Applying Cauchy-Schwarz directly, \[64\cdot25=(3^2+4^2)((x-7)^2 + (y-3)^2) \ge (3(x-7)+4(y-3))^2.\] \[40 \ge 3x+4y-33\] \[3x+4y \le 73.\] Thus our answer is $\boxed{73}$ | B | 73 |
48cb3b38af65ce8b3febd28bb25df7de | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | The first equation is a circle , so we find its center and radius by completing the square $x^2 - 14x + y^2 - 6y = 6$ , so \[(x-7)^2 + (y-3)^2 = (x^2- 14x + 49) + (y^2 - 6y + 9) = 6 + 49 + 9 = 64.\]
So we have a circle centered at $(7,3)$ with radius $8$ , and we want to find the max of $3x + 4y$
The set of lines $3x + 4y = A$ are all parallel , with slope $-\frac{3}{4}$ . Increasing $A$ shifts the lines up and/or to the right.
We want to shift this line up high enough that it's tangent to the circle, but not so high that it misses the circle altogether. This means $3x + 4y = A$ will be tangent to the circle.
Imagine that this line hits the circle at point $(a,b)$ . The slope of the radius connecting the center of the circle, $(7,3)$ , to tangent point $(a,b)$ will be $\frac{4}{3}$ , since the radius is perpendicular to the tangent line.
So we have a point, $(7,3)$ , and a slope of $\frac{4}{3}$ that represents the slope of the radius to the tangent point. Let's start at the point $(7,3)$ . If we go $4k$ units up and $3k$ units right from $(7,3)$ , we would arrive at a point that's $5k$ units away. But in reality we want $5k = 8$ to reach the tangent point, since the radius of the circle is $8$
Thus, $k = \frac{8}{5}$ , and we want to travel $4\cdot \frac{8}{5}$ up and $3\cdot \frac{8}{5}$ over from the point $(7,3)$ to reach our maximum. This means the maximum value of $3x + 4y$ occurs at $\left(7 +3\cdot \frac{8}{5}, 3 + 4\cdot \frac{8}{5}\right)$ , which is $\left(\frac{59}{5}, \frac{47}{5}\right).$
Plug in those values for $x$ and $y$ , and you get the maximum value of $3x + 4y = 3\cdot\frac{59}{5} + 4\cdot\frac{47}{5} = \boxed{73}$ | B | 73 |
48cb3b38af65ce8b3febd28bb25df7de | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | Let the tangent point be $P$ , and the tangent line's x-intercept be $Q$ . Consider the horizontal line starting from center of circle (O) meeting the tangent line at K. Now triangle $OPK$ is 3-4-5, $OP=8$ , so $OK = \frac{5}{3}*8 = \frac{40}{3}$ . Note that the horizontal distance from $O$ to the origin is $7$ , and the horizontal distance from K to Q is 4, ( $\frac{4}{3}$ of its y coordinate), so the x-intercept is $7+4+OK = 73/3$ . The value of $3x+4y$ is 73 at point $Q$ . Note that this value is constant on the tangent line, so there is no need to calculate the coordinate of $P$ $\boxed{73}$ | B | 73 |
48cb3b38af65ce8b3febd28bb25df7de | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | Let $z = 3x + 4y$ . Solving for $y$ , we get $y = (z - 3x)/4$ . Substituting into the given equation, we get \[x^2 + \left( \frac{z - 3x}{4} \right)^2 = 14x + 6 \cdot \frac{z - 3x}{4} + 6,\] which simplifies to \[25x^2 - (6z + 152)x + (z^2 - 24z - 96) = 0.\]
This quadratic equation has real roots in $x$ if and only if its discriminant is nonnegative, so \[(6z + 152)^2 - 4 \cdot 25 \cdot (z^2 - 24z - 96) \ge 0,\] which simplifies to \[-64z^2 + 4224z + 32704 \ge 0,\] which can be factored as \[-64(z + 7)(z - 73) \ge 0.\] The largest value of $z$ that satisfies this inequality is $\boxed{73}$ | B | 73 |
48cb3b38af65ce8b3febd28bb25df7de | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | First, we move all the non-constant terms of the constraint to one side and assign it to the function $g(x,y)$ \[g(x,y)=x^2+y^2-14x-6y.\] Since we are trying to maximize $f(x,y)=3x+4y$ , we need to solve for $x$ and $y$ in the system \[\begin{cases}x^2+y^2-14x-6y=6,\\\nabla g(x,y)=\lambda\nabla f(x,y).\end{cases}\] We have that \begin{align*}\nabla f(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}3x+4y\\\dfrac{\partial}{\partial y}3x+4y\end{pmatrix}\\&=\begin{pmatrix}3\\4\end{pmatrix},\end{align*} and \begin{align*}\nabla g(x,y)&=\begin{pmatrix}\dfrac{\partial}{\partial x}x^2+y^2-14x-6y\\\dfrac{\partial}{\partial y}x^2+y^2-14x-6y\end{pmatrix}\\&=\begin{pmatrix}2x-14\\2y-6\end{pmatrix}\end{align*}. To solve the original system, we can solve for $x$ and $y$ in terms of $\lambda$ using our equations from the gradients, then substitute them into the first equation. We have that $x=\frac{3}{2}\lambda+7$ and $y=2\lambda+3$ . Substituting into the first equation, we have that \begin{align*}\left(\frac{3}{2}\lambda+7\right)^2+(2\lambda+3)^2-14\left(\frac{3}{2}\lambda+7\right)-6(2\lambda+3)&=6\\\frac{25}{4}\lambda^2&=64\\\lambda&=\pm\frac{16}{5}\end{align*} Using the solutions of $x$ and $y$ in terms of $\lambda$ that we found earlier, we have that \[3x+4y=\frac{25}{2}\lambda+33.\] Because we are trying to maximize this function, we will use the positive solution for $\lambda$ . Therefore, after substituting, we have that the largest value of $g(x,y)$ that satisfies $f(x,y)=6$ is $\boxed{73}$ | B | 73 |
48cb3b38af65ce8b3febd28bb25df7de | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | Completing the square gives the equation of a circle, $(x-7)^2+(y-3)^2=64.$ Seeing that we would like to maximize $3x+4y,$ we parameterize the circle using polar coordinates: \begin{align*} x&=7+8\cos\theta&y&=3+8\sin\theta. \end{align*} Then, we have $3x+4y=3(7+8\cos\theta)+4(3+8\sin\theta)=33+24\cos\theta+32\sin\theta.$ Since $a\cos\theta+b\sin\theta\le\sqrt{a^2+b^2},$ the desired answer is $33+\sqrt{24^2+32^2}=33+40=\boxed{73}.$ - Ultroid999OCPN | null | 73 |
48cb3b38af65ce8b3febd28bb25df7de | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_25 | Given that $x^2 + y^2 = 14x + 6y + 6$ , what is the largest possible value that $3x + 4y$ can have?
$\text{(A)}\ 72\qquad\text{(B)}\ 73\qquad\text{(C)}\ 74\qquad\text{(D)}\ 75\qquad\text{(E)}\ 76$ | The equation $x^2 + y^2 = 14x + 6y + 6$ can be written\[
(x-7)^2 + (y-3)^2 = 8^2,
\]which defines a circle of radius 8 centered at $(7,3)$ . If $k$ is a possible value of $3x + 4y$ for $(x,y)$ on the circle, then the line $3x + 4y = k$ must intersect the circle in at least one point. The largest value of $k$ occurs when the line is tangent to the circle, and is therefore perpendicular to the radius at the point of tangency. Because the slope of the tangent line is $-3/4$ the slope of the radius is $\ 4/3$ . It follows that the point on the circle that yields the maximum value of $3x + 4y$ is one of the two points of tangency,\[
x = 7 + \frac{3 \cdot 8}{5} = \frac{59}{5}, \hspace{.3in} y = 3 +
\frac{4 \cdot 8}{5} = \frac{47}{5},
\]or\[
x = 7 - \frac{3 \cdot 8}{5} = \frac{11}{5}, \hspace{.3in} y = 3 -
\frac{4 \cdot 8}{5} = - \frac{17}{5}.
\][asy]
import olympiad; import geometry; size(150); defaultpen(linewidth(0.8));
draw((0,15)--origin--(15,0));
dot(" $(7,3)$ ",(7,3),S);
draw(Circle((7,3),8));
line a = line((0,9),(12,0));
line b = line((0,12),(16,0));
line c = line((0,18.3),(18.3*4/3,0));
draw(a^^b^^c);
[/asy]
The first point of tangency gives\[
3x + 4y = 3 \cdot \frac{59}{5} + 4 \cdot \frac{47}{5} =
\frac{177}{5}+\frac{188}{5}= 73,
\]and the second one gives \[\, 3x + 4y = \frac{33}{5}-\frac{68}{5} = -7.\] Thus, $\boxed{73}$ is the desired maximum, while $-7$ is the minimum. | null | 73 |
894a9b642e308ebaea8fc7a1d3d0d85b | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_26 | An urn contains marbles of four colors: red, white, blue, and green. When four marbles are drawn without replacement, the following events are equally likely:
(a) the selection of four red marbles;
(b) the selection of one white and three red marbles;
(c) the selection of one white, one blue, and two red marbles; and
(d) the selection of one marble of each color.
What is the smallest number of marbles satisfying the given condition?
$\text{(A)}\ 19\qquad\text{(B)}\ 21\qquad\text{(C)}\ 46\qquad\text{(D)}\ 69\qquad\text{(E)}\ \text{more than 69}$ | Let the bag contain $n$ marbles total, with $r, w, b, g$ representing the number of red, white, blue, and green marbles, respectively. Note that $r + w + b + g = n$
The number of ways to select four red marbles out of the set of marbles without replacement is:
\[\binom{r}{4} = \frac{r!}{24\cdot (r -4)!}\]
The number of ways to select one white and three red marbles is:
\[\binom{w}{1}\binom{r}{3} = \frac{w\cdot r!}{6\cdot (r - 3)!}\]
The number of ways to select one white, one blue, and two red marbles is:
\[\binom{w}{1}\binom{b}{1} \binom{r}{2} = \frac{wb\cdot r!}{2(r-2)!}\]
The number of ways to select one marble of each colors is:
\[\binom{w}{1}\binom{b}{1} \binom{g}{1}\binom{r}{1} = wbgr\] Setting the first and second statements equal, we find:
\[\frac{r!}{24\cdot (r -4)!} = \frac{w\cdot r!}{6\cdot (r - 3)!}\]
\[r - 3 = 4w\]
Setting the first and third statements equal, we find:
\[\frac{r!}{24\cdot (r -4)!} = \frac{wb\cdot r!}{2(r-2)!}\]
\[(r-3)(r-2) = 12wb\]
Setting the last two statements equal, we find:
\[\frac{wb\cdot r!}{2(r-2)!} = wbgr\]
\[r - 1 = 2g\]
These are all the "linking equations" that are needed; the transitive property of equality makes the other three equalities unnecessary.
From the first equation, we know that $r$ must be $3$ more than a multiple of $4$ , or that $r \equiv 3 \mod 4$
Putting the first equation into the second equation, we find $r-2 = 3b$ . Therefore, $r \equiv 2 \mod 3$ . Using the Chinese Remainder Theorem , we find that $r \equiv 11 \mod 12$
The third equation gives no new restrictions on $r$ ; it is already odd by the first equation.
Thus, the minimal positive value of $r$ is $11$ . This requires $g=\frac{r - 1}{2} = 5$ by the third equation, and $w = \frac{r-3}{4} = 2$ by the first equation. Finally, the second equation gives $b = \frac{(r-3)(r-2)}{12w} = 3$
The minimal total number of marbles is $11 + 5 + 2 + 3 = \boxed{21}$ | B | 21 |
0d44dc0427955a132e766bdee2b095fd | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_27 | Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ with radius $6$ , and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$ . How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$ | The two equations of the balls are
\[x^2 + y^2 + \left(z - \frac{21}{2}\right)^2 \le 36\]
\[x^2 + y^2 + (z - 1)^2 \le \frac{81}{4}\]
Note that along the $z$ axis, the first ball goes from $10.5 \pm 6$ , and the second ball goes from $1 \pm 4.5$ . The only integer value that $z$ can be is $z=5$
Plugging that in to both equations, we get:
\[x^2 + y^2 \le \frac{23}{4}\]
\[x^2 + y^2 \le \frac{17}{4}\]
The second inequality implies the first inequality, so the only condition that matters is the second inequality.
From here, we do casework, noting that $|x|, |y| \le 3$
For $x=\pm 2$ , we must have $y=0$ . This gives $2$ points.
For $x = \pm 1$ , we can have $y\in \{-1, 0, 1\}$ . This gives $2\cdot 3 = 6$ points.
For $x = 0$ , we can have $y \in \{-2, -1, 0, 1, 2\}$ . This gives $5$ points.
Thus, there are $\boxed{13}$ | D | 13 |
0d44dc0427955a132e766bdee2b095fd | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_27 | Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ with radius $6$ , and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$ . How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$ | Because both spheres have their centers on the x-axis, we can simplify the graph a bit by looking at a 2-dimensional plane (the previous z-axis is the new x-axis while the y-axis remains the same).
The spheres now become circles with centers at $(1,0)$ and $(\frac{21}{2},0)$ . They have radii $\frac{9}{2}$ and $6$ , respectively.
Let circle $A$ be the circle centered on $(1,0)$ and circle $B$ be the one centered on $(\frac{21}{2},0)$
The point on circle $A$ closest to the center of circle $B$ is $(\frac{11}{2},0)$ . The point on circle B closest to the center of circle $A$ is $(\frac{11}{2},0)$
Taking a look back at the 3-dimensional coordinate grid with the spheres, we can see that their intersection appears to be a circle with congruent "dome" shapes on either end. Because the tops of the "domes" are at $(0,0,\frac{9}{2})$ and $(0,0,\frac{11}{2})$ , respectively, the lattice points inside the area of intersection must have z-value $5$ (because $5$ is the only integer between $\frac{9}{2}$ and $\frac{11}{2}$ ). Thus, the lattice points in the area of intersection must all be on the 2-dimensional circle. The radius of the circle will be the distance from the z-axis.
Now, looking at the 2-dimensional coordinate plane, we see that the radius of the circle (now the distance from the x-axis, because there is no more z-axis) is the altitude of a triangle with two points on centers of circles $A$ and $B$ and third point at the first quadrant intersection of the circles. Let's call that altitude $h$
We know all three side lengths of this triangle: $\frac{9}{2}$ (the radius of circle $A$ ), $6$ (the radius of circle $B$ ), and $\frac{19}{2}$ (the distance between the centers of circles $A$ and $B$ ). We can now find the area of the triangle using Heron's formula:
\[s=\frac{\frac{9}{2}+6+\frac{19}{2}}{2}=20\]
\[A=\sqrt{(20)(20-\frac{9}{2})(20-6)(20-\frac{19}{2})}=\sqrt{110}\]
Using the area of the triangle, we can find that altitude $h$ from the x-axis:
\[\sqrt{110}=\frac{h*\frac{19}{2}}{2}\]
\[h=\frac{4*\sqrt{110}}{19}\]
Remember, the altitude $h$ is also the radius of the circle containing all the solutions to the problem.
Going back to the 3-dimensional grid and looking at the circle, we can again make the figure 2-dimensional.
The radius $h$ of the circle in a 2-dimensional plane is $\frac{4\sqrt{110}}{19}$ , a little greater than $2$ . We know that $h>2$ because $4\sqrt{110}>4*10$ , and $\frac{40}{19}>2$
Finally, looking at a circle with radius slightly larger than $2$ , we see that there are $\boxed{13}$ | D | 13 |
0d44dc0427955a132e766bdee2b095fd | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_27 | Consider two solid spherical balls, one centered at $\left(0, 0,\frac{21}{2}\right)$ with radius $6$ , and the other centered at $(0, 0, 1)$ with radius $\frac{9}{2}$ . How many points with only integer coordinates (lattice points) are there in the intersection of the balls?
$\text{(A)}\ 7\qquad\text{(B)}\ 9\qquad\text{(C)}\ 11\qquad\text{(D)}\ 13\qquad\text{(E)}\ 15$ | Note that the spheres are on the same $x$ and $y$ axis. Therefore, we can draw the spheres so that only the $x$ and $z$ axis are featured.
$A = (1, 0, 5)$ $\sqrt{1^2+4^2} = \sqrt{17}<\frac92$ $A$ is inside the smaller sphere. $\sqrt{1^2+(\frac{11}{2})^2} = \frac{\sqrt{126}}{2}<6$ $A$ is inside the larger sphere. Point $A$ is inside the intersection.
$B = (2, 0, 5)$ $\sqrt{2^2+4^2} = \sqrt{20}<\frac92$ $B$ is inside the smaller sphere. $\sqrt{2^2+(\frac{11}{2})^2} = \frac{\sqrt{137}}{2}<6$ $B$ is inside the larger sphere. Point $B$ is inside the intersection.
$C = (3, 0, 5)$ $\sqrt{3^2+4^2} = \sqrt{25}>\frac92$ $C$ is outside the smaller sphere. $\sqrt{3^2+(\frac{11}{2})^2} = \frac{\sqrt{157}}{2}>6$ $C$ is outside the larger sphere. Point $C$ is outside the intersection.
The intersection of $2$ spheres is a circle, by drawing the circle flat we can see that there are $4$ more points within the intersection.
We can see that points with the same $x$ -coordinates with $A$ are still inside the circle. $B$ is very close to the circle, the points with the same $x$ -coordinates with $B$ are not necessarily inside the circle.
For example, $D = (2,1,5)$ $\sqrt{2^2+1^2+4^2}=\sqrt{21}>\frac92$ $D$ is outside the smaller sphere. $\sqrt{2^2 + 1^2 + (\frac{11}{2})^2} = \frac{\sqrt{141}}{2}<6$ $D$ is inside the larger sphere. Point $D$ is outside the intersection.
Therefore, the answer is $9+4 = \boxed{13}$ | D | 13 |
9e7a8d12ca38d275230b7d86177e347b | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_28 | On a $4\times 4\times 3$ rectangular parallelepiped, vertices $A$ $B$ , and $C$ are adjacent to vertex $D$ . The perpendicular distance from $D$ to the plane containing $A$ $B$ , and $C$ is closest to
[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, S); label("$D$", D, E); label("$4$", (4,2,0), SW); label("$4$", (2,4,0), SE); label("$3$", (0, 4, 1.5), E); [/asy]
$\text{(A)}\ 1.6\qquad\text{(B)}\ 1.9\qquad\text{(C)}\ 2.1\qquad\text{(D)}\ 2.7\qquad\text{(E)}\ 2.9$ | By placing the cube in a coordinate system such that $D$ is at the origin, $A(0,0,3)$ $B(4,0,0)$ , and $C(0,4,0)$ , we find that the equation of plane $ABC$ is:
\[\frac{x}{4} + \frac{y}{4} + \frac{z}{3} = 1,\] so $3x + 3y + 4z - 12 = 0.$ The equation for the distance of a point $(a,b,c)$ to a plane $Ax + By + Cz + D = 0$ is given by:
\[\frac{|Aa + Bb + Cc + D|}{\sqrt{A^2 + B^2 + C^2}}.\]
Note that the capital letters are coefficients, while the lower case is the point itself. Thus, the distance from the origin (where $a=b=c=0$ ) to the plane is given by:
\[\frac{D}{\sqrt{A^2 + B^2 + C^2}} = \frac{12}{\sqrt{9 + 9 + 16}} = \frac{12}{\sqrt{34}}.\]
Since $\sqrt{34} < 6$ , this number should be just a little over $2$ , and the correct answer is $\boxed{2.1}$ | C | 2.1 |
9e7a8d12ca38d275230b7d86177e347b | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_28 | On a $4\times 4\times 3$ rectangular parallelepiped, vertices $A$ $B$ , and $C$ are adjacent to vertex $D$ . The perpendicular distance from $D$ to the plane containing $A$ $B$ , and $C$ is closest to
[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, S); label("$D$", D, E); label("$4$", (4,2,0), SW); label("$4$", (2,4,0), SE); label("$3$", (0, 4, 1.5), E); [/asy]
$\text{(A)}\ 1.6\qquad\text{(B)}\ 1.9\qquad\text{(C)}\ 2.1\qquad\text{(D)}\ 2.7\qquad\text{(E)}\ 2.9$ | Let $x$ be the desired distance. Recall that the volume of a pyramid is given by $\frac{1}{3}\cdot h \cdot B$ , where $B$ is the area of the base and $h$ is the height. Consider pyramid $ABCD$ . Letting $ABC$ be the base, the volume of $ABCD$ is given by $\frac{1}{3} \cdot x \cdot [ABC]$ , but if we let $BCD$ be the base, the volume is given by $\frac{1}{3} \cdot [BCD]\cdot [AD] = \frac{1}{3} \cdot [\frac{1}{2} \cdot 4 \cdot 4] \cdot 3 = 8$ . Clearly, these two volumes must be equal, so we get the equation $\frac{1}{3}\cdot x \cdot[ABC]=8$ . Thus, to find $x$ , we just need to find $[ABC]$
By the Pythagorean Theorem, $AB=\sqrt{AD^2+DB^2}=5$ $AC=\sqrt{AD^2+DC^2}=5$ $BC=\sqrt{BD^2+DC^2}=4\sqrt{2}$
The altitude to $BC$ in triangle $ABC$ has length $\sqrt{AC^2-\frac{BC}{2}^2}=\sqrt{17}$ , so $[ABC]=\frac{1}{2}\cdot 4\sqrt{2} \cdot \sqrt{17} = 2\sqrt{34}$ . Then $x=\frac{24}{[ABC]}=\frac{24}{2\sqrt{34}}=\frac{6\sqrt{34}}{17}$ or about $2.1$ . The answer is $\boxed{2.1}$ | C | 2.1 |
03c0866d5bac9beb097f08092474f0d5 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_29 | If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$ | Working with the second part of the problem first, we know that $3n$ has $30$ divisors. We try to find the various possible prime factorizations of $3n$ by splitting $30$ into various products of $1, 2$ or $3$ integers.
$30 \rightarrow p^{29}$
$2 \cdot 15 \rightarrow pq^{14}$
$3\cdot 10 \rightarrow p^2q^9$
$5\cdot 6 \rightarrow p^4q^5$
$2\cdot 3\cdot 5 \rightarrow pq^2r^4$
The variables $p, q, r$ are different prime factors, and one of them must be $3$ . We now try to count the factors of $2n$ , to see which prime factorization is correct and has $28$ factors.
In the first case, $p=3$ is the only possibility. This gives $2n = 2\cdot p^{28}$ , which has $2\cdot {29}$ factors, which is way too many.
In the second case, $p=3$ gives $2n = 2q^{14}$ . If $q=2$ , then there are $16$ factors, while if $q\neq 2$ , there are $2\cdot 15 = 30$ factors.
In the second case, $q=3$ gives $2n = 2p3^{13}$ . If $p=2$ , then there are $3\cdot 13$ factors, while if $p\neq 2$ , there are $2\cdot 2 \cdot 13$ factors.
In the third case, $p=3$ gives $2n = 2\cdot 3\cdot q^9$ . If $q=2$ , then there are $11\cdot 2 = 22$ factors, while if $q \neq 2$ , there are $2\cdot 2\cdot 10$ factors.
In the third case, $q=3$ gives $2n = 2\cdot p^2\cdot 3^8$ . If $p=2$ , then there are $4\cdot 9$ factors, while if $p \neq 2$ , there are $2\cdot 3\cdot 9$ factors.
In the fourth case, $p=3$ gives $2n = 2\cdot 3^3\cdot q^5$ . If $q=2$ , then there are $7\cdot 4= 28$ factors. This is the factorization we want.
Thus, $3n = 3^4 \cdot 2^5$ , which has $5\cdot 6 = 30$ factors, and $2n = 3^3 \cdot 2^6$ , which has $4\cdot 7 = 28$ factors.
In this case, $6n = 3^4\cdot 2^6$ , which has $5\cdot 7 = 35$ factors, and the answer is $\boxed{35}$ | C | 35 |
03c0866d5bac9beb097f08092474f0d5 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_29 | If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$ | Because $2n$ has $28$ factors and $3n$ has $30$ factors, we should rewrite the number $n = 2^{e_1}3^{e_2}... p_n^{e_n}$ As the formula for the number of divisors for such a number gives: $(e_1+1)(e_2+1)... (e_n+1)$ We plug in the variations we need to make for the cases $2n$ and $3n$ $2n$ has $(e_1+2)(e_2+1)(e_3+1)... (e_n+1) = 28$ $3n$ has $(e_1+1)(e_2+2)(e_3+1)...(e_n+1) = 30$
If we take the top and divide by the bottom, we get the following equation: $\frac{(e_1+2)(e_2+1)}{(e_1+1)(e_2+2)} = \frac{14}{15}$ . Letting $e_1=x$ and $e_2 = y$ for convenience and expanding this out gives us: $xy-13x+16y+2=0$
We can use Simon's Favorite Factoring Trick (SFFT) to turn this back into: $(x+16)(y-13) +2 + 208 = 0$ or $(x+16)(y-13) = - 210$
As we want to be dealing with rather reasonable numbers for $x$ and $y$ , we try to make the $x+16$ term the slightly larger term and the $y-13$ term the slightly smaller term. This effect is achieved when $x+16 = 21$ and $y-13 = -10$ . Therefore, $x = 5, y = 3$ . We get that this already satisfies the requirements for the number we are looking for, and we take $(5+2)(3+2) = \boxed{35}$ | null | 35 |
03c0866d5bac9beb097f08092474f0d5 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_29 | If $n$ is a positive integer such that $2n$ has $28$ positive divisors and $3n$ has $30$ positive divisors, then how many positive divisors does $6n$ have?
$\text{(A)}\ 32\qquad\text{(B)}\ 34\qquad\text{(C)}\ 35\qquad\text{(D)}\ 36\qquad\text{(E)}\ 38$ | Let $\, 2^{e_1} 3^{e_2} 5^{e_3} \cdots \,$ be the prime factorization of $\, n$ . Then the number of positive divisors of $\, n \,$ is $\, (e_1 + 1)(e_2 + 1)(e_3 + 1) \cdots \;$ . In view of the given information, we have \[28 = (e_1 + 2)(e_2 + 1)P\] and \[30 = (e_1 + 1)(e_2 + 2)P,\] where $\, P = (e_3 + 1)(e_4 + 1) \cdots \;$ . Subtracting the first equation from the second, we obtain $\, 2 = (e_1 - e_2)P, \,$ so either $\, e_1 - e_2 = 1 \,$ and $\, P = 2, \,$ or $\, e_1 - e_2 = 2 \,$ and $\, P = 1$ . The first case yields $\, 14 = (e_1 + 2)e_1 \,$ and $\, (e_1 + 1)^2 = 15$ ; since $\, e_1 \,$ is a nonnegative integer, this is impossible. In the second case, $\, e_2 = e_1 - 2 \,$ and $\, 30 = (e_1 + 1)e_1, \,$ from which we find $\, e_1 = 5 \,$ and $\, e_2 = 3$ . Thus $\, n = 2^5 3^3, \,$ so $\, 6n = 2^6 3^4 \,$ has $\, (6+1)(4+1) = \boxed{35} \,$ positive divisors. | null | 35 |
6a24a5d58b3973804758305f6b8645c4 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_30 | A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$ | In hexagon $ABCDEF$ , let $AB=BC=CD=3$ and let $DE=EF=FA=5$ . Since arc $BAF$ is one third of the circumference of the circle, it follows that $\angle BCF = \angle BEF=60^{\circ}$ . Similarly, $\angle CBE =\angle CFE=60^{\circ}$ . Let $P$ be the intersection of $\overline{BE}$ and $\overline{CF}$ $Q$ that of $\overline{BE}$ and $\overline{AD}$ , and $R$ that of $\overline{CF}$ and $\overline{AD}$ . Triangles $EFP$ and $BCP$ are equilateral, and by symmetry, triangle $PQR$ is isosceles and thus also equilateral. [asy] import olympiad; import geometry; size(150); defaultpen(linewidth(0.8)); real angleUnit = 15; draw(Circle(origin,1)); pair D = dir(22.5); pair C = dir(3*angleUnit + degrees(D)); pair B = dir(3*angleUnit + degrees(C)); pair A = dir(3*angleUnit + degrees(B)); pair F = dir(5*angleUnit + degrees(A)); pair E = dir(5*angleUnit + degrees(F)); draw(A--B--C--D--E--F--cycle); dot("$A$",A,A); dot("$B$",B,B); dot("$C$",C,C); dot("$D$",D,D); dot("$E$",E,E); dot("$F$",F,F); draw(A--D^^B--E^^C--F); label("$3$",D--C,SW); label("$3$",B--C,S); label("$3$",A--B,SE); label("$5$",A--F,NE); label("$5$",F--E,N); label("$5$",D--E,NW); [/asy]
Furthermore, $\angle BAD$ and $\angle BED$ subtend the same arc, as do $\angle ABE$ and $\angle ADE$ . Hence triangles $ABQ$ and $EDQ$ are similar. Therefore, \[\frac{AQ}{EQ}=\frac{BQ}{DQ}=\frac{AB}{ED}=\frac{3}{5}.\] It follows that \[\frac{\frac{AD-PQ}{2}}{PQ+5} =\frac{3}{5}\quad \mbox {and}\quad \frac{3-PQ}{\frac{AD+PQ}{2}}=\frac{3}{5}.\] Solving the two equations simultaneously yields $AD=360/49,$ so $m+n=\boxed{409}. \blacksquare$ | null | 409 |
6a24a5d58b3973804758305f6b8645c4 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_30 | A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$ | All angle measures are in degrees.
Let the first trapezoid be $ABCD$ , where $AB=BC=CD=3$ . Then the second trapezoid is $AFED$ , where $AF=FE=ED=5$ . We look for $AD$
Since $ABCD$ is an isosceles trapezoid, we know that $\angle BAD=\angle CDA$ and, since $AB=BC$ , if we drew $AC$ , we would see $\angle BCA=\angle BAC$ . Anyway, $\widehat{AB}=\widehat{BC}=\widehat{CD}$ $\widehat{AB}$ means arc AB). Using similar reasoning, $\widehat{AF}=\widehat{FE}=\widehat{ED}$
Let $\widehat{AB}=2\phi$ and $\widehat{AF}=2\theta$ . Since $6\theta+6\phi=360$ (add up the angles), $2\theta+2\phi=120$ and thus $\widehat{AB}+\widehat{AF}=\widehat{BF}=120$ . Therefore, $\angle FAB=\frac{1}{2}\widehat{BDF}=\frac{1}{2}(240)=120$ $\angle CDE=120$ as well.
Now I focus on triangle $FAB$ . By the Law of Cosines, $BF^2=3^2+5^2-30\cos{120}=9+25+15=49$ , so $BF=7$ . Seeing $\angle ABF=\theta$ and $\angle AFB=\phi$ , we can now use the Law of Sines to get: \[\sin{\phi}=\frac{3\sqrt{3}}{14}\;\text{and}\;\sin{\theta}=\frac{5\sqrt{3}}{14}.\]
Now I focus on triangle $AFD$ $\angle AFD=3\phi$ and $\angle ADF=\theta$ , and we are given that $AF=5$ , so \[\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}.\] We know $\sin{\theta}=\frac{5\sqrt{3}}{14}$ , but we need to find $\sin{3\phi}$ . Using various identities, we see \begin{align*}\sin{3\phi}&=\sin{(\phi+2\phi)}=\sin{\phi}\cos{2\phi}+\cos{\phi}\sin{2\phi}\\ &=\sin{\phi}(1-2\sin^2{\phi})+2\sin{\phi}\cos^2{\phi}\\ &=\sin{\phi}\left(1-2\sin^2{\phi}+2(1-\sin^2{\phi})\right)\\ &=\sin{\phi}(3-4\sin^2{\phi})\\ &=\frac{3\sqrt{3}}{14}\left(3-\frac{27}{49}\right)=\frac{3\sqrt{3}}{14}\left(\frac{120}{49}\right)=\frac{180\sqrt{3}}{343} \end{align*} Returning to finding $AD$ , we remember \[\frac{\sin{\theta}}{5}=\frac{\sin{3\phi}}{AD}\;\text{so}\;AD=\frac{5\sin{3\phi}}{\sin{\theta}}.\] Plugging in and solving, we see $AD=\frac{360}{49}$ . Thus, the answer is $360 + 49 = 409$ , which is answer choice $\boxed{409}$ | E | 409 |
6a24a5d58b3973804758305f6b8645c4 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_30 | A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$ | Note that minor arc $\overarc{AB}$ is a third of the circumference, therefore, $\angle AOB = 120^{\circ}$ . Major arc $\overarc{AB}$ $=240^{\circ}$ $\angle ACB = 120^{\circ}$
By the Law of Cosine, $AB = \sqrt{ 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^{\circ}} = \sqrt{ 9 + 25 + 15 } = 7$
$\frac{\angle AOB}{2} = 60^{\circ}$ , therefore, $r = \frac{\frac{AB}{2}}{\sin 60^{\circ}} = \frac{\frac{7}{2}}{ \frac{\sqrt{3}}{2} } = \frac{7\sqrt{3}}{3}$
$\sin \frac{\theta}{2} = \frac{\frac32}{r} = \frac{3}{2r} = \frac{3}{2 \cdot \frac{7\sqrt{3}}{3}} = \frac{3 \sqrt{3}}{14}$
Let $x$ be the length of the chord, $\sin \frac{3 \theta}{2} = \frac{\frac{x}{2}}{r}$
By the triple angle formula, $\sin \frac{3 \theta}{2} = 3 \cdot \sin \frac{\theta}{2} - 4 \cdot \sin(\frac{ \theta}{2})^3 = 3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3$
$x = 2 \cdot \frac{7\sqrt{3}}{3} \cdot [3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3] = 2 \cdot \frac{7\sqrt{3}}{3} \cdot (\frac{9\sqrt{3}}{14} - \frac{82\sqrt{3}}{2 \cdot 7^3}) = 9 - \frac{81}{49} = \frac{360}{49}$
Therefore, the answer is $\boxed{409}$ | E | 409 |
6a24a5d58b3973804758305f6b8645c4 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_30 | A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$ | Note that minor arc $\overarc{AB}$ is a third of the circumference, therefore, $\angle AOB = 120^{\circ}$
$\sin \frac{\alpha}{2} = \frac{\frac32}{r}$ $\sin \frac{\alpha}{2} = \frac{3}{2r}$
$\sin \frac{120^{\circ}-\alpha}{2} = \frac{\frac52}{r}$ $\sin (60^{\circ} - \frac{\alpha}{2}) = \frac{5}{2r}$
$\frac{\sin \frac{\alpha}{2}}{\sin (60^{\circ} - \frac{\alpha}{2}) } = \frac{\frac{3}{2r}}{\frac{5}{2r}} = \frac35$ $5 \cdot \sin \frac{\alpha}{2} = 3 \cdot \sin (60^{\circ} - \frac{\alpha}{2})$
$5 \cdot \sin \frac{\alpha}{2} = 3 ( \sin 60^{\circ} \cos \frac{\alpha}{2} - \sin \frac{\alpha}{2} \cos 60^{\circ}) = 3 ( \frac{\sqrt{3}}{2} \cdot \cos \frac{\alpha}{2} - \frac12 \cdot \sin \frac{\alpha}{2})$
$13 \cdot \sin \frac{\alpha}{2} = \frac{3\sqrt{3}}{2} \cdot \cos \frac{\alpha}{2}$
Let $\sin \frac{\alpha}{2} = a$ $\cos \frac{\alpha}{2} = \sqrt{1-a^2}$ $13a = \frac{3\sqrt{3}}{2} \cdot \sqrt{1-a^2}$
$169a^2 = 27-27a^2$ $196a^2=27$ $\sin \frac{\alpha}{2} = a = \sqrt{\frac{27}{196}} = \frac{3 \sqrt{3}}{14}$
Let $x$ be the length of the chord, $\sin \frac{3 \alpha}{2} = \frac{\frac{x}{2}}{r}$
By the triple angle formula, $\sin \frac{3 \alpha}{2} = 3 \cdot \sin \frac{\alpha}{2} - 4 \cdot \sin(\frac{ \alpha}{2})^3 = 3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3$
$x = 2 \cdot \frac{7\sqrt{3}}{3} \cdot [3 \cdot \frac{3 \sqrt{3}}{14} - 4 \cdot (\frac{3 \sqrt{3}}{14})^3] = 2 \cdot \frac{7\sqrt{3}}{3} \cdot (\frac{9\sqrt{3}}{14} - \frac{82\sqrt{3}}{2 \cdot 7^3}) = 9 - \frac{81}{49} = \frac{360}{49}$
Therefore, the answer is $\boxed{409}$ | E | 409 |
6a24a5d58b3973804758305f6b8645c4 | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_30 | A hexagon inscribed in a circle has three consecutive sides each of length 3 and three consecutive sides each of length 5. The chord of the circle that divides the hexagon into two trapezoids, one with three sides each of length 3 and the other with three sides each of length 5, has length equal to $m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$
$\textbf{(A)}\ 309 \qquad \textbf{(B)}\ 349 \qquad \textbf{(C)}\ 369 \qquad \textbf{(D)}\ 389 \qquad \textbf{(E)}\ 409$ | Note that major arc $\overarc{AE}$ is two thirds of the circumference, therefore, $\angle AFE = 120^{\circ}$
By the Law of Cosine, $AE= \sqrt{ 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^{\circ}} = \sqrt{ 9 + 25 + 15 } = 7$
By the Ptolemy's theorem of quadrilateral $ABDE$ $AD \cdot BE = AB \cdot DE + BD \cdot AE$ $AD = BE$ $AD^2= 3 \cdot 5 + 7^2 = 64$ $AD = 8$
By the Ptolemy's theorem of quadrilateral $ABCD$ $AC \cdot BD = BC \cdot AD + AB \cdot CD$ $7AC = 3 \cdot 8 + 3 \cdot 5 = 39$ $AC = \frac{39}{7}$
By the Ptolemy's theorem of quadrilateral $ABCF$ $AC \cdot BF = AB \cdot CF + BC \cdot AF$ $AC = BF$ $(\frac{39}{7})^2 = 3 \cdot CF + 3 \cdot 3$ $CF = \frac{360}{49}$
Therefore, the answer is $\boxed{409}$ | E | 409 |
f9f41f728e95dd275ca0ac97cfc38225 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_13 | The addition below is incorrect. The display can be made correct by changing one digit $d$ , wherever it occurs, to another digit $e$ . Find the sum of $d$ and $e$
$\begin{tabular}{ccccccc} & 7 & 4 & 2 & 5 & 8 & 6 \\ + & 8 & 2 & 9 & 4 & 3 & 0 \\ \hline 1 & 2 & 1 & 2 & 0 & 1 & 6 \end{tabular}$
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ \text{more than 10} }$ | If we change $0$ , the units column would be incorrect.
If we change $1$ , then the leading $1$ in the sum would be incorrect.
However, looking at the $2$ in the hundred-thousands column, it would be possible to change the $2$ to either a $5$ (no carry) or a $6$ (carry) to create a correct statement.
Changing the $2$ to a $5$ would give $745586 + 859430$ on top, which equals $1605016$ . This does not match up to the bottom.
Changing the $2$ to a $6$ gives $746586 + 869430$ on top, which has a sum of $1616016$ . This is the number on the bottom if the $2$ s were changed to $6$ s.
Thus $d=2$ and $e=6$ . so $d+e= 8 \boxed{8}$ | C | 8 |
997c457a4a2352ef8ab26fb481663e55 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_14 | If $f(x) = ax^4 - bx^2 + x + 5$ and $f( - 3) = 2$ , then $f(3) =$
$\mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }$ | Substituting $x = -3$ , we get \[f(-3) = 81a - 9b - 3 + 5 = 81a - 9b + 2.\] But $f(-3) = 2$ , so $81a - 9b + 2 = 2$ , which means $81a - 9b = 0$ . Then \[f(3) = 81a - 9b + 3 + 5 = 0 + 3 + 5 = \boxed{8}.\] | null | 8 |
653741ed776eb6512f66d0788c17bad2 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_22 | pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths $13, 19, 20, 25$ and $31$ , although this is not necessarily their order around the pentagon. The area of the pentagon is
$\mathrm{(A) \ 459 } \qquad \mathrm{(B) \ 600 } \qquad \mathrm{(C) \ 680 } \qquad \mathrm{(D) \ 720 } \qquad \mathrm{(E) \ 745 }$ | Since the pentagon is cut from a rectangle, the cut-off triangle must be right. Since all of the lengths given are integers, it follows that this triangle is a Pythagorean Triple . We know that $31$ and either $25,\, 20$ must be the dimensions of the rectangle, since they are the largest lengths. With some trial-and-error, if we assign the shortest side, $13$ , to be the hypotenuse of the triangle, we see the $5-12-13$ triple. Indeed this works, by placing the $31$ side opposite from the $19$ side and the $25$ side opposite from the $20$ side, leaving the cutaway side to be, as before, $13$
To find the area of the pentagon, we subtract the area of the triangle from that of the big rectangle: $31\cdot 25-\frac{12\cdot5}{2}=775-30=745\Longrightarrow \boxed{745}$ | E | 745 |
0df648a97bb52e00ba99771614c04f27 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_25 | A list of five positive integers has mean $12$ and range $18$ . The mode and median are both $8$ . How many different values are possible for the second largest element of the list?
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$ | Let $a$ be the smallest element, so $a+18$ is the largest element. Since the mode is $8$ , at least two of the five numbers must be $8$ . The last number we denote as $b$
Then their average is $\frac{a + (8) + (8) + b + (a+18)}5 = 12 \Longrightarrow 2a + b = 26$ . Clearly $a \le 8$ . Also we have $b \le a + 18 \Longrightarrow 26-2a \le a + 18 \Longrightarrow 8/3 < 3 \le a$ . Thus there are a maximum of $6$ values of $a$ which corresponds to $6$ values of $b$ ; listing shows that all such values work. The answer is $\boxed{6}$ | B | 6 |
549b5b0c8964aef75b8863c327e7d781 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_29 | For how many three-element sets of distinct positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$
$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$ | $2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11$ . We wish to figure out the number of ways to distribute these prime factors amongst 3 different integers, without over counting triples which are simply permutations of one another.
We can account for permutations by assuming WLOG that $a$ contains the prime factor 2. Thus, there are $3^4$ ways to position the other 4 prime numbers. Note that, with the exception of when all of the prime factors belong to $a$ , we have over counted each case twice, as for when we put certain prime factors into $b$ and the rest into $c$ , we count the exact same case when we put those prime factors which were in $b$ into $c$
Thus, our total number of cases is $\frac{3^4 - 1}{2} = 40 \Rightarrow \boxed{40}.$ | C | 40 |
549b5b0c8964aef75b8863c327e7d781 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_29 | For how many three-element sets of distinct positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$
$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$ | The prime factorization of $2310$ is $2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11.$ Therefore, we have the equation \[abc = 2310 = 2 \cdot 3 \cdot 5 \cdot 7 \cdot 11,\] where $a, b, c$ must be distinct positive integers and order does not matter. There are $3$ ways to assign each prime number on the right-hand side to one of the variables $a, b, c,$ which gives $3^5 = 243$ solutions for $(a, b, c).$ However, three of these solutions have two $1$ s and one $2310,$ which contradicts the fact that $a, b, c$ must be distinct. Because each prime factor appears only once, all other solutions have $a, b, c$ distinct. Correcting for this, we get $243 - 3 = 240$ ordered triples $(a, b, c)$ where $a, b, c$ are all distinct.
Finally, since order does not matter, we must divide by $3!,$ the number of ways to order $a, b, c.$ This gives the final answer, \[\frac{240}{3!} = \frac{240}{6} = \boxed{40}.\] | null | 40 |
0974983f6ad697b1a5e04eacf41a7470 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_30 | A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is
$\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }$ | Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.
Now consider the diagonal from $(0,0,0)$ to $(3,3,3)$ . The midpoint of this diagonal is at $\left(\frac 32,\frac 32,\frac 32\right)$ . The plane that passes through this point and is orthogonal to the diagonal has the equation $x+y+z=\frac 92$
The unit cube with opposite corners at $(x,y,z)$ and $(x+1,y+1,z+1)$ is intersected by this plane if and only if $x+y+z < \frac 92 < (x+1)+(y+1)+(z+1)=(x+y+z)+3$ . Therefore the cube is intersected by this plane if and only if $x+y+z\in\{2,3,4\}$
There are six cubes such that $x+y+z=2$ : permutations of $(1,1,0)$ and $(2,0,0)$ Symmetrically, there are six cubes such that $x+y+z=4$ Finally, there are seven cubes such that $x+y+z=3$ : permutations of $(2,1,0)$ and the central cube $(1,1,1)$
That gives a total of $\boxed{19}$ intersected cubes. | null | 19 |
0974983f6ad697b1a5e04eacf41a7470 | https://artofproblemsolving.com/wiki/index.php/1995_AHSME_Problems/Problem_30 | A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is
$\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }$ | Place the cube so that its space diagonal is perpendicular to the ground. The space diagonal has length of $3\sqrt{3}$ , the altitude of the top vertex of the newly placed cube is $3\sqrt{3}$ . The plane perpendicular and bisecting the space diagonal is now parallel to the ground and also bisecting the space diagonal into $\frac{3\sqrt{3}}{2}$ , so that the height of the plane is $\frac{3\sqrt{3}}{2}$
By symmetry, the space diagonal is trisected by the pyramid at the top of the cube and the pyramid at the bottom of the cube.
We can prove that the space diagonal is trisected. Let the altitude of the pyramid at the top of the cube be $h$ . The base of the pyramid is an equilateral triangle with side length of $\sqrt{3^2+3^2}=3\sqrt{2}$ . The height of the triangle is $\frac{ \sqrt{3} }{2} \cdot 3\sqrt{2}$ . The distance of the center of the triangle to the vertex is $\frac23 \cdot \frac{ \sqrt{3} }{2} \cdot 3\sqrt{2} = \sqrt {6}$ . Therefore, $h = \sqrt{3^2-(\sqrt {6})^2} = \sqrt{9-6}=\sqrt{3}$ . The altitude of the pyramid at the bottom of the cube is also $h$ . The altitude in the middle is $3\sqrt{3}-\sqrt{3}-\sqrt{3}=\sqrt{3}$
The altitude of the vertex at the top is $3\sqrt{3}$ . The altitude of the second highest $3$ vertices are all $2\sqrt{3}$ . The altitude of the third highest $3$ vertices are all $\sqrt{3}$ . The altitude of the bottom-most vertex is $0$
By scale, for the unit cube, place the cube so that its space diagonal is perpendicular to the ground. The altitude of the vertex at the top is $\sqrt{3}$ . The altitude of the second highest $3$ vertices are all $\frac{2\sqrt{3}}{3}$ . The altitude of the third highest $3$ vertices are all $\frac{\sqrt{3}}{3}$ . The altitude of the bottom-most vertex is $0$
The length of the space diagonal of a unit cube is $\sqrt{3}$ . The highest vertex of the bottom-most unit cube has an altitude of $\sqrt{3}$ . As $\sqrt{3} < \frac{3\sqrt{3}}{2}$ , therefore, the plane will not pass through the unit cube at the bottom.
For the next $3$ cubes from the bottom, the altitude of their highest vertex is $\frac{\sqrt{3}}{3} + \sqrt{3} = \frac{4\sqrt{3}}{3}$ . As $\frac{4\sqrt{3}}{3} < \frac{3\sqrt{3}}{2}$ , therefore, the plane will not pass through the next $3$ unit cubes.
For the next $3$ cubes from the bottom, the altitude of their highest vertex is $\frac{2\sqrt{3}}{3} + \frac{\sqrt{3}}{3} = \frac{5\sqrt{3}}{3}$ . As $\frac{5\sqrt{3}}{3} > \frac{3\sqrt{3}}{2}$ , therefore, the plane will pass through the next $3$ unit cubes.
So at the bottom half of the cube, there are only $1+3 = 4$ unit cubes that the plane does not passes through. By symmetry, the plane will not pass through $4$ unit cubes at the top half of the cube.
Thus, the plane does not pass through $4+4 = 8$ unit cubes, it passes through $27-8=\boxed{19}$ unit cubes. | D | 19 |
d717e0c97706181fd4d0b4def1dfc7c1 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_2 | A large rectangle is partitioned into four rectangles by two segments parallel to its sides. The areas of three of the resulting rectangles are shown. What is the area of the fourth rectangle? [asy] draw((0,0)--(10,0)--(10,7)--(0,7)--cycle); draw((0,5)--(10,5)); draw((3,0)--(3,7)); label("6", (1.5,6)); label("?", (1.5,2.5)); label("14", (6.5,6)); label("35", (6.5,2.5)); [/asy]
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 15 \qquad\textbf{(C)}\ 20 \qquad\textbf{(D)}\ 21 \qquad\textbf{(E)}\ 25$ | [asy] pair A=(0,0),B=(10,0),C=(10,7),D=(0,7),EE=(0,5),F=(10,5),G=(3,0),H=(3,7); path BG=shift(0,-0.5)*(B--G); path BF=shift(0.5,0)*(B--F); path FC=shift(0.5,0)*(F--C); path DH=shift(0,0.5)*(D--H); draw(A--B--C--D--cycle); draw(EE--F); draw(G--H); draw(BG,L=Label("$7$",position=MidPoint,align=(0,-1)),arrow=Arrows(),bar=Bars,red); draw(BF,L=Label("$5$",position=MidPoint,align=(1,0)),arrow=Arrows(),bar=Bars,red); draw(FC,L=Label("$2$",position=MidPoint,align=(1,0)),arrow=Arrows(),bar=Bars,red); draw(DH,L=Label("$3$",position=MidPoint,align=(0,1)),arrow=Arrows(),bar=Bars,red); label("$6$", (1.5,6)); label("$15$", (1.5,2.5),blue); label("$14$", (6.5,6)); label("$35$", (6.5,2.5)); [/asy]
We can easily see the dimensions of each small rectangle. So the area of the last rectangle is $3\times 5=\boxed{15}$ | B | 15 |
eba7daaf888ae0202948818b05f897eb | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_3 | How many of the following are equal to $x^x+x^x$ for all $x>0$
$\textbf{I:}\ 2x^x \qquad\textbf{II:}\ x^{2x} \qquad\textbf{III:}\ (2x)^x \qquad\textbf{IV:}\ (2x)^{2x}$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$ | We look at each statement individually.
$\textbf{I:}\ 2x^x$ . We note that $x^x+x^x=x^x(1+1)=2x^x$ . So statement $\textbf{I}$ is true.
$\textbf{II:}\ x^{2x}$ . We find a counter example which is $x=1$ $2\neq 1$ . So statement $\textbf{II}$ is false.
$\textbf{III:}\ (2x)^x$ . We see that this statement is equal to $2^xx^x$ $x=2$ is a counter example. $8\neq 16$ . So statement $\textbf{III}$ is false.
$\textbf{IV:}\ (2x)^{2x}$ . We see that $x=1$ is again a counter example. $2\neq 4$ . So statement $\textbf{IV}$ is false.
Therefore, our answer is $\boxed{1}$ | B | 1 |
7d40628893393d69718ce8baec455a32 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_4 | In the $xy$ -plane, the segment with endpoints $(-5,0)$ and $(25,0)$ is the diameter of a circle. If the point $(x,15)$ is on the circle, then $x=$
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20$ | We see that the center of this circle is at $\left(\frac{-5+25}{2},0\right)=(10,0)$ . The radius is $\frac{30}{2}=15$ . So the equation of this circle is \[(x-10)^2+y^2=225.\] Substituting $y=15$ yields $(x-10)^2=0$ so $x=\boxed{10}$ | A | 10 |
8cb97b00626f026fbbdd3cfc586db795 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_5 | Pat intended to multiply a number by $6$ but instead divided by $6$ . Pat then meant to add $14$ but instead subtracted $14$ . After these mistakes, the result was $16$ . If the correct operations had been used, the value produced would have been
$\textbf{(A)}\ \text{less than 400} \qquad\textbf{(B)}\ \text{between 400 and 600} \qquad\textbf{(C)}\ \text{between 600 and 800} \\ \textbf{(D)}\ \text{between 800 and 1000} \qquad\textbf{(E)}\ \text{greater than 1000}$ | We reverse the operations that he did and then use the correct operations. His end result is $16$ . Before that, he subtracted $14$ which means that his number after the first operation was $30$ . He divided by $6$ so his number was $180$
Now, we multiply $180$ by $6$ to get $1080$ . Finally, $1080+14=1094$ . Since $1094>1000$ , our answer is $\boxed{1000}$ | E | 1000 |
23c6a72e4de5247019166830d868dbbf | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_6 | In the sequence \[..., a, b, c, d, 0, 1, 1, 2, 3, 5, 8,...\] each term is the sum of the two terms to its left. Find $a$
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 3$ | We work backwards to find $a$
\[d+0=1\implies d=1\] \[c+1=0\implies c=-1\] \[b+(-1)=1\implies b=2\] \[a+2=-1\implies a=\boxed{3.}\] | A | 3. |
336f7c591d8879dbd2d94b8ed5391549 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_7 | Squares $ABCD$ and $EFGH$ are congruent, $AB=10$ , and $G$ is the center of square $ABCD$ . The area of the region in the plane covered by these squares is [asy] draw((0,0)--(10,0)--(10,10)--(0,10)--cycle); draw((5,5)--(12,-2)--(5,-9)--(-2,-2)--cycle); label("A", (0,0), W); label("B", (10,0), E); label("C", (10,10), NE); label("D", (0,10), NW); label("G", (5,5), N); label("F", (12,-2), E); label("E", (5,-9), S); label("H", (-2,-2), W); dot((-2,-2)); dot((5,-9)); dot((12,-2)); dot((0,0)); dot((10,0)); dot((10,10)); dot((0,10)); dot((5,5)); [/asy] $\textbf{(A)}\ 75 \qquad\textbf{(B)}\ 100 \qquad\textbf{(C)}\ 125 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 175$ | The area of the entire region in the plane is the area of the figure. However, we cannot simply add the two areas of the squares. We find the area of $\triangle ABG$ and subtract this from $200$ , the total area of the two squares.
Since $G$ is the center of $ABCD$ $BG$ is half of the diagonal of the square. The diagonal of $ABCD$ is $10\sqrt{2}$ so $BG=5\sqrt{2}$ . Since $EFGH$ is a square, $\angle G=90^\circ$ . So $\triangle ABG$ is an isosceles right triangle. Its area is $\frac{(5\sqrt{2})^2}{2}=\frac{50}{2}=25$ . Therefore, the area of the region is $200-25=\boxed{175.}$ | E | 175. |
15a700101468c51f6816b47a3099abf6 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_8 | In the polygon shown, each side is perpendicular to its adjacent sides, and all 28 of the sides are congruent. The perimeter of the polygon is $56$ . The area of the region bounded by the polygon is [asy] draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle); [/asy] $\textbf{(A)}\ 84 \qquad\textbf{(B)}\ 96 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 112 \qquad\textbf{(E)}\ 196$ | Since the perimeter is $56$ and all of the sides are congruent, the length of each side is $2$ . We break the figure into squares as shown below.
[asy] unitsize(0.8cm); draw(shift(3,4)*((0,-1)--(1,-1))); draw(shift(3,4)*((-1,-2)--(2,-2))); draw(shift(3,4)*((-2,-3)--(3,-3))); draw(shift(3,4)*((-2,-4)--(3,-4))); draw(shift(3,4)*((-1,-5)--(2,-5))); draw(shift(3,4)*((0,-6)--(1,-6))); draw((0,0)--(1,0)--(1,-1)--(2,-1)--(2,-2)--(3,-2)--(3,-3)--(4,-3)--(4,-2)--(5,-2)--(5,-1)--(6,-1)--(6,0)--(7,0)--(7,1)--(6,1)--(6,2)--(5,2)--(5,3)--(4,3)--(4,4)--(3,4)--(3,3)--(2,3)--(2,2)--(1,2)--(1,1)--(0,1)--cycle); draw((1,0)--(1,1)); draw((2,-1)--(2,2)); draw((3,-2)--(3,3)); draw((4,-2)--(4,3)); draw((5,-1)--(5,2)); draw((6,0)--(6,1)); [/asy]
We see that there are a total of $2(1+3+5)+7=25$ squares with side length $2$ . Therefore, the total area is $4\cdot 25=\boxed{100.}$ | C | 100. |
60293f5aef8e2d16bf859b52e9908549 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_9 | If $\angle A$ is four times $\angle B$ , and the complement of $\angle B$ is four times the complement of $\angle A$ , then $\angle B=$
$\textbf{(A)}\ 10^{\circ} \qquad\textbf{(B)}\ 12^{\circ} \qquad\textbf{(C)}\ 15^{\circ} \qquad\textbf{(D)}\ 18^{\circ} \qquad\textbf{(E)}\ 22.5^{\circ}$ | Let $\angle A=x$ and $\angle B=y$ . From the first condition, we have $x=4y$ . From the second condition, we have \[90-y=4(90-x).\] Substituting $x=4y$ into the previous equation and solving yields \begin{align*}90-y=4(90-4y)&\implies 90-y=360-16y\\&\implies 15y=270\\&\implies y=\boxed{18} | D | 18 |
64e92fab358180c27c4dfcc496985e03 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_12 | If $i^2=-1$ , then $(i-i^{-1})^{-1}=$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ -2i \qquad\textbf{(C)}\ 2i \qquad\textbf{(D)}\ -\frac{i}{2} \qquad\textbf{(E)}\ \frac{i}{2}$ | We simplify step by step as follows: \begin{align*}(i-i^{-1})^{-1}&=\frac{1}{i-i^{-1}}\\&=\frac{1}{i-\frac{1}{i}}\\&=\frac{1}{\left(\frac{i^2-1}{i}\right)}\\&=\frac{i}{i^2-1}\\&=\boxed{2} | D | 2 |
c5ee8c1d4a5257b66e71671769cda020 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_14 | Find the sum of the arithmetic series \[20+20\frac{1}{5}+20\frac{2}{5}+\cdots+40\] $\textbf{(A)}\ 3000 \qquad\textbf{(B)}\ 3030 \qquad\textbf{(C)}\ 3150 \qquad\textbf{(D)}\ 4100 \qquad\textbf{(E)}\ 6000$ | Our first term is $20$ and our last term is $40$ . To find the number of terms, $n$ , we note that the common difference between each term is $\frac{1}{5}$ . So we have \[20+\frac{1}{5}(n-1)=40\implies n-1=100\implies n=101.\] Using our formula, our sum is \[101\left(\frac{20+40}{2}\right)=101\times 30=\boxed{3030.}\] | B | 3030. |
03bfd48c8088e2e7be88270006be0e3b | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_15 | For how many $n$ in $\{1, 2, 3, ..., 100 \}$ is the tens digit of $n^2$ odd?
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 20 \qquad\textbf{(C)}\ 30 \qquad\textbf{(D)}\ 40 \qquad\textbf{(E)}\ 50$ | Let $n=10a+b$ . So $n^2=(10a+b)^2=100a^2+20ab+b^2$ . The term $100a^2$ only contributes digits starting at the hundreds place, so this does not affect whether the tens digit is odd. The term $20ab$ only contributes digits starting at the tens place, and the tens digit contributed will be the ones digit of $2ab$ which is even. So we see this term also does not affect whether the tens digit is odd. This means only $b^2$ can affect whether the tens digit is odd. We can quickly check $1^2=1, \dots, 9^2=81$ and discover only for $b=4$ or $b=6$ does $b^2$ have an odd tens digit. The total number of positive integers less than or equal to $100$ that have $4$ or $6$ as the units digit is \[10\times 2=\boxed{20.}\] | B | 20. |
6312bd081d0b8d5a1a3fa8035ea15c32 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_16 | Some marbles in a bag are red and the rest are blue. If one red marble is removed, then one-seventh of the remaining marbles are red. If two blue marbles are removed instead of one red, then one-fifth of the remaining marbles are red. How many marbles were in the bag originally?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 22 \qquad\textbf{(C)}\ 36 \qquad\textbf{(D)}\ 57 \qquad\textbf{(E)}\ 71$ | Let $r$ and $b$ be the number of red and blue marbles originally in the bag respectively. After $1$ red marble is removed, there are $r+b-1$ marbles left in the bag and $r-1$ red marbles left. So \[\frac{r-1}{r+b-1}=\frac{1}{7}.\] When $2$ blue marbles are removed, there are $r$ red marbles and $r+b-2$ total marbles left in the bag. So \[\frac{r}{r+b-2}=\frac{1}{5}.\] Cross multiplying for each yields \begin{align*}7r-7=r+b-1&\implies 7r-6=r+b\\ 5r=r+b-2&\implies 5r+2=r+b.\end{align*} We can equate each of these expressions to yields \[7r-6=5r+2\implies 2r=8\implies r=4\implies b=18.\] Therefore, the total number of marbles is \[r+b=4+18=\boxed{22.}\] | B | 22. |
c570bd49f181da2b390a3244b1ec3b16 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_18 | Triangle $ABC$ is inscribed in a circle, and $\angle B = \angle C = 4\angle A$ . If $B$ and $C$ are adjacent vertices of a regular polygon of $n$ sides inscribed in this circle, then $n=$ [asy] draw(Circle((0,0), 5)); draw((0,5)--(3,-4)--(-3,-4)--cycle); label("A", (0,5), N); label("B", (-3,-4), SW); label("C", (3,-4), SE); dot((0,5)); dot((3,-4)); dot((-3,-4)); [/asy] $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18$ | We solve for $\angle A$ as follows: \[4\angle A+4\angle A+\angle A=180\implies 9\angle A=180\implies \angle A=20.\] That means that minor arc $\widehat{BC}$ has measure $40^\circ$ . We can fit a maximum of $\frac{360}{40}=\boxed{9}$ of these arcs in the circle. | C | 9 |
bb2eea4e4fdadd4395ed8d5016c58d01 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_19 | Label one disk " $1$ ", two disks " $2$ ", three disks " $3$ $, ...,$ fifty disks " $50$ ". Put these $1+2+3+ \cdots+50=1275$ labeled disks in a box. Disks are then drawn from the box at random without replacement. The minimum number of disks that must be drawn to guarantee drawing at least ten disks with the same label is
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 51 \qquad\textbf{(C)}\ 415 \qquad\textbf{(D)}\ 451 \qquad\textbf{(E)}\ 501$ | We can solve this problem by thinking of the worst case scenario, essentially an adaptation of the Pigeon-hole principle.
We can start by picking up all the disks numbered 1 to 9 since even if we have all those disks we won't have 10 of any one disk. This gives us 45 disks.
From disks numbered from 10 to 50, we can pick up at most 9 disks to prevent picking up 10. There are 50-10+1 = 41 different numbers from 10 to 50. We pick up 9 from each number, therefore, we multiply $41 \cdot 9 = 369$ . In total, the maximum number we can pick up without picking up 10 of the same kind is $369+45=414$ . We need one more disk to guarantee a complete set of 10. Therefore, the answer is $\boxed{415}$ | null | 415 |
ff9cb0a422bc804ac29e2e4ae2980e8a | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_24 | A sample consisting of five observations has an arithmetic mean of $10$ and a median of $12$ . The smallest value that the range (largest observation minus smallest) can assume for such a sample is
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 10$ | The minimum range occurs in the set $\{7,7,12,12,12\}$ , so the answer is $\boxed{5}$ | C | 5 |
6efb48640b221a1b17cf7252958e5269 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_25 | If $x$ and $y$ are non-zero real numbers such that \[|x|+y=3 \qquad \text{and} \qquad |x|y+x^3=0,\] then the integer nearest to $x-y$ is
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5$ | We have two cases to consider: $x$ is positive or $x$ is negative. If $x$ is positive, we have $x+y=3$ and $xy+x^3=0$
Solving for $y$ in the top equation gives us $3-x$ . Plugging this in gives us: $x^3-x^2+3x=0$ . Since we're told $x$ is not zero, we can divide by $x$ , giving us: $x^2-x+3=0$
The discriminant of this is $(-1)^2-4(1)(3)=-11$ , which means the equation has no real solutions.
We conclude that $x$ is negative. In this case $-x+y=3$ and $-xy+x^3=0$ . Negating the top equation gives us $x-y=-3$ . We seek $x-y$ , so the answer is $\boxed{3}$ | A | 3 |
6895b5bb98bd007824f80be8767f937c | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_26 | A regular polygon of $m$ sides is exactly enclosed (no overlaps, no gaps) by $m$ regular polygons of $n$ sides each. (Shown here for $m=4, n=8$ .) If $m=10$ , what is the value of $n$ [asy] size(200); defaultpen(linewidth(0.8)); draw(unitsquare); path p=(0,1)--(1,1)--(1+sqrt(2)/2,1+sqrt(2)/2)--(1+sqrt(2)/2,2+sqrt(2)/2)--(1,2+sqrt(2))--(0,2+sqrt(2))--(-sqrt(2)/2,2+sqrt(2)/2)--(-sqrt(2)/2,1+sqrt(2)/2)--cycle; draw(p); draw(shift((1+sqrt(2)/2,-sqrt(2)/2-1))*p); draw(shift((0,-2-sqrt(2)))*p); draw(shift((-1-sqrt(2)/2,-sqrt(2)/2-1))*p);[/asy] $\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 6 \qquad\textbf{(C)}\ 14 \qquad\textbf{(D)}\ 20 \qquad\textbf{(E)}\ 26$ | To find the number of sides on the regular polygons that surround the decagon, we can find the interior angles and work from there. Knowing that the measure of the interior angle of any regular polygon is $\frac{(n-2)*180}{n}$ , the measure of the decagon's interior angle is $\frac{8*180}{10} = 144$ degrees.
The regular polygons meet at every vertex such that the angle outside of the decagon is divided evenly in two. With this, we know that the angle of the regular polygon is $216/2=108$ degrees. Using the previous formula, $n=5$ $\boxed{5}$ | A | 5 |
a43416f0263fe461c90d934aede70724 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_28 | In the $xy$ -plane, how many lines whose $x$ -intercept is a positive prime number and whose $y$ -intercept is a positive integer pass through the point $(4,3)$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$ | The line with $x$ -intercept $a$ and $y$ -intercept $b$ is given by the equation $\frac{x}{a} + \frac{y}{b} = 1$ . We are told $(4,3)$ is on the line so
\[\frac{4}{a} + \frac{3}{b} = 1 \implies ab - 4b - 3a = 0 \implies (a-4)(b-3)=12\]
Since $a$ and $b$ are integers, this equation holds only if $(a-4)$ is a factor of $12$ . The factors are $1, 2, 3, 4, 6, 12$ which means $a$ must be one of $5, 6, 7, 8, 10, 16$ . The only members of this list which are prime are $a=5$ and $a=7$ , so the number of solutions is $\boxed{2}$ | C | 2 |
a43416f0263fe461c90d934aede70724 | https://artofproblemsolving.com/wiki/index.php/1994_AHSME_Problems/Problem_28 | In the $xy$ -plane, how many lines whose $x$ -intercept is a positive prime number and whose $y$ -intercept is a positive integer pass through the point $(4,3)$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$ | Let $C = (4,3)$ $DF=a$ , and $AD=b$ . As stated in the problem, the $x$ -intercept $DF=a$ is a positive prime number, and the $y$ -intercept $AD=b$ is a positive integer.
Through similar triangles, $\frac{AB}{BC}=\frac{CE}{EF}$ $\frac{b-3}{4}=\frac{3}{a-4}$ $(a-4)(b-3)=12$
The only cases where $a$ is prime are: \[\begin{cases} a-4=1 & a=5 \\ b-3=12 & b=15 \end{cases}\]
\[and\]
\[\begin{cases} a-4=3 & a=7 \\ b-3=4 & b=5 \end{cases}\]
So the number of solutions are $\boxed{2}$ | C | 2 |
96d84a3934988835446139373f90fc71 | https://artofproblemsolving.com/wiki/index.php/1993_AHSME_Problems/Problem_30 | Given $0\le x_0<1$ , let \[x_n=\left\{ \begin{array}{ll} 2x_{n-1} &\text{ if }2x_{n-1}<1 \\ 2x_{n-1}-1 &\text{ if }2x_{n-1}\ge 1 \end{array}\right.\] for all integers $n>0$ . For how many $x_0$ is it true that $x_0=x_5$
$\text{(A) 0} \quad \text{(B) 1} \quad \text{(C) 5} \quad \text{(D) 31} \quad \text{(E) }\infty$ | We are going to look at this problem in binary.
$x_0 = (0.a_1 a_2 \cdots )_2$
$2x_0 = (a_1.a_2 a_3 \cdots)_2$
If $2x_0 < 1$ , then $x_0 < \frac{1}{2}$ which means that $a_1 = 0$ and so $x_1 = (.a_2 a_3 a_4 \cdots)_2$
If $2x_0 \geq 1$ then $x \geq \frac{1}{2}$ which means that $x_1 = 2x_0 - 1 = (.a_2 a_3 a_4 \cdots)_2$
Using the same logic, we notice that this sequence cycles and that since $x_0 = x_5$ we notice that $a_n = a_{n+5}$
We have $2$ possibilities for each of $a_1$ to $a_5$ but we can't have $a_1 = a_2 = a_3 = a_4 = a_5 = 1$ so we have $2^5 - 1 = \boxed{31}$ | D | 31 |
48c31b8fd7f722f04c8b1a04ae168f4a | https://artofproblemsolving.com/wiki/index.php/1992_AHSME_Problems/Problem_12 | Let $y=mx+b$ be the image when the line $x-3y+11=0$ is reflected across the $x$ -axis. The value of $m+b$ is
$\text{(A) -6} \quad \text{(B) } -5\quad \text{(C) } -4\quad \text{(D) } -3\quad \text{(E) } -2$ | $\fbox{C}$ First we want to put this is slope-intercept form, so we get $y=\dfrac{1}{3}x+\dfrac{11}{3}$ . When we reflect a line across the x-axis, the line's x-intercept stays the same, while the y-intercept and the slope are additive inverses of the original line. Since $m+b$ is the sum of the slope and the y-intercept, we get $-\dfrac{1}{3}-\dfrac{11}{3}=\boxed{4}$ | null | 4 |
b27b513163474656c4d76ae1c24b6cc4 | https://artofproblemsolving.com/wiki/index.php/1992_AHSME_Problems/Problem_16 | If \[\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\] for three positive numbers $x,y$ and $z$ , all different, then $\frac{x}{y}=$
$\text{(A) } \frac{1}{2}\quad \text{(B) } \frac{3}{5}\quad \text{(C) } \frac{2}{3}\quad \text{(D) } \frac{5}{3}\quad \text{(E) } 2$ | Since \[\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} = \frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n},\] we can say that \[\frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} = \frac {y+(x+y)+x}{(x-z)+z+y} = \boxed{2}$ | E | 2 |
8c64f0f4cf1d17bb6dc8e9a7bff81986 | https://artofproblemsolving.com/wiki/index.php/1992_AHSME_Problems/Problem_21 | For a finite sequence $A=(a_1,a_2,...,a_n)$ of numbers, the Cesáro sum of A is defined to be $\frac{S_1+\cdots+S_n}{n}$ , where $S_k=a_1+\cdots+a_k$ and $1\leq k\leq n$ . If the Cesáro sum of
the 99-term sequence $(a_1,...,a_{99})$ is 1000, what is the Cesáro sum of the 100-term sequence $(1,a_1,...,a_{99})$
$\text{(A) } 991\quad \text{(B) } 999\quad \text{(C) } 1000\quad \text{(D) } 1001\quad \text{(E) } 1009$ | Let us define the Cesáro total of a particular sequence to be n * Cesáro sum. We can see that the Cesáro total is \[S_1 + S_2 + S_3 + ... + S_n\] \[= a_1 + (a_1 + a_2) + (a_1 + a_2 + a_3) + \ldots + (a_1 + a_2 + \ldots + a_n)\] \[= n \cdot a_1 + (n - 1) \cdot a_2 + \ldots + a_n.\] If we take this to be the Cesáro total for the second sequence, we can see that all the terms but the first term make up the first sequence. Since we know that the Cesáro total of the original sequence is $1000 * 99 = 99,000,$ than the Cesáro total of the second sequence is $n \cdot a_1 + 99,000 = 100 \cdot 1 + 99,000 = 99,100.$ Thus the Cesáro sum of the second sequence is $\frac{99,100}{100} = \boxed{991,}\, .$ | A | 991, |
e6d0b5c137f5cd3535cd685f4a9744ce | https://artofproblemsolving.com/wiki/index.php/1992_AHSME_Problems/Problem_24 | Let $ABCD$ be a parallelogram of area $10$ with $AB=3$ and $BC=5$ . Locate $E,F$ and $G$ on segments $\overline{AB},\overline{BC}$ and $\overline{AD}$ , respectively, with $AE=BF=AG=2$ . Let the line through $G$ parallel to $\overline{EF}$ intersect $\overline{CD}$ at $H$ . The area of quadrilateral $EFHG$ is
$\text{(A) } 4\quad \text{(B) } 4.5\quad \text{(C) } 5\quad \text{(D) } 5.5\quad \text{(E) } 6$ | We note that $ABFG$ is a parallelogram because $AG = BF = 2$ and $AG \parallel BF$ . Using the same reasoning, $GFCD$ is also a parallelogram.
Assume that the height of parallelogram $ABFG$ with respect to base $AB$ is $x$ . Then, the area of parallelogram $ABFG$ is $AB * x$ . The area of triangle $EFG$ is $\frac{AB * x}{2}$ , which is half of the area of parallelogram $ABFG$
Likewise, the area of triangle $FGH$ is half the area of parallelogram $GFCD$
Thus, $[EFHG] = [EFG] + [FGH] = 1/2[ABFG] + 1/2[GFCD] = 1/2[ABCD] = 1/2(10) = \boxed{5}$ | null | 5 |
7df11c02ed32302a732b076d16130689 | https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_10 | Point $P$ is $9$ units from the center of a circle of radius $15$ . How many different chords of the circle contain $P$ and have integer lengths?
(A) 11 (B) 12 (C) 13 (D) 14 (E) 29 | Let $O$ be the center of the circle, and let the chord passing through $P$ that is perpendicular to $OP$ intersect the circle at $Q$ and $R$ . Then $OP = 9$ and $OQ = 15$ , so by the Pythagorean Theorem, $PQ = 12$ . By symmetry, $PR = 12$
[asy] import graph; unitsize(0.15 cm); pair O, P, Q, R; O = (0,0); P = (9,0); Q = (9,12); R = (9,-12); draw(Circle(O,15)); draw((-15,0)--(15,0)); draw(O--Q); draw(Q--R,red); dot("$O$", O, S); dot("$P$", P, NW); dot("$Q$", Q, NE); dot("$R$",R,SE); label("$15$", (-15/2,0), S); label("$9$", (O + P)/2, S); label("$6$", (12,0), S); label("$15$", (O + Q)/2, NW); label("$12$", (P + Q)/2, E); [/asy] Let $AB$ be the diameter passing through $P$. [asy] import graph; unitsize(0.15 cm); pair A, B, O, P, Q, R; A = (-15,0); B = (15,0); O = (0,0); P = (9,0); Q = (9,12); R = (9,-12); draw(Circle(O,15)); draw(A--B,red); draw(Q--R); dot("$A$", A, W); dot("$B$", B, E); dot("$O$", O, S); dot("$P$", P, NW); dot("$Q$", Q, NE); dot("$R$", R, SE); label("$15$", (-15/2,0), S); label("$9$", (O + P)/2, S); label("$6$", (12,0), S); label("$12$", (P + Q)/2, E); label("$12$", (P + R)/2, E); [/asy] Then the longest chord of the circle that passes through $P$ is $AB$, which has length 30, and the shortest chord is $QR$, which has length 24. If we rotate the red chord (while ensuring it passes through $P$), we can create all possible lengths between 24 and 30. Indeed, we see that for each positive integer $n=25,26,27,28,29$, there are two chords of length $n$ passing through $P$, as seen in this picture: [asy] import graph; unitsize(0.15 cm); pair O, P,A,B,A2,B2; O = (0,0); P = (9,0); A = 15*dir(20); B = 15*dir(250); A2 = 15*dir(-20); B2 = 15*dir(-250); draw(Circle(O,15)); draw(A--B,red); draw(A2--B2,red); draw(O--(15,0)); dot("$O$", O, S); dot("$P$", P, N); [/asy]
Therefore, there are $2 + 2(29 - 25 + 1) = \boxed{12}$ chords of integer length passing through $P$ | null | 12 |
e7f5f2774bc908de8992ca928195172e | https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_14 | If $x$ is the cube of a positive integer and $d$ is the number of positive integers that are divisors of $x$ , then $d$ could be
$\text{(A) } 200\quad\text{(B) } 201\quad\text{(C) } 202\quad\text{(D) } 203\quad\text{(E) } 204$ | Solution by e_power_pi_times_i
Notice that if $x$ is expressed in the form $a^b$ , then the number of positive divisors of $x^3$ is $3b+1$ . Checking through all the answer choices, the only one that is in the form $3b+1$ is $\boxed{202}$ | C | 202 |
e7f5f2774bc908de8992ca928195172e | https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_14 | If $x$ is the cube of a positive integer and $d$ is the number of positive integers that are divisors of $x$ , then $d$ could be
$\text{(A) } 200\quad\text{(B) } 201\quad\text{(C) } 202\quad\text{(D) } 203\quad\text{(E) } 204$ | Solution by e_power_pi_times_i
Since the divisors are from $x^3$ , then the answer must be something in (mod $3$ ). Since $200$ and $203$ are the same (mod $3$ ), as well as $201$ and $204$ $\boxed{202}$ is the only answer left. | C | 202 |
0f4582d09295cfbb1039a9d9d1d8776a | https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_16 | One hundred students at Century High School participated in the AHSME last year, and their mean score was 100. The number of non-seniors taking the AHSME was $50\%$ more than the number of seniors, and the mean score of the seniors was $50\%$ higher than that of the non-seniors. What was the mean score of the seniors?
(A) $100$ (B) $112.5$ (C) $120$ (D) $125$ (E) $150$ | Solution by e_power_pi_times_i
Let $s$ and $\dfrac{3}{2}s$ denote the numbers of seniors and non-seniors, respectively. Then $\dfrac{5}{2}s = 100$ , so $s = 40$ $\dfrac{3}{2}s = 60$ . Let $m$ and $\dfrac{2}{3}m$ denote the mean score of seniors and non-seniors, respectively. Then $40m + 60(\dfrac{2}{3}m) = 40m + 40m = 10000$ . The answer is $\boxed{125}$ | D | 125 |
a567b4816c25acda5c85493beeea4a05 | https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_17 | A positive integer $N$ is a palindrome if the integer obtained by reversing the sequence of digits of $N$ is equal to $N$ . The year 1991 is the only year in the current century with the following 2 properties:
(a) It is a palindrome
(b) It factors as a product of a 2-digit prime palindrome and a 3-digit prime palindrome.
How many years in the millenium between 1000 and 2000 have properties (a) and (b)?
$\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$ | Solution by e_power_pi_times_i
Notice that all four-digit palindromes are divisible by $11$ , so that is our two-digit prime. Because the other factor is a three-digit number, we are looking at palindromes between $1100$ and $2000$ , which also means that the last digit of the three-digit number is $1$ . Checking through the three-digit numbers $101, 111, 121,\dots, 181$ , we find out that there are $\boxed{4}$ three-digit prime numbers, which when multiplied by $11$ , result in palindromes. | D | 4 |
52c4dc18c65a7c86b6976610c8d8a9c6 | https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_19 | [asy] draw((0,0)--(0,3)--(4,0)--cycle,dot); draw((4,0)--(7,0)--(7,10)--cycle,dot); draw((0,3)--(7,10),dot); MP("C",(0,0),SW);MP("A",(0,3),NW);MP("B",(4,0),S);MP("E",(7,0),SE);MP("D",(7,10),NE); [/asy]
Triangle $ABC$ has a right angle at $C, AC=3$ and $BC=4$ . Triangle $ABD$ has a right angle at $A$ and $AD=12$ . Points $C$ and $D$ are on opposite sides of $\overline{AB}$ . The line through $D$ parallel to $\overline{AC}$ meets $\overline{CB}$ extended at $E$ . If \[\frac{DE}{DB}=\frac{m}{n},\] where $m$ and $n$ are relatively prime positive integers, then $m+n=$
$\text{(A) } 25\quad \text{(B) } 128\quad \text{(C) } 153\quad \text{(D) } 243\quad \text{(E) } 256$ | Solution by e_power_pi_times_i
Let $F$ be the point such that $DF$ and $CF$ are parallel to $CE$ and $DE$ , respectively, and let $DE = x$ and $BE^2 = 169-x^2$ . Then, $[FDEC] = x(4+\sqrt{169-x^2}) = [ABC] + [BED] + [ABD] + [AFD] = 6 + \dfrac{x\sqrt{169-x^2}}{2} + 30 + \dfrac{(x-3)(4+\sqrt{169-x^2})}{2}$ . So, $4x+x\sqrt{169-x^2} = 60 + x\sqrt{169-x^2} - 3\sqrt{169-x^2}$ . Simplifying $3\sqrt{169-x^2} = 60 - 4x$ , and $1521 - 9x^2 = 16x^2 - 480x + 3600$ . Therefore $25x^2 - 480x + 2079 = 0$ , and $x = \dfrac{48\pm15}{5}$ . Checking, $x = \dfrac{63}{5}$ is the answer, so $\dfrac{DE}{DB} = \dfrac{\dfrac{63}{5}}{13} = \dfrac{63}{65}$ . The answer is $\boxed{128}$ | B | 128 |
52c4dc18c65a7c86b6976610c8d8a9c6 | https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_19 | [asy] draw((0,0)--(0,3)--(4,0)--cycle,dot); draw((4,0)--(7,0)--(7,10)--cycle,dot); draw((0,3)--(7,10),dot); MP("C",(0,0),SW);MP("A",(0,3),NW);MP("B",(4,0),S);MP("E",(7,0),SE);MP("D",(7,10),NE); [/asy]
Triangle $ABC$ has a right angle at $C, AC=3$ and $BC=4$ . Triangle $ABD$ has a right angle at $A$ and $AD=12$ . Points $C$ and $D$ are on opposite sides of $\overline{AB}$ . The line through $D$ parallel to $\overline{AC}$ meets $\overline{CB}$ extended at $E$ . If \[\frac{DE}{DB}=\frac{m}{n},\] where $m$ and $n$ are relatively prime positive integers, then $m+n=$
$\text{(A) } 25\quad \text{(B) } 128\quad \text{(C) } 153\quad \text{(D) } 243\quad \text{(E) } 256$ | Solution by Arjun Vikram
Extend lines $AD$ and $CE$ to meet at a new point $F$ . Now, we see that $FAC\sim FDE \sim ACB$ . Using this relationship, we can see that $AF=\frac{15}4$ , (so $FD=\frac{63}4$ ), and the ratio of similarity between $FDE$ and $FAC$ is $\frac{63}{15}$ . This ratio gives us that $\frac{63}5$ . By the Pythagorean Theorem, $DB=13$ . Thus, $\frac{DE}{DB}=\frac{63}{65}$ , and the answer is $63+65=\boxed{128}$ | B | 128 |
c738ad96b945440f65bd4f15838592a9 | https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_20 | The sum of all real $x$ such that $(2^x-4)^3+(4^x-2)^3=(4^x+2^x-6)^3$ is
$\textbf{(A) } \frac32 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } \frac52 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \frac72$ | Note that $(2^x-4)+(4^x-2)=4^x+2^x-6,$ so we let $a=2^x-4$ and $b=4^x-2.$ The original equation becomes \[a^3+b^3=(a+b)^3.\] We expand the right side, then rearrange: \begin{align*} a^3+b^3 &= a^3+3a^2b+3ab^2+b^3 \\ 0 &= 3a^2b+3ab^2 \\ 0 &= 3ab(a+b). \end{align*}
Together, the answer is $2+\frac12+1=\boxed{72}.$ | E | 72 |
be40f6955b3086a22a5601aeac0c21d7 | https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_27 | If \[x+\sqrt{x^2-1}+\frac{1}{x-\sqrt{x^2-1}}=20,\] then \[x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}=\]
$\textbf{(A) } 5.05 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 51.005 \qquad \textbf{(D) } 61.25 \qquad \textbf{(E) } 400$ | We rationalize the denominator in the given equation, then solve for $x:$ \begin{align*} x+\sqrt{x^2-1}+\frac{x+\sqrt{x^2-1}}{\left(x-\sqrt{x^2-1}\right)\left(x+\sqrt{x^2-1}\right)} &= 20 \\ x+\sqrt{x^2-1}+x+\sqrt{x^2-1} &= 20 \\ x+\sqrt{x^2-1} &= 10 \\ \sqrt{x^2-1} &= 10-x \\ x^2-1 &= 100-20x+x^2 \\ 20x &= 101 \\ x &= 5.05. \end{align*} We rationalize the denominator in the requested expression, then simplify the result: \begin{align*} x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}} &= x^2+\sqrt{x^4-1}+\frac{x^2-\sqrt{x^4-1}}{\left(x^2+\sqrt{x^4-1}\right)\left(x^2-\sqrt{x^4-1}\right)} \\ &= x^2+\sqrt{x^4-1}+x^2-\sqrt{x^4-1} \\ &= 2x^2 \\ &= \boxed{51.005} | C | 51.005 |
11b6fb54b95d21acbca83665d711f8e3 | https://artofproblemsolving.com/wiki/index.php/1991_AHSME_Problems/Problem_30 | For any set $S$ , let $|S|$ denote the number of elements in $S$ , and let $n(S)$ be the number of subsets of $S$ , including the empty set and the set $S$ itself. If $A$ $B$ , and $C$ are sets for which $n(A)+n(B)+n(C)=n(A\cup B\cup C)$ and $|A|=|B|=100$ , then what is the minimum possible value of $|A\cap B\cap C|$
$(A) 96 \ (B) 97 \ (C) 98 \ (D) 99 \ (E) 100$ | As $|A|=|B|=100$ $n(A)=n(B)=2^{100}$
As $n(A)+n(B)+n(C)=n(A \cup B \cup C)$ $2^{|A|}+2^{|B|}+2^{|C|}=2^{|A \cup B \cup C|}$ $2^{100}+2^{100}+2^{|C|}=2^{|A \cup B \cup C|}$
$2^{101}+2^{|C|}=2^{|A \cup B \cup C|}$ as $|C|$ and $|A \cup B \cup C|$ are integers, $|C|=101$ and $|A \cup B \cup C| = 102$
By the Principle of Inclusion-Exclusion $|A \cup B| = |A| + |B| - |A \cap B| = 200 - |A \cap B|$
$|A|=|B| \le |A \cup B| \le |A \cup B \cup C|$ $100 \le |A \cup B| \le 102$ $98 \le |A \cap B| \le 100$
By the Principle of Inclusion-Exclusion $|A \cup C| = |A| + |C| - |A \cap C| = 201 - |A \cap C|$
$|C| \le |A \cup C| \le |A \cup B \cup C|$ $101 \le |A \cup C| \le 102$ $99 \le |A \cap C| \le 100$
By the Principle of Inclusion-Exclusion $|B \cup C| = |B| + |C| - |B \cap C| = 201 - |B \cap C|$
$|C| \le |B \cup C| \le |A \cup B \cup C|$ $101 \le |B \cup C| \le 102$ $99 \le |B \cap C| \le 100$
By the Principle of Inclusion-Exclusion $|A \cap B \cap C|=|A \cup B \cup C|- |A| - |B| - |C| + |A \cap B| + |A \cap C|+|B \cap C| = 102-100-100-101+ |A \cap B| + |A \cap C|$ $+|B \cap C|=|A \cap B| + |A \cap C|+|B \cap C| -199$
\[98 + 99 + 99 - 199 \le |A \cap B \cap C| \le 100+100+100-199\]
\[\boxed{97} \le |A \cap B \cap C| \le 101\] | null | 97 |
6c063993a8b401ade028af93349fa871 | https://artofproblemsolving.com/wiki/index.php/1990_AHSME_Problems/Problem_1 | If $\dfrac{\frac{x}{4}}{2}=\dfrac{4}{\frac{x}{2}}$ , then $x=$
$\text{(A)}\ \pm\frac{1}{2}\qquad\text{(B)}\ \pm 1\qquad\text{(C)}\ \pm 2\qquad\text{(D)}\ \pm 4\qquad\text{(E)}\ \pm 8$ | Cross-multiplying leaves
\begin{align*}\dfrac{x^2}{8} &= 8\\ x^2 &= 64\\ \sqrt{x^2} &= \sqrt{64}\\ x &= \pm 8\end{align*}
So the answer is $\boxed{8}$ | E | 8 |
b808e1fc59c5ea8ed7956a49f3f2a27a | https://artofproblemsolving.com/wiki/index.php/1990_AHSME_Problems/Problem_26 | Ten people form a circle. Each picks a number and tells it to the two neighbors adjacent to them in the circle. Then each person computes and announces the average of the numbers of their two neighbors. The figure shows the average announced by each person ( not the original number the person picked.) [asy] unitsize(2 cm); for(int i = 1; i <= 10; ++i) { label("``" + (string) i + "''", dir(90 - 360/10*(i - 1))); } [/asy] The number picked by the person who announced the average $6$ was
$\textbf{(A) } 1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 10 \qquad \textbf{(E) }\text{not uniquely determined from the given information}$ | For $i\in\{1,2,3,\ldots,10\},$ suppose Person $i$ picks the number $a_i$ and announces the number $i.$ We wish to find $a_6.$
Taking the indices modulo $10,$ we are given that $\frac{a_{i-1}+a_{i+1}}{2}=i,$ from which $a_{i-1}+a_{i+1}=2i.$
We have ten equations: five with odd-numbered indices and five with even-numbered indices. Note that these two sets of equations are independent. The set that involves $a_6$ is \begin{align*} a_2 + a_4 & = 6, &&(1) \\ a_4 + a_6 & = 10, &&(2) \\ a_6 + a_8 & = 14, &&(3) \\ a_8 + a_{10} & = 18, &&(4) \\ a_{10} + a_2 & = 2. &&(5) \end{align*} Summing these five equations, we get $2(a_2 + a_4 + a_6 + a_8 + a_{10}) = 50,$ from which \[a_2 + a_4 + a_6 + a_8 + a_{10} = 25. \hspace{15mm} (\bigstar)\] Subtracting $(1)+(4)$ from $(\bigstar),$ we obtain $a_6=\boxed{1}.$ | A | 1 |
b808e1fc59c5ea8ed7956a49f3f2a27a | https://artofproblemsolving.com/wiki/index.php/1990_AHSME_Problems/Problem_26 | Ten people form a circle. Each picks a number and tells it to the two neighbors adjacent to them in the circle. Then each person computes and announces the average of the numbers of their two neighbors. The figure shows the average announced by each person ( not the original number the person picked.) [asy] unitsize(2 cm); for(int i = 1; i <= 10; ++i) { label("``" + (string) i + "''", dir(90 - 360/10*(i - 1))); } [/asy] The number picked by the person who announced the average $6$ was
$\textbf{(A) } 1 \qquad \textbf{(B) } 5 \qquad \textbf{(C) } 6 \qquad \textbf{(D) } 10 \qquad \textbf{(E) }\text{not uniquely determined from the given information}$ | For $i\in\{1,2,3,\ldots,10\},$ suppose Person $i$ announces the number $i.$
Let $x$ be the number picked by Person $6.$ We construct the following table: \[\begin{array}{c|c|c||l} & & & \\ [-2.5ex] \textbf{People} & \textbf{Average of \#s Picked} & \textbf{Sum of \#s Picked} & \multicolumn{1}{c}{\textbf{Conclusion}} \\ [0.5ex] \hline & & & \\ [-2ex] 6\text{ and }8 & 7 & 14 & \text{Person 8 picks } 14-x. \\ 8\text{ and }10 & 9 & 18 & \text{Person 10 picks } 4+x. \\ 10\text{ and }2 & 1 & 2 & \text{Person 2 picks } -2-x \\ 2\text{ and }4 & 3 & 6 & \text{Person 4 picks } 8+x \\ 4\text{ and }6 & 5 & 10 & \text{Person 6 picks } 2-x \\ \end{array}\] We have $x=2-x,$ from which $x=\boxed{1}.$ | A | 1 |
e3cbc8b21d76470ed1146e51fc747514 | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_4 | In the figure, $ABCD$ is an isosceles trapezoid with side lengths $AD=BC=5$ $AB=4$ , and $DC=10$ . The point $C$ is on $\overline{DF}$ and $B$ is the midpoint of hypotenuse $\overline{DE}$ in right triangle $DEF$ . Then $CF=$
[asy] defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0); draw(B--C--F--E--B--A--D--B^^C--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F); pair point=(5,3); label("$A$", A, N); label("$B$", B, N); label("$C$", C, S); label("$D$", D, S); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); markscalefactor=0.05; draw(rightanglemark(E,F,D), linewidth(0.7));[/asy]
$\textrm{(A)}\ 3.25\qquad\textrm{(B)}\ 3.5\qquad\textrm{(C)}\ 3.75\qquad\textrm{(D)}\ 4.0\qquad\textrm{(E)}\ 4.25$ | Drop perpendiculars from $A$ and $B$ ; then the triangle $DBY$ is similar to $DEF$ but with corresponding sides of half the length. [asy] defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0); draw(B--C--F--E--B--A--D--B^^C--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F); pair point=(5,3); label("$A$", A, N); label("$B$", B, N); label("$C$", C, S); label("$D$", D, S); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); markscalefactor=0.05; draw(rightanglemark(E,F,D), linewidth(0.7)); pair X=(3,0), Y=(7,0); draw(A--X^^B--Y,dotted); dot(X^^Y); label("$X$", X, S); label("$Y$", Y, S); [/asy] $XY=AB=4$ and $DX=YC=3$ , hence $DY=7\implies DF=14\implies CF=\boxed{4.0}$ | null | 4.0 |
f0bb2e0b734ec6be84066ab0211fb331 | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_5 | Toothpicks of equal length are used to build a rectangular grid as shown. If the grid is 20 toothpicks high and 10 toothpicks wide, then the number of toothpicks used is
[asy] real xscl = 1.2; int[] x = {0,1,2,4,5},y={0,1,3,4,5}; for(int a:x){ for(int b:y) { dot((a*xscl,b)); } } for(int a:x) { pair prev = (a,y[0]); for(int i = 1;i<y.length;++i) { pair p = (a,y[i]); pen pen = linewidth(.7); if(y[i]-prev.y!=1){ pen+=dotted; } draw((xscl*prev.x,prev.y)--(xscl*p.x,p.y),pen); prev = p; } }for(int a:y) { pair prev = (x[0],a); for(int i = 1;i<x.length;++i) { pair p = (x[i],a); pen pen = linewidth(.7); if(x[i]-prev.x!=1){ pen+=dotted; } draw((xscl*prev.x,prev.y)--(p.x*xscl,p.y),pen); prev = p; } } path lblx = (0,-.7)--(5*xscl,-.7); draw(lblx); label("$10$",lblx); path lbly = (5*xscl+.7,0)--(5*xscl+.7,5); draw(lbly); label("$20$",lbly);[/asy]
$\textrm{(A)}\ 30\qquad\textrm{(B)}\ 200\qquad\textrm{(C)}\ 410\qquad\textrm{(D)}\ 420\qquad\textrm{(E)}\ 430$ | There are 21 horizontal lines made of 10 matches, and 11 vertical lines made of 20 matches, and $21\cdot10+11\cdot20=\boxed{430}$ | null | 430 |
4ffcce3dbd5b0a9fa9a518dac3f7d18c | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_8 | For how many integers $n$ between 1 and 100 does $x^2+x-n$ factor into the product of two linear factors with integer coefficients?
$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$ | For $x^2+x-n$ to factor into a product of two linear factors, we must have $x^2+x-n = (x + a)(x + b)$ , where $a$ and $b$ are integers.
By expansion of the product of the linear factors and comparison to the original quadratic,
$ab = -n$
$a + b = 1$
The only way for this to work if $n$ is a positive integer is if $a = -b +1$
Here are the possible pairs:
This gives us 9 integers for $n$ $\boxed{9}$ | D | 9 |
4ffcce3dbd5b0a9fa9a518dac3f7d18c | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_8 | For how many integers $n$ between 1 and 100 does $x^2+x-n$ factor into the product of two linear factors with integer coefficients?
$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$ | For $x^2+x-n$ to factor into a product of two linear factors, the discriminant must be a perfect square. The discriminant equals $\sqrt{1^2-4(1)(-n)}$ or $\sqrt{4n+1}$
Since $n$ must be a positive integer, $4n+1$ must be odd because if it were even, $4n$ would have to be both odd and divisible by 4, which is a contradiction. Therefore, $n$ must be even.
The maximum value of $4n+1$ is 401, and the minimum is 5. Therefore, we must find the number of odd perfect squares between 5 and 401.
There are 9: $9,25,49,81,121,169,225,289,$ and $361$ . Therefore, the answer is $\boxed{9}$ | D | 9 |
c704562094cd2e7ed6016c20637bda11 | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_9 | Mr. and Mrs. Zeta want to name their baby Zeta so that its monogram (first, middle, and last initials) will be in alphabetical order with no letter repeated. How many such monograms are possible?
$\textrm{(A)}\ 276\qquad\textrm{(B)}\ 300\qquad\textrm{(C)}\ 552\qquad\textrm{(D)}\ 600\qquad\textrm{(E)}\ 15600$ | We see that for any combination of two distinct letters other than Z (as the last name will automatically be Z), there is only one possible way to arrange them in alphabetical order, thus the answer is just $\dbinom{25}{2}=\boxed{300}$ | B | 300 |
d89e2d8fad77dd2d180b67e78e80fdab | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_10 | Consider the sequence defined recursively by $u_1=a$ (any positive number), and $u_{n+1}=-1/(u_n+1)$ $n=1,2,3,...$ For which of the following values of $n$ must $u_n=a$
$\mathrm{(A) \ 14 } \qquad \mathrm{(B) \ 15 } \qquad \mathrm{(C) \ 16 } \qquad \mathrm{(D) \ 17 } \qquad \mathrm{(E) \ 18 }$ | Repeatedly applying the function, and simplifying, we get \[a,\quad-\frac1{a+1},\quad-\frac{a+1}a,\] and then $a$ again. So $a$ must appear at every third term after $u_1$ . The only option given of the form $1+3k$ is $\boxed{16}$ | C | 16 |
a21d9decf3440a524acdcb41918b2a28 | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_16 | A lattice point is a point in the plane with integer coordinates. How many lattice points are on the line segment whose endpoints are $(3,17)$ and $(48,281)$ ? (Include both endpoints of the segment in your count.)
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 46$ | The difference in the $y$ -coordinates is $281 - 17 = 264$ , and the difference in the $x$ -coordinates is $48 - 3 = 45$ .
The gcd of 264 and 45 is 3, so the line segment joining $(3,17)$ and $(48,281)$ has slope \[\frac{88}{15}.\] The points on the line have coordinates \[\left(3+t,17+\frac{88}{15}t\right).\] If $t$ is an integer, the $y$ -coordinate of this point is an integer if and only if $t$ is a multiple of 15. The points where $t$ is a multiple of 15 on the segment $3\leq x\leq 48$ are $3$ $3+15$ $3+30$ , and $3+45$ . There are 4 lattice points on this line. Hence the answer is $\boxed{4}$ | B | 4 |
450c28be35c78bb9a48617e610caae12 | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_21 | A square flag has a red cross of uniform width with a blue square in the center on a white background as shown. (The cross is symmetric with respect to each of the diagonals of the square.) If the entire cross (both the red arms and the blue center) takes up 36% of the area of the flag, what percent of the area of the flag is blue?
[asy] unitsize(2.5 cm); pair[] A, B, C; real t = 0.2; A[1] = (0,0); A[2] = (1,0); A[3] = (1,1); A[4] = (0,1); B[1] = (t,0); B[2] = (1 - t,0); B[3] = (1,t); B[4] = (1,1 - t); B[5] = (1 - t,1); B[6] = (t,1); B[7] = (0,1 - t); B[8] = (0,t); C[1] = extension(B[1],B[4],B[7],B[2]); C[2] = extension(B[3],B[6],B[1],B[4]); C[3] = extension(B[5],B[8],B[3],B[6]); C[4] = extension(B[7],B[2],B[5],B[8]); fill(C[1]--C[2]--C[3]--C[4]--cycle,blue); fill(A[1]--B[1]--C[1]--C[4]--B[8]--cycle,red); fill(A[2]--B[3]--C[2]--C[1]--B[2]--cycle,red); fill(A[3]--B[5]--C[3]--C[2]--B[4]--cycle,red); fill(A[4]--B[7]--C[4]--C[3]--B[6]--cycle,red); draw(A[1]--A[2]--A[3]--A[4]--cycle); draw(B[1]--B[4]); draw(B[2]--B[7]); draw(B[3]--B[6]); draw(B[5]--B[8]); [/asy]
$\text{(A)}\ 0.5\qquad\text{(B)}\ 1\qquad\text{(C)}\ 2\qquad\text{(D)}\ 3\qquad\text{(E)}\ 6$ | The diagram can be quartered as shown: [asy] draw((0,0)--(0,5)--(5,5)--(5,0)--(0,0)); draw((0,1)--(4,5)); draw((1,0)--(5,4)); draw((0,4)--(4,0)); draw((1,5)--(5,1)); draw((0,0)--(5,5),dotted); draw((0,5)--(5,0),dotted); [/asy] and reassembled into two smaller squares of side $k$ , each of which looks like this: [asy] draw((0,0)--(0,5)--(5,5)--(5,0)--(0,0)); draw((0,1)--(4,1)--(4,5)); draw((1,0)--(1,4)--(5,4)); label("blue",(0.5,0.5)); label("blue",(4.5,4.5)); label("red",(0.5,4.5)); label("red",(4.5,0.5)); label("white",(2.5,2.5)); [/asy] The border in this figure is the former cross, which still occupies 36% of the area. Therefore the inner square occupies 64% of the area, from which we deduce that it is $0.8k \times 0.8k$ , and that one blue square must be $0.1k\times 0.1k=0.01k^2$ or 1% each. Thus the blue area is $\boxed{2}$ of the total. | null | 2 |
9d0f8adce9fd280ec2f83b1ad77b7ad7 | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_22 | A child has a set of $96$ distinct blocks. Each block is one of $2$ materials (plastic, wood), $3$ sizes (small, medium, large), $4$ colors (blue, green, red, yellow), and $4$ shapes (circle, hexagon, square, triangle). How many blocks in the set differ from the 'plastic medium red circle' in exactly $2$ ways? (The 'wood medium red square' is such a block)
(A) 29 (B) 39 (C) 48 (D) 56 (E) 62 | The process of choosing a block can be represented by a generating function. Each choice we make can match the 'plastic medium red circle' in one of its qualities $(1)$ or differ from it in $k$ different ways $(kx)$ . Choosing the material is represented by the factor $(1+1x)$ , choosing the size by the factor $(1+2x)$ , etc: \[(1+x)(1+2x)(1+3x)^2\] Expanding out the first two factors and the square: \[(1+3x+2x^2)(1+6x+9x^2)\] By expanding further we can find the coefficient of $x^2$ , which represents the number of blocks differing from the original block in exactly two ways. We don't have to expand it completely, but choose the terms which will be multiplied together to result in a constant multiple of $x^2$ \[1\cdot9+3\cdot6+2\cdot1=\boxed{29}\] | null | 29 |
9d0f8adce9fd280ec2f83b1ad77b7ad7 | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_22 | A child has a set of $96$ distinct blocks. Each block is one of $2$ materials (plastic, wood), $3$ sizes (small, medium, large), $4$ colors (blue, green, red, yellow), and $4$ shapes (circle, hexagon, square, triangle). How many blocks in the set differ from the 'plastic medium red circle' in exactly $2$ ways? (The 'wood medium red square' is such a block)
(A) 29 (B) 39 (C) 48 (D) 56 (E) 62 | The amount of ways we can differ in the material, sizes, colors, and shapes category is 1, 2, 3, and 3 respectively (we take away one because we cannot choose the characteristic already chosen). We can choose to differ two of these characteristics, so our answer is $1\cdot2 + 1\cdot3 + 1\cdot3 + 2\cdot3 + 2\cdot3 + 3\cdot3 = \boxed{29}$ | null | 29 |
1536879be5f866916728912c6e0432d9 | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_23 | A particle moves through the first quadrant as follows. During the first minute it moves from the origin to $(1,0)$ . Thereafter, it continues to follow the directions indicated in the figure, going back and forth between the positive x and y axes, moving one unit of distance parallel to an axis in each minute. At which point will the particle be after exactly 1989 minutes?
[asy] import graph; Label f; f.p=fontsize(6); xaxis(0,3.5,Ticks(f, 1.0)); yaxis(0,4.5,Ticks(f, 1.0)); draw((0,0)--(1,0)--(1,1)--(0,1)--(0,2)--(2,2)--(2,0)--(3,0)--(3,3)--(0,3)--(0,4)--(1.5,4),blue+linewidth(2)); arrow((2,4),dir(180),blue); [/asy]
$\text{(A)}\ (35,44)\qquad\text{(B)}\ (36,45)\qquad\text{(C)}\ (37,45)\qquad\text{(D)}\ (44,35)\qquad\text{(E)}\ (45,36)$ | Squares of size $1\times1,\ 2\times2,\ 3\times3,\ ...$ are successively enclosed between the path and the axes.
[asy] import graph; axialshade((0,0)--(0,3)--(3,3)--(3,0)--cycle,yellow,(-6,-6),white,(3,3)); axialshade((0,0)--(0,2)--(2,2)--(2,0)--cycle,orange,(-3,-3),white,(2,2)); axialshade((0,0)--(0,1)--(1,1)--(1,0)--cycle,red,(-2,-2),white,(1,1)); Label f; f.p=fontsize(6); xaxis(0,3.5,Ticks(f, 1.0)); yaxis(0,4.5,Ticks(f, 1.0)); draw((0,0)--(1,0)--(1,1)--(0,1),red+linewidth(2)); draw((0,1)--(0,2)--(2,2)--(2,0),orange+linewidth(2)); draw((2,0)--(3,0)--(3,3)--(0,3),yellow+linewidth(2)); draw((0,3)--(0,4)--(1.5,4),green+linewidth(2)); arrow((2,4),dir(180),green); dot((0,1),red);dot((2,0),orange);dot((0,3),yellow); [/asy]
It takes $1+1+1$ minutes to enclose the first square, $1+2+2$ minutes to enclose the second, $1+3+3$ minutes to enclose the third, and so on. After odd squares, the particle is on the Y axis; after even squares, the particle is on the X axis.
First we find the highest integer $n$ such that $\sum_{k=1}^n1+2k\le1989$ . The sum is equal to \[\sum_{k=1}^n1+2\sum_{k=1}^nk=n+n(n+1)=n(n+2)=(n+1)^2-1\] so the highest value is $n=43$ for which the sum is $1935$
After $1935$ minutes the 43rd square is enclosed and the particle is at the point $(0,43)$ . During the 1936th minute it moves up to $(0,44)$ . At the end of the 1980th minute it has moved right to $(44,44)$ . After this it moves downward, and at the end of the 1989th minute it is at $(44,35)$ . The answer is $\boxed{44,35}$ | D | 44,35 |
a5a2916842015ec7c74472be6b4b80ff | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_24 | Five people are sitting at a round table. Let $f\geq 0$ be the number of people sitting next to at least 1 female and $m\geq0$ be the number of people sitting next to at least one male. The number of possible ordered pairs $(f,m)$ is
$\mathrm{(A) \ 7 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 9 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 11 }$ | Suppose there are more men than women; then there are between zero and two women.
If there are no women, the pair is $(0,5)$ . If there is one woman, the pair is $(2,5)$
If there are two women, there are two arrangements: one in which they are together, and one in which they are apart, giving the pairs $(4,5)$ and $(3,4)$
All four pairs are asymmetrical; therefore by symmetry there are eight pairs altogether, so $\boxed{8}$ | B | 8 |
a5a2916842015ec7c74472be6b4b80ff | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_24 | Five people are sitting at a round table. Let $f\geq 0$ be the number of people sitting next to at least 1 female and $m\geq0$ be the number of people sitting next to at least one male. The number of possible ordered pairs $(f,m)$ is
$\mathrm{(A) \ 7 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 9 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 11 }$ | Denote $T_n$ as the number of such pairs for $n$ people. Then for $T_{n-1}$ , when we add an extra spot, we can either have a male or female giving two options. Note that these two options however double the value of $T_{n-1}$ . Now if we note that $T_2=1$ , we have that $T_5=8$ , so that the answer is $\boxed{8}$ | B | 8 |
06538951a6a1b3aeb1a517e27687aecc | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_25 | In a certain cross country meet between 2 teams of 5 runners each, a runner who finishes in the $n$ th position contributes $n$ to his teams score. The team with the lower score wins. If there are no ties among the runners, how many different winning scores are possible?
(A) 10 (B) 13 (C) 27 (D) 120 (E) 126 | The scores of all ten runners must sum to $55$ . So a winning score is anything between $1+2+3+4+5=15$ and $\lfloor\tfrac{55}{2}\rfloor=27$ inclusive. It is easy to check that this range is covered by considering $1+2+3+4+x$ $1+2+3+x+10$ and $1+2+x+9+10$ , so the answer is $\boxed{13}$ | null | 13 |
a4a6034ecf3ee0a2558cab02f7844278 | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_30 | Suppose that 7 boys and 13 girls line up in a row. Let $S$ be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row $\text{GBBGGGBGBGGGBGBGGBGG}$ we have that $S=12$ . The average value of $S$ (if all possible orders of these 20 people are considered) is closest to
$\text{(A)}\ 9\qquad\text{(B)}\ 10\qquad\text{(C)}\ 11\qquad\text{(D)}\ 12\qquad\text{(E)}\ 13$ | We approach this problem using Linearity of Expectation. Consider a pair of two people standing next to each other. Ignoring all other people, the probability that a boy is standing on the left position and a girl is standing on the right position is $\frac7{20}\cdot\frac{13}{19}$ . Similarly, if a girl is standing on the left position and a boy is standing on the right position the probability is also $\frac{7\cdot 13}{20\cdot 19}$ . Thus, the total probability of the two people being one boy and one girl is $\frac{91}{190}$
There are a total of 19 different adjacent pairs, so by Linearity of Expectation, we have that the expected value of $S$ is $\frac{91}{10} \to \boxed{9}$ | A | 9 |
a4a6034ecf3ee0a2558cab02f7844278 | https://artofproblemsolving.com/wiki/index.php/1989_AHSME_Problems/Problem_30 | Suppose that 7 boys and 13 girls line up in a row. Let $S$ be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row $\text{GBBGGGBGBGGGBGBGGBGG}$ we have that $S=12$ . The average value of $S$ (if all possible orders of these 20 people are considered) is closest to
$\text{(A)}\ 9\qquad\text{(B)}\ 10\qquad\text{(C)}\ 11\qquad\text{(D)}\ 12\qquad\text{(E)}\ 13$ | Suppose that the class tried every configuration. Boy $i$ and girl $j$ would stand next to each other in $2$ different orders, in $19$ different positions, $18!$ times each. Summing over all $i,j$ gives $7\cdot13\cdot2\cdot19\cdot18!=\tfrac{91}{10}\cdot20!$ , so the average value of $S$ is $\boxed{9}$ | A | 9 |
a8a003efddee9216f7747c6658d37f36 | https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_3 | [asy] draw((0,0)--(1,0)--(1,4)--(0,4)--(0,0)--(0,1)--(-1,1)--(-1,2)); draw((-1,2)--(0,2)--(0,4)--(-1,4)--(-1,5)--(1,5)--(1,6)--(0,6)); draw((0,6)--(0,5)--(3,5)--(3,6)--(4,6)--(4,2)--(5,2)); draw((5,2)--(5,1)--(1,1)--(3,1)--(3,0)--(4,0)--(4,1)); draw((1,4)--(3,4)--(3,2)--(1,2)--(4,2)--(3,2)--(3,6)); draw((3,6)--(4,6)--(4,5)--(5,5)--(5,4)--(4,4)); [/asy]
Four rectangular paper strips of length $10$ and width $1$ are put flat on a table and overlap perpendicularly as shown. How much area of the table is covered?
$\text{(A)}\ 36 \qquad \text{(B)}\ 40 \qquad \text{(C)}\ 44 \qquad \text{(D)}\ 98 \qquad \text{(E)}\ 100$ | We first notice that the paper strips cover up part of the others. Since the width of the overlap is $1$ and the length of the overlap is $1$ , the area of
each of the strips with the overlap is $(10\cdot 1)-1=9$ . Since there are 4 strips, $4\cdot 9=36 \implies \boxed{36}$ | A | 36 |
fd6c85b4cf12a7f11ac0c6bc25cbf811 | https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_5 | If $b$ and $c$ are constants and $(x + 2)(x + b) = x^2 + cx + 6$ , then $c$ is
$\textbf{(A)}\ -5\qquad \textbf{(B)}\ -3\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 5$ | We first start out by expanding the left side of the equation, $(x+2)(x+b)=x^{2}+bx+2x+2b=x^2+(2+b)x+2b=x^2+cx+6$ . We know the constant
terms have to be equal so we have $2b=6$ , so $b=3$ . Plugging $b=3$ back in yields $x^2+(2+3)x+6=x^2+cx+6$ . Thus, $c=5 \implies \boxed{5}$ | E | 5 |
c2847e0635d2724b4f8babc95b77b659 | https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_7 | Estimate the time it takes to send $60$ blocks of data over a communications channel if each block consists of $512$ "chunks" and the channel can transmit $120$ chunks per second.
$\textbf{(A)}\ 0.04 \text{ seconds}\qquad \textbf{(B)}\ 0.4 \text{ seconds}\qquad \textbf{(C)}\ 4 \text{ seconds}\qquad \textbf{(D)}\ 4\text{ minutes}\qquad \textbf{(E)}\ 4\text{ hours}$ | We want to figure out the number of chunks in $60$ blocks, so we have $60\cdot 512 \approx 30000$ . We divide this by $120$ to determine the number of seconds necessary to transmit. $30000/120 \approx 250$ , which means that it takes approximately $4$ minutes to transmit. Thus, the answer is $\boxed{4}$ | D | 4 |
c2847e0635d2724b4f8babc95b77b659 | https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_7 | Estimate the time it takes to send $60$ blocks of data over a communications channel if each block consists of $512$ "chunks" and the channel can transmit $120$ chunks per second.
$\textbf{(A)}\ 0.04 \text{ seconds}\qquad \textbf{(B)}\ 0.4 \text{ seconds}\qquad \textbf{(C)}\ 4 \text{ seconds}\qquad \textbf{(D)}\ 4\text{ minutes}\qquad \textbf{(E)}\ 4\text{ hours}$ | This solution is if you are running out of time and just want to write down an answer. So, this is quite unreliable. You can logic it out. It doesn't make sense for the first three options to be the answer since that is way too quick. The last option is way too long. That just leaves $\boxed{4}$ | D | 4 |
72cb38b649a71ae5e9b22a33626ca414 | https://artofproblemsolving.com/wiki/index.php/1988_AHSME_Problems/Problem_9 | [asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(0,0), B=(16,0), C=(16,16), D=(0,16), E=(32,0), F=(48,0), G=(48,16), H=(32,16), I=(0,8), J=(10,8), K=(10,16), L=(32,6), M=(40,6), N=(40,16); draw(A--B--C--D--A^^E--F--G--H--E^^I--J--K^^L--M--N); label("S", (18,8)); label("S", (50,8)); label("Figure 1", (A+B)/2, S); label("Figure 2", (E+F)/2, S); label("10'", (I+J)/2, S); label("8'", (12,12)); label("8'", (L+M)/2, S); label("10'", (42,11)); label("table", (5,12)); label("table", (36,11)); [/asy]
An $8'\times 10'$ table sits in the corner of a square room, as in Figure $1$ below.
The owners desire to move the table to the position shown in Figure $2$ .
The side of the room is $S$ feet. What is the smallest integer value of $S$ for which the table can be moved as desired without tilting it or taking it apart?
$\textbf{(A)}\ 11\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 13\qquad \textbf{(D)}\ 14\qquad \textbf{(E)}\ 15$ | Small correction: The writer below has maximized the area of the rectangle (with sides parallel to the walls) that fits around the table, but there is a larger single dimension we can find in the table. The height or width is maximized when the diagonal of the table is horizontal or vertical. By the Pythagorean Theorem, this diagonal is $\sqrt{8^2+10^2} = \sqrt{164},$ which is between $\sqrt{144}$ and $\sqrt{169},$ so the answer is still $\textbf{(C)}.$ -hailstone
We begin by thinking about the motion of the table. As it moves, the table will have it's maximum height and width when the rectangle's sides form $45$ degree angles relative to the sides of the square. Therefore, by the Pythagorean Theorem, we have that $S= 4\sqrt{2}+5\sqrt{2}$ , with $4\sqrt{2}$ being the length of the leg formed by the side of the square with length $8$ and $5\sqrt{2}$ being the length of the leg formed by the side of the square with length $10$ . Adding these up yields $9\sqrt{2}$
We have that $\sqrt{2}\approx 1.414\approx 1.4$ . That means that $9\sqrt{2}\approx 12.6$ , which rounds up to $\boxed{13}$ | C | 13 |