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5a853380761c42487ee1f6ee5955bafe | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_2 | $90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$ | The finite arithmetic sequence formula states that the sum in the sequence is equal to $\frac{n}{2}\cdot(a_1+a_n)$ where $n$ is the number of terms in the sequence, $a_1$ is the first term and $a_n$ is the last term.
Applying the formula, we have: \[\frac{10}{2}\cdot(90+99)=\boxed{945}\] | B | 945 |
5a853380761c42487ee1f6ee5955bafe | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_2 | $90+91+92+93+94+95+96+97+98+99=$
$\text{(A)}\ 845 \qquad \text{(B)}\ 945 \qquad \text{(C)}\ 1005 \qquad \text{(D)}\ 1025 \qquad \text{(E)}\ 1045$ | The expression is equal to the sum of integers from $1$ to $99$ minus the sum of integers from $1$ to $89$ , so it is equal to $\frac{99(100)}{2} - \frac{89(90)}{2} = 4950 - 4005 = \boxed{945}$ | B | 945 |
a3ffe1a30f9d9e3982dd228f559bee0c | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_3 | $\frac{10^7}{5\times 10^4}=$
$\text{(A)}\ .002 \qquad \text{(B)}\ .2 \qquad \text{(C)}\ 20 \qquad \text{(D)}\ 200 \qquad \text{(E)}\ 2000$ | We immediately see some canceling. We see powers of ten on the top and on the bottom of the fraction, and we make quick work of this: \[\frac{10^7}{5 \times 10^4} = \frac{10^3}{5}\]
We know that $10^3 = 10 \times 10 \times 10$ , so
\begin{align*} \frac{10^3}{5} &= \frac{10\times 10\times 10}{5} \\ &= 2\times 10\times 10 \\ &= 200 \\ \end{align*}
So the answer is $\boxed{200}$ | D | 200 |
97aed6b4474cfa7a0e3aa600642c7d64 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_4 | The area of polygon $ABCDEF$ , in square units, is
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \text{(E)}\ 74$
[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy] | [asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((2,4)--(6,4),dashed); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("G",(6,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]
Obviously, there are no formulas to find the area of such a messed up shape, but we do recognize some shapes we do know how to find the area of.
If we continue segment $\overline{FE}$ until it reaches the right side at $G$ , we create two rectangles - one on the top and one on the bottom.
We know how to find the area of a rectangle, and we're given the sides! We can easily find that the area of $ABGF$ is $6\times5 = 30$ . For the rectangle on the bottom, we do know the length of one of its sides, but we don't know the other.
Note that $GC+GB=9$ , and $GB=AF=5$ , so we must have \[GC+5=9\Rightarrow GC=4\]
The area of the bottom rectangle is then \[(DC)(GC)=4\times 4=16\]
Finally, we just add the areas of the rectangles together to get $16 + 30 = 46$
$\boxed{46}$ | C | 46 |
97aed6b4474cfa7a0e3aa600642c7d64 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_4 | The area of polygon $ABCDEF$ , in square units, is
$\text{(A)}\ 24 \qquad \text{(B)}\ 30 \qquad \text{(C)}\ 46 \qquad \text{(D)}\ 66 \qquad \text{(E)}\ 74$
[asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy] | [asy] draw((0,9)--(6,9)--(6,0)--(2,0)--(2,4)--(0,4)--cycle); draw((0,4)--(0,0),dashed); draw((0,0)--(2,0),dashed); label("A",(0,9),NW); label("B",(6,9),NE); label("C",(6,0),SE); label("D",(2,0),SW); label("E",(2,4),NE); label("F",(0,4),SW); label("G",(0,0),SW); label("6",(3,9),N); label("9",(6,4.5),E); label("4",(4,0),S); label("5",(0,6.5),W); [/asy]
Let $\langle ABCDEF \rangle$ be the area of polygon $ABCDEF$ . Also, let $G$ be the intersection of $DC$ and $AF$ when both are extended.
Clearly, \[\langle ABCDEF \rangle = \langle ABCG \rangle - \langle GFED \rangle\]
Since $AB=6$ and $BC=9$ $\langle ABCG \rangle =6\times 9=54$
To compute the area of $GFED$ , note that \[AB=GD+DC\] \[BC=GF+FA\]
We know that $AB=6$ $DC=4$ $BC=9$ , and $FA=5$ , so \[6=GD+4\Rightarrow GD=2\] \[9=GF+5\Rightarrow GF=4\]
Thus $\langle GFED \rangle = 4\times 2=8$
Finally, we have \begin{align*} \langle ABCDEF \rangle &= \langle ABCG \rangle - \langle GFED \rangle \\ &= 54-8 \\ &= 46 \\ \end{align*}
This is answer choice $\boxed{46}$ | C | 46 |
1d0946487dce2b423dc0e0cbe6025dc0 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_6 | A ream of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high?
$\text{(A)}\ 250 \qquad \text{(B)}\ 550 \qquad \text{(C)}\ 667 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1250$ | We could solve the first equation for the thickness of one sheet of paper, and divide into the 2nd equation (which is one way to do the problem), but there are other ways, too.
Let's say that $500\text{ sheets}=5\text{ cm}\Rightarrow \frac{500 \text{ sheets}}{5 \text{ cm}} = 1$ . So by multiplying $7.5 \text{ cm}$ by this fraction, we SHOULD get the number of sheets in 7.5 cm. Solving gets
\begin{align*} \frac{7.5 \times 500}{5} &= 7.5 \times 100 \\ &= 750 \text{ sheets} \\ \end{align*}
$750$ is $\boxed{750}$ | D | 750 |
1d0946487dce2b423dc0e0cbe6025dc0 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_6 | A ream of paper containing $500$ sheets is $5$ cm thick. Approximately how many sheets of this type of paper would there be in a stack $7.5$ cm high?
$\text{(A)}\ 250 \qquad \text{(B)}\ 550 \qquad \text{(C)}\ 667 \qquad \text{(D)}\ 750 \qquad \text{(E)}\ 1250$ | We can set up a direct proportion relating the amount of sheets to the thickness because according to the problem, all the papers have the same thickness.
Our proportion is \[\frac{5}{500}=\frac{7.5}{x}\] where $x$ is the number we are looking for.
Next, we cross-multiply to get $5x=500 \times 7.5$ so $x=750$ which is $\boxed{750}$ | D | 750 |
b598e72f2eaf5b35432c4868be17ee79 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_7 | A "stair-step" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\text{th}$ row is
[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]
$\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38$ | The best way to solve this problem is to find patterns and to utilize them to our advantage. For example, we can't really do anything without knowing how many squares there are in the 37th row. But who wants to continue the diagram for 37 rows? And what if the problem said 100,000th row? It'll still be possible - but not if your method is to continue the diagram...
So hopefully there's a pattern. We find a pattern by noticing what is changing from row 1 to row 2. Basically, for the next row, we are just adding 2 squares (1 on each side) to the number of squares we had in the previous row. So each time we're adding 2. So how can we find $N$ , if $N$ is the number of squares in the $a^\text{th}$ row of this diagram? We can't just say that $N = 1 + 2a$ , because it doesn't work for the first row. But since 1 is the first term, we have to EXCLUDE the first term, meaning that we must subtract 1 from a. Thus, $N = 1 + 2\times(a - 1) = 2a - 1$ . So in the 37th row we will have $2 \times 37 - 1 = 74 - 1 = 73$
You may now be thinking - aha, we're finished. But we're only half finished. We still need to find how many black squares there are in these 73 squares. Well let's see - they alternate white-black-white-black... but we can't divide by two - there aren't exactly as many white squares as black squares... there's always 1 more white square... aha! If we subtract 1 from the number of squares (1 white square), we will have exactly 2 times the number of black squares.
Thus, the number of black squares is $\frac{73 - 1}{2} = \frac{72}{2} = 36$
36 is $\boxed{36}$ | C | 36 |
b598e72f2eaf5b35432c4868be17ee79 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_7 | A "stair-step" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\text{th}$ row is
[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]
$\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38$ | Note that each row adds one black and one white square to the end of the previous one, and then shifts the new row over a little. Since the first row had no black squares, the number of black squares in any row is one less than the row number.
Now that this has been established, we just have $37-1=36\rightarrow \boxed{36}$ | C | 36 |
b598e72f2eaf5b35432c4868be17ee79 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_7 | A "stair-step" figure is made of alternating black and white squares in each row. Rows $1$ through $4$ are shown. All rows begin and end with a white square. The number of black squares in the $37\text{th}$ row is
[asy] draw((0,0)--(7,0)--(7,1)--(0,1)--cycle); draw((1,0)--(6,0)--(6,2)--(1,2)--cycle); draw((2,0)--(5,0)--(5,3)--(2,3)--cycle); draw((3,0)--(4,0)--(4,4)--(3,4)--cycle); fill((1,0)--(2,0)--(2,1)--(1,1)--cycle,black); fill((3,0)--(4,0)--(4,1)--(3,1)--cycle,black); fill((5,0)--(6,0)--(6,1)--(5,1)--cycle,black); fill((2,1)--(3,1)--(3,2)--(2,2)--cycle,black); fill((4,1)--(5,1)--(5,2)--(4,2)--cycle,black); fill((3,2)--(4,2)--(4,3)--(3,3)--cycle,black); [/asy]
$\text{(A)}\ 34 \qquad \text{(B)}\ 35 \qquad \text{(C)}\ 36 \qquad \text{(D)}\ 37 \qquad \text{(E)}\ 38$ | We can easily spot the pattern. In the first row, there are $0$ black squares. In the second row, there is $1$ black square. So the pattern is that in order to find the number of black squares in a row, you need to subtract $1$ from the row number. Therefore, the number of black squares in the $37\text{th}$ row we subtract 1. $37-1=36\rightarrow \boxed{36}$ | C | 36 |
a0bf41760c52b81382af845caba8a75d | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_8 | If $a = - 2$ , the largest number in the set $\{ - 3a, 4a, \frac {24}{a}, a^2, 1\}$ is
$\text{(A)}\ -3a \qquad \text{(B)}\ 4a \qquad \text{(C)}\ \frac {24}{a} \qquad \text{(D)}\ a^2 \qquad \text{(E)}\ 1$ | Since all the numbers are small, we can just evaluate the set to be \[\{ (-3)(-2), 4(-2), \frac{24}{-2}, (-2)^2, 1 \}= \{ 6, -8, -12, 4, 1 \}\]
The largest number is $6$ , which corresponds to $-3a$
$\boxed{3}$ | A | 3 |
5d7fd1e73891a954ae79d6a0ed07fdd1 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_15 | How many whole numbers between $100$ and $400$ contain the digit $2$
$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148$ | This is a very common type of counting problem that you'll see quite often. Doing this the simple way would take too long, and might even make lots of mistakes.
If you ever learned about complementary counting , this would be the best time to utilize it. Instead of counting how many DO have 2's, why don't we count how many that DON'T?
So let's find the number of numbers. Obviously, we'd start by subtracting 100 from 400, getting us 300, but we're not done. Since just subtracting includes the number 400, we must subtract one (because 400 isn't allowed - it says between), getting us 299.
So how many numbers are there that DON'T have a 2? Well, we have 2 possibilities for the hundreds digit (1, 3, note that 2 is not allowed), 9 possibilities for the tens digit (1, 3, 4, 5, ... , 9, 0), and 9 possibilities for the ones digit. $2 \times 9 \times 9 = 162$ . However, one of the numbers we counted is $100$ , which isn't allowed, so there are $162-1=161$ numbers without a 2.
Since there are 299 numbers in total and 161 that DON'T have any 2's, $299 - 161 = 138$ numbers WILL have at least one two.
$\boxed{138}$ | C | 138 |
5d7fd1e73891a954ae79d6a0ed07fdd1 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_15 | How many whole numbers between $100$ and $400$ contain the digit $2$
$\text{(A)}\ 100 \qquad \text{(B)}\ 120 \qquad \text{(C)}\ 138 \qquad \text{(D)}\ 140 \qquad \text{(E)}\ 148$ | As in the previous solution, get rid of the $400$ to make things easier. This gives us the numbers from $100$ to $399$
Let $A$ be the event that the first digit is $2$ $B$ be the event that the second digit is $2$ , and $C$ be the event that the third digit is $2$ . Then, PIE says that our answer will be $|A|+|B|+|C|-|A\cup B|-|A\cup C|-|B\cup C|+|A\cup B\cup C|$
We have that $|A|$ is just $1\times10\times10=100$ $|B|=3\times1\times10=30$ , and $|C|=3\times10\times1=30$
Next, $|A\cup B|$ is just having something in the form $22\_$ , so there are $10$ ways. Similarly, $|A\cup C|$ means that we have $2\_2$ , so there are again $10$ ways. Finally, $|B\cup C|$ means that our number is like $\_22$ , so there are $3$ ways.
Finally, $|A\cup B\cup C|$ counts the number of three digit numbers with all three digits $2$ , which there is only $1$ of: $222$
Putting this together, our answer will be $100+30+30-10-10-3+1=\boxed{138}$ | C | 138 |
8bb4619fff5729bf23319bfe94dfc150 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_16 | The ratio of boys to girls in Mr. Brown's math class is $2:3$ . If there are $30$ students in the class, how many more girls than boys are in the class?
$\text{(A)}\ 1 \qquad \text{(B)}\ 3 \qquad \text{(C)}\ 5 \qquad \text{(D)}\ 6 \qquad \text{(E)}\ 10$ | Let the number of boys be $2x$ . It follows that the number of girls is $3x$ . These two values add up to $30$ students, so \[2x+3x=5x=30\Rightarrow x=6\]
The difference between the number of girls and the number of boys is $3x-2x=x$ , which is $6$ , so the answer is $\boxed{6}$ | D | 6 |
07d768ff3c3dccb094caded1383afcf2 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_17 | If your average score on your first six mathematics tests was $84$ and your average score on your first seven mathematics tests was $85$ , then your score on the seventh test was
$\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92$ | If the average score of the first six is $84$ , then the sum of those six scores is $6\times 84=504$
The average score of the first seven is $85$ , so the sum of the seven is $7\times 85=595$
Taking the difference leaves us with just the seventh score, which is $595-504=91$ , so the answer is $\boxed{91}$ | D | 91 |
07d768ff3c3dccb094caded1383afcf2 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_17 | If your average score on your first six mathematics tests was $84$ and your average score on your first seven mathematics tests was $85$ , then your score on the seventh test was
$\text{(A)}\ 86 \qquad \text{(B)}\ 88 \qquad \text{(C)}\ 90 \qquad \text{(D)}\ 91 \qquad \text{(E)}\ 92$ | Let's remove the condition that the average of the first seven tests is $85$ , and say the 7th test score was a $0$ . Then, the average of the first seven tests would be \[\frac{6\times 84}{7}=72\]
If we increase the seventh test score by $7$ , the average will increase by $\frac{7}{7}=1$ . We need the average to increase by $85-72=13$ , so the seventh test score is $7\times 13=91$ more than $0$ , which is clearly $91$ . This is choice $\boxed{91}$ | D | 91 |
fea55e466a9107d3de314c6379cbafdc | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_19 | If the length and width of a rectangle are each increased by $10\%$ , then the perimeter of the rectangle is increased by
$\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\%$ | Let the width be $w$ and the length be $l$ . Then, the original perimeter is $2(w+l)$
After the increase, the new width and new length are $1.1w$ and $1.1l$ , so the new perimeter is $2(1.1w+1.1l)=2.2(w+l)$
Therefore, the percent change is \begin{align*} \frac{2.2(w+l)-2(w+l)}{2(w+l)} &= \frac{.2(w+l)}{2(w+l)} \\ &= \frac{.2}{2} \\ &= 10\% \\ \end{align*}
$\boxed{10}$ | B | 10 |
fea55e466a9107d3de314c6379cbafdc | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_19 | If the length and width of a rectangle are each increased by $10\%$ , then the perimeter of the rectangle is increased by
$\text{(A)}\ 1\% \qquad \text{(B)}\ 10\% \qquad \text{(C)}\ 20\% \qquad \text{(D)}\ 21\% \qquad \text{(E)}\ 40\%$ | Assume WLOG that the rectangle is a square with length $10$ and width $10$ . Thus, the square has a perimeter of $40$ . Increasing the length and width by $10\%$ will increase the dimensions of the square to 11x11. Thus, the new square's perimeter is $44$ , and because $44$ is $110\%$ of $40$ , our answer is $\boxed{10}$ | B | 10 |
ee62cf6f97633cefdd68ce3cb02fbe6c | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_21 | Mr. Green receives a $10\%$ raise every year. His salary after four such raises has gone up by what percent?
$\text{(A)}\ \text{less than }40\% \qquad \text{(B)}\ 40\% \qquad \text{(C)}\ 44\% \qquad \text{(D)}\ 45\% \qquad \text{(E)}\ \text{more than }45\%$ | Assume his salary is originally $100$ dollars. Then, in the next year, he would have $110$ dollars, and in the next, he would have $121$ dollars. The next year he would have $133.1$ dollars and in the final year, he would have $146.41$ . As the total increase is greater than $45\%$ , the answer is $\boxed{45}$ | E | 45 |
e0dc0eca3766d57e65ff3bb22b662d4a | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_23 | King Middle School has $1200$ students. Each student takes $5$ classes a day. Each teacher teaches $4$ classes. Each class has $30$ students and $1$ teacher. How many teachers are there at King Middle School?
$\text{(A)}\ 30 \qquad \text{(B)}\ 32 \qquad \text{(C)}\ 40 \qquad \text{(D)}\ 45 \qquad \text{(E)}\ 50$ | If each student has $5$ classes, and there are $1200$ students, then they have a total of $5\times 1200=6000$ classes among them.
Each class has $30$ students, so there must be $\frac{6000}{30}=200$ classes. Each class has $1$ teacher, so the teachers have a total of $200$ classes among them.
Each teacher teaches $4$ classes, so if there are $t$ teachers, they have $4t$ classes among them. This was found to be $200$ , so \[4t=200\Rightarrow t=50\]
This is answer choice $\boxed{50}$ | E | 50 |
68930ee0523a817d0c9c826ec971b197 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_24 | In a magic triangle, each of the six whole numbers $10-15$ is placed in one of the circles so that the sum, $S$ , of the three numbers on each side of the triangle is the same. The largest possible value for $S$ is
[asy] draw(circle((0,0),1)); draw(dir(60)--6*dir(60)); draw(circle(7*dir(60),1)); draw(8*dir(60)--13*dir(60)); draw(circle(14*dir(60),1)); draw((1,0)--(6,0)); draw(circle((7,0),1)); draw((8,0)--(13,0)); draw(circle((14,0),1)); draw(circle((10.5,6.0621778264910705273460621952706),1)); draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176)); draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788)); [/asy]
$\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40$ | Let the number in the top circle be $a$ and then $b$ $c$ $d$ $e$ , and $f$ , going in clockwise order. Then, we have \[S=a+b+c\] \[S=c+d+e\] \[S=e+f+a\]
Adding these equations together, we get
\begin{align*} 3S &= (a+b+c+d+e+f)+(a+c+e) \\ &= 75+(a+c+e) \\ \end{align*}
where the last step comes from the fact that since $a$ $b$ $c$ $d$ $e$ , and $f$ are the numbers $10-15$ in some order, their sum is $10+11+12+13+14+15=75$
The left hand side is divisible by $3$ and $75$ is divisible by $3$ , so $a+c+e$ must be divisible by $3$ . The largest possible value of $a+c+e$ is then $15+14+13=42$ , and the corresponding value of $S$ is $\frac{75+42}{3}=39$ , which is choice $\boxed{39}$ | D | 39 |
68930ee0523a817d0c9c826ec971b197 | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_24 | In a magic triangle, each of the six whole numbers $10-15$ is placed in one of the circles so that the sum, $S$ , of the three numbers on each side of the triangle is the same. The largest possible value for $S$ is
[asy] draw(circle((0,0),1)); draw(dir(60)--6*dir(60)); draw(circle(7*dir(60),1)); draw(8*dir(60)--13*dir(60)); draw(circle(14*dir(60),1)); draw((1,0)--(6,0)); draw(circle((7,0),1)); draw((8,0)--(13,0)); draw(circle((14,0),1)); draw(circle((10.5,6.0621778264910705273460621952706),1)); draw((13.5,0.86602540378443864676372317075294)--(11,5.1961524227066318805823390245176)); draw((10,6.9282032302755091741097853660235)--(7.5,11.258330249197702407928401219788)); [/asy]
$\text{(A)}\ 36 \qquad \text{(B)}\ 37 \qquad \text{(C)}\ 38 \qquad \text{(D)}\ 39 \qquad \text{(E)}\ 40$ | To make the sum the greatest, put the three largest numbers $(13,14$ and $15)$ in the corners. Then, balance the sides by putting the least integer $(10)$ between the greatest sum $(14$ and $15)$ . Then put the next least integer $(11)$ between the next greatest sum ( $13 +15$ ). Fill in the last integer $(12)$ and you can see that the sum of any three numbers on a side is (for example) $14 +10 + 15 = 39$ $\boxed{39}$ .
-by goldenn | D | 39 |
a6767506693504ff6076ef8634b79a0d | https://artofproblemsolving.com/wiki/index.php/1985_AJHSME_Problems/Problem_25 | Five cards are lying on a table as shown.
\[\begin{matrix} & \qquad & \boxed{\tt{P}} & \qquad & \boxed{\tt{Q}} \\ \\ \boxed{\tt{3}} & \qquad & \boxed{\tt{4}} & \qquad & \boxed{\tt{6}} \end{matrix}\] | Using the answer choices, we see that P and Q are logically equivalent (both are non-vowel letters) and so are $4$ and $6$ (both are even numbers). Thus, if one of these is correct, then the other option in its pair must also be correct. There can't be 2 answers. So, the only remaining answer choice is $\boxed{3}$ | A | 3 |
7ae8cbbe8497fe6c2a52d8155be19b1c | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_4 | Find the sum of all prime numbers between $1$ and $100$ that are simultaneously $1$ greater than a multiple of $4$ and $1$ less than a multiple of $5$
$\mathrm{(A) \ } 118 \qquad \mathrm{(B) \ }137 \qquad \mathrm{(C) \ } 158 \qquad \mathrm{(D) \ } 187 \qquad \mathrm{(E) \ } 245$ | Numbers that are $1$ less than a multiple of $5$ all end in $4$ or $9$
No prime number ends in $4$ , since all numbers that end in $4$ are divisible by $2$ . Thus, we are only looking for numbers that end in $9$
Writing down the ten numbers that so far qualify, we get $9, 19, 29, 39, 49, 59, 69, 79, 89, 99$
Crossing off multiples of $3$ gives $19, 29, 49, 59, 79, 89$
Crossing off numbers that are not $1$ more than a multiple of $4$ (in other words, numbers that are $1$ less than a multiple of $4$ , since all numbers are odd), we get:
$29, 49, 89.$
Noting that $49$ is not prime, we have only $29$ and $89$ , which give a sum of $118$ , so the answer is $\boxed{118}$ | A | 118 |
da5a4c74e02871ef236b242ee2c90f85 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_5 | The marked price of a book was 30% less than the suggested retail price. Alice purchased the book for half the marked price at a Fiftieth Anniversary sale. What percent of the suggested retail price did Alice pay?
$\mathrm{(A)\ }25\%\qquad\mathrm{(B)\ }30\%\qquad\mathrm{(C)\ }35\%\qquad\mathrm{(D)\ }60\%\qquad\mathrm{(E)\ }65\%$ | Without loss of generality, let's assume that the retail price was $100$ USD.
The marked price of the book is $30 \%$ off of $100$ which is equal to $100-100(0.3)=70.$
Half of that marked price is $0.5(70)=35.$
Therefore the percent Alice payed of the suggested retail price is $35/100=\boxed{35}.$ | C | 35 |
6d2b4bceab02e6d121958fecaef83051 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_6 | What is the sum of the digits of the decimal form of the product $2^{1999}\cdot 5^{2001}$
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10$ | $2^{1999}\cdot5^{2001}=2^{1999}\cdot5^{1999}\cdot5^{2}=25\cdot10^{1999}$ , a number with the digits "25" followed by 1999 zeros. The sum of the digits in the decimal form would be $2+5=7$ , thus making the answer $\boxed{7}$ | D | 7 |
802f68319e8e398dd75ef5c7b691ff04 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_7 | What is the largest number of acute angles that a convex hexagon can have?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$ | The sum of the interior angles of a hexagon is $720$ degrees. In a convex polygon, each angle must be strictly less than $180$ degrees.
Six acute angles can only sum to less than $90\cdot 6 = 540$ degrees, so six acute angles could not form a hexagon.
Five acute angles and one obtuse angle can only sum to less than $90\cdot 5 + 180 = 630$ degrees, so these angles could not form a hexagon.
Four acute angles and two obtuse angles can only sum to less than $90\cdot 4 + 180\cdot 2 = 720$ degrees. This is a strict inequality, so these angles could not form a hexagon. (The limiting figure would be four right angles and two straight angles, which would really be a square with two "extra" points on two sides to form the straight angles.)
Three acute angles and three obtuse angles work. For example, if you pick three acute angles of $80$ degrees, the three obtuse angles would be $160$ degrees and give a sum of $80\cdot 3 + 160\cdot 3 = 720$ degrees, which is a genuine hexagon. Thus, the answer is $\boxed{3}$ | B | 3 |
376146a4402385756e6d65e205bee145 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_8 | At the end of $1994$ , Walter was half as old as his grandmother. The sum of the years in which they were born was $3838$ . How old will Walter be at the end of $1999$
$\textbf{(A)}\ 48 \qquad \textbf{(B)}\ 49\qquad \textbf{(C)}\ 53\qquad \textbf{(D)}\ 55\qquad \textbf{(E)}\ 101$ | In $1994$ , if Water is $x$ years old, then Walter's grandmother is $2x$ years old.
This means that Walter was born in $1994 - x$ , and Walter's grandmother was born in $1994 - 2x$
The sum of those years is $3838$ , so we have:
$1994 - x + 1994 - 2x = 3838$
$3988 - 3x = 3838$
$x = 50$
If Walter is $50$ years old in $1994$ , then he will be $55$ years old in $1999$ , thus giving answer $\boxed{55}$ | D | 55 |
2b4829ee4058015fe395f2e537291385 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_11 | The student locker numbers at Olympic High are numbered consecutively beginning with locker number $1$ . The plastic digits used to number the lockers cost two cents apiece. Thus, it costs two cents to label locker number $9$ and four centers to label locker number $10$ . If it costs $137.94 to label all the lockers, how many lockers are there at the school?
$\textbf{(A)}\ 2001 \qquad \textbf{(B)}\ 2010 \qquad \textbf{(C)}\ 2100 \qquad \textbf{(D)}\ 2726 \qquad \textbf{(E)}\ 6897$ | Since all answers are over $2000$ , work backwards and find the cost of the first $1999$ lockers. The first $9$ lockers cost $0.18$ dollars, while the next $90$ lockers cost $0.04\cdot 90 = 3.60$ . Lockers $100$ through $999$ cost $0.06\cdot 900 = 54.00$ , and lockers $1000$ through $1999$ inclusive cost $0.08\cdot 1000 = 80.00$
This gives a total cost of $0.18 + 3.60 + 54.00 + 80.00 = 137.78$ . There are $137.94 - 137.78 = 0.16$ dollars left over, which is enough for $8$ digits, or $2$ more four digit lockers. These lockers are $2000$ and $2001$ , leading to answer $\boxed{2001}$ | A | 2001 |
56a64f72cd236c74e2fc698b869b03f0 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_12 | What is the maximum number of points of intersection of the graphs of two different fourth degree polynomial functions $y=p(x)$ and $y=q(x)$ , each with leading coefficient 1?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 8$ | The intersections of the two polynomials, $p(x)$ and $q(x)$ , are precisely the roots of the equation $p(x)=q(x) \rightarrow p(x) - q(x) = 0$ . Since the leading coefficients of both polynomials are $1$ , the degree of $p(x) - q(x) = 0$ is at most three, and the maximum point of intersection is three, because a third degree polynomial can have at most three roots. Thus, the answer is $\boxed{3}$ | C | 3 |
52b03eaf574709bbc85568f4f2032f08 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_14 | Four girls — Mary, Alina, Tina, and Hanna — sang songs in a concert as trios, with one girl sitting out each time. Hanna sang $7$ songs, which was more than any other girl, and Mary sang $4$ songs, which was fewer than any other girl. How many songs did these trios sing?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 9 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 11$ | Alina and Tina must sing more than $4$ , but less than $7$ , songs. Therefore, Alina sang $5$ or $6$ songs, and Tina sang $5$ or $6$ songs, with $4$ possible combinations.
However, since every song is a trio, if you add up all the numbers of songs a person sang for all four singers, it must be divisible by $3$ . Thus, $7 + 4 + a + t$ must be divisible by $3$ , and $a+t - 1$ must be divisible by $3$
Since $10 \le a+t \le 12$ by the bounds of $5 \le a, t \le 6$ , we have $9 \le a + t - 1 \le 11$ . Because the middle number must be a multiple of $3$ , we set $a + t - 1 = 9$ , leading to $a + t = 10$ and $a = t = 5$
So, the four girls sang $4, 5, 5$ and $7$ songs in trios. That means there were $\frac{4 + 5 + 5 + 7}{3} = 7$ songs sung, and the answer is $\boxed{7}$ | A | 7 |
5316400109acdba785da7ec186b81f29 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_15 | Let $x$ be a real number such that $\sec x - \tan x = 2$ . Then $\sec x + \tan x =$
$\textbf{(A)}\ 0.1 \qquad \textbf{(B)}\ 0.2 \qquad \textbf{(C)}\ 0.3 \qquad \textbf{(D)}\ 0.4 \qquad \textbf{(E)}\ 0.5$ | $(\sec x - \tan x)(\sec x + \tan x) = \sec^{2} x - \tan^{2} x = 1$ , so $\sec x + \tan x = \boxed{0.5}$ | E | 0.5 |
5316400109acdba785da7ec186b81f29 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_15 | Let $x$ be a real number such that $\sec x - \tan x = 2$ . Then $\sec x + \tan x =$
$\textbf{(A)}\ 0.1 \qquad \textbf{(B)}\ 0.2 \qquad \textbf{(C)}\ 0.3 \qquad \textbf{(D)}\ 0.4 \qquad \textbf{(E)}\ 0.5$ | Note that $\sec x - \tan x = (1-\sin x)/\cos x$ and $\sec x + \tan x = (1+\sin x)/\cos x$ . Let $(1+\sin x)/\cos x = y$ . Multiplying, we get $(1-\sin^{2}x)/\cos^{2}x = 1$ .Then, $2y = 1$ $\sec x + \tan x = \boxed{0.5}$ . ~songmath20
Edited 5.1.2023 | E | 0.5 |
a49422ded7ba9da55938e594f17915b5 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_17 | Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x-19$ , the remainder is $99$ , and when $P(x)$ is divided by $x - 99$ , the remainder is $19$ . What is the remainder when $P(x)$ is divided by $(x-19)(x-99)$
$\mathrm{(A) \ } -x + 80 \qquad \mathrm{(B) \ } x + 80 \qquad \mathrm{(C) \ } -x + 118 \qquad \mathrm{(D) \ } x + 118 \qquad \mathrm{(E) \ } 0$ | According to the problem statement, there are polynomials $Q(x)$ and $R(x)$ such that $P(x) = Q(x)(x-19) + 99 = R(x)(x-99) + 19$
From the last equality we get $Q(x)(x-19) + 80 = R(x)(x-99)$
The value $x=99$ is a root of the polynomial on the right hand side, therefore it must be a root of the one on the left hand side as well. Substituting, we get $Q(99)(99-19) + 80 = 0$ , from which $Q(99)=-1$ . This means that $99$ is a root of the polynomial $Q(x)+1$ . In other words, there is a polynomial $S(x)$ such that $Q(x)+1 = S(x)(x-99)$
Substituting this into the original formula for $P(x)$ we get \[P(x) = Q(x)(x-19) + 99 = (S(x)(x-99) - 1)(x-19) + 99 =\] \[= S(x)(x-99)(x-19) - (x-19) + 99\]
Therefore when $P(x)$ is divided by $(x-19)(x-99)$ , the remainder is $\boxed{118}$ | null | 118 |
a49422ded7ba9da55938e594f17915b5 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_17 | Let $P(x)$ be a polynomial such that when $P(x)$ is divided by $x-19$ , the remainder is $99$ , and when $P(x)$ is divided by $x - 99$ , the remainder is $19$ . What is the remainder when $P(x)$ is divided by $(x-19)(x-99)$
$\mathrm{(A) \ } -x + 80 \qquad \mathrm{(B) \ } x + 80 \qquad \mathrm{(C) \ } -x + 118 \qquad \mathrm{(D) \ } x + 118 \qquad \mathrm{(E) \ } 0$ | Since the divisor $(x-19)(x-99)$ is a quadratic, the degree of the remainder is at most linear. We can write $P(x)$ in the form \[P(x) = Q(x)(x-19)(x-99) + cx+d\] where $cx+d$ is the remainder.
By the Remainder Theorem, plugging in $19$ and $99$ gives us a system of equations. \[99c+d = 19\] \[19c+d = 99\]
Solving gives us $c=-1$ and $d = 118$ , thus, our answer is $\boxed{118}$ | C | 118 |
f3aa0c4c36f16e3980d3e7d1266b29a7 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_20 | The sequence $a_{1},a_{2},a_{3},\ldots$ satisfies $a_{1} = 19,a_{9} = 99$ , and, for all $n\geq 3$ $a_{n}$ is the arithmetic mean of the first $n - 1$ terms. Find $a_2$
$\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179$ | Let $m$ be the arithmetic mean of $a_1$ and $a_2$ . We can then write $a_1=m-x$ and $a_2=m+x$ for some $x$
By definition, $a_3=m$
Next, $a_4$ is the mean of $m-x$ $m+x$ and $m$ , which is again $m$
Realizing this, one can easily prove by induction that $\forall n\geq 3;~ a_n=m$
It follows that $m=a_9=99$ . From $19=a_1=m-x$ we get that $x=80$ . And thus $a_2 = m+x = \boxed{179}$ | E | 179 |
f3aa0c4c36f16e3980d3e7d1266b29a7 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_20 | The sequence $a_{1},a_{2},a_{3},\ldots$ satisfies $a_{1} = 19,a_{9} = 99$ , and, for all $n\geq 3$ $a_{n}$ is the arithmetic mean of the first $n - 1$ terms. Find $a_2$
$\textrm{(A)} \ 29 \qquad \textrm{(B)} \ 59 \qquad \textrm{(C)} \ 79 \qquad \textrm{(D)} \ 99 \qquad \textrm{(E)} \ 179$ | Let $a_1=a$ and $a_2=b$ . Then, $a_3=\frac{a+b}{2}$ $a_4=\frac{a+b+\frac{a+b}{2}}{3}=\frac{a+b}{2},$ and so on.
Since $a_3=a_4$ $a_n=a_3$ for all $n\geq3.$
Hence, $a_9=\frac{a_1+a_2}{2}=\frac{a+b}{2}=99, a+b=198.$ We also know that $a_1=a=19.$
Subtracting $a_1$ from $198,$ we get $b=a_2=\boxed{179}.$ | E | 179 |
f8ebbd067b93ed4f5c913a709d579594 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_22 | The graphs of $y = -|x-a| + b$ and $y = |x-c| + d$ intersect at points $(2,5)$ and $(8,3)$ . Find $a+c$
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 18$ | Each of the graphs consists of two orthogonal half-lines. In the first graph both point downwards at a $45^\circ$ angle, in the second graph they point upwards. One can easily find out that the only way how to get these graphs to intersect in two points is the one depicted below:
Obviously, the maximum of the first graph is achieved when $x=a$ , and its value is $-0+b=b$ . Similarly, the minimum of the other graph is $(c,d)$ . Therefore the two remaining vertices of the area between the graphs are $(a,b)$ and $(c,d)$
As the area has four right angles, it is a rectangle. Without actually computing $a$ and $c$ we can therefore conclude that $a+c=2+8=\boxed{10}$ | null | 10 |
867eebf0a3bc5d4b19487e27cc7b78c0 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_25 | There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ . Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$ | Multiply out the $7!$ to get
\[5 \cdot 6! = (3 \cdot 4 \cdots 7)a_2 + (4 \cdots 7)a_3 + (5 \cdot 6 \cdot 7)a_4 + 42a_5 + 7a_6 + a_7 .\]
By Wilson's Theorem (or by straightforward division), $a_7 + 7(a_6 + 6a_5 + \cdots) \equiv 5 \cdot 6! \equiv -5 \equiv 2 \pmod{7}$ , so $a_7 = 2$ . Then we move $a_7$ to the left and divide through by $7$ to obtain
\[\frac{5 \cdot 6!-2}{7} = 514 = 360a_2 + 120a_3 + 30a_4 + 6a_5 + a_6.\]
We then repeat this procedure $\pmod{6}$ , from which it follows that $a_6 \equiv 514 \equiv 4 \pmod{6}$ , and so forth. Continuing, we find the unique solution to be $(a_2, a_3, a_4, a_5, a_6, a_7) = (1,1,1,0,4,2)$ (uniqueness is assured by the Division Theorem ). The answer is $9 \Longrightarrow \boxed{9}$ | B | 9 |
867eebf0a3bc5d4b19487e27cc7b78c0 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_25 | There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ . Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$ | We start by multiplying both sides by $7!$ , and we get: \[3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7\] After doing some guess and check, we find that the answer is $\boxed{9}$ | B | 9 |
867eebf0a3bc5d4b19487e27cc7b78c0 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_25 | There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ . Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$ | Let's clear up the fractions: \[\frac{5}{7}=\frac{2520a_2+840a_3+210a_4+42a_5+7a_6+a_7}{7!}\] \[3600=2520a_2+840a_3+210a_4+42a_5+7a_6+a_7\] Notice that if we divide everything by $7$ then we would have: \[514+\frac{2}{7}=360a_2+120a_3+30a_4+6a_5+a_6+\frac{1}{7}a_7\] Since $0 \le a_7<7$ and $a_7$ must be an integer, then we have $\frac{2}{7}=\frac{1}{7}a_7$ , so $a_7=2$
Similarly, if we divide everything by $6$ , then we would have: \[85+\frac{4}{6}=60a_2+20a_3+5a_4+a_5+\frac{1}{6}a_6\] Again, since $0 \le a_6<6$ and $a_6$ must be an integer, we have $\frac{4}{6}=\frac{1}{6}a_6$ , so $a_6=4$
The pattern repeats itself, so in the end we have $a_2=1$ $a_3=1$ $a_4=1$ $a_5=0$ $a_6=4$ $a_7=2$ . So $a_2+a_3+a_4+a_5+a_6+a_7=\boxed{9}$ | B | 9 |
867eebf0a3bc5d4b19487e27cc7b78c0 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_25 | There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$ . Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$
$\textrm{(A)} \ 8 \qquad \textrm{(B)} \ 9 \qquad \textrm{(C)} \ 10 \qquad \textrm{(D)} \ 11 \qquad \textrm{(E)} \ 12$ | By multiplying both sides by $7$ we get
\[5 = \frac72 a_2 + \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]
since $0 \le a_2 < 2$ , if $a_2 = 0$ the rest of the right hand side will not add up to be $5$ , so $a_2 = 1$
\[\frac32 = \frac76 a_3+ \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]
If $a_3 = 2$ $\frac76 a_3 = \frac73 > \frac32$ , so $a_3 = 1$
\[\frac13 = \frac{7}{24} a_4 + \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]
If $a_4 = 2$ $\frac{7}{24} a_4 = \frac{7}{12} > \frac13$ , so $a_4 = 1$
\[\frac{1}{24} = \frac{7}{120} a_5 + \frac{7}{720} a_6 + \frac{a_7}{720}\]
If $a_5 = 1$ $\frac{7}{120} a_5 = \frac{7}{120} > \frac{1}{24}$ , so $a_5 = 0$
\[\frac{1}{24} = \frac{7}{720} a_6 + \frac{a_7}{720}\]
$30 = 7 a_6 + a_7$ . Since $a_7 < 7$ $a_7 = 2$ and $a_6=4$
Therefore, $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} = 1 + 1 + 1 + 0 + 4 + 2= \boxed{9}$ | B | 9 |
83574b9d5ab8532dd9dfb1ea000af6e2 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_26 | Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $1$ . The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^{\circ}$ . Thus the three polygons form a new polygon with $A$ as an interior point. What is the largest possible perimeter that this polygon can have?
$\mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24$ | We are looking for three regular polygons such that the sum of their internal angle sizes is exactly $360^{\circ}$
Let the number of sides in our polygons be $3\leq a,b,c$ . From each of the polygons, two sides touch the other two, and the remaining sides are on the perimeter. Therefore the answer to our problem is the value $(a-2)+(b-2)+(c-2) = (a+b+c)-6$
The integral angle of a regular $k$ -gon is $180 \frac{k-2}k$ . Therefore we are looking for integer solutions to:
\[360 = 180\left( \frac{a-2}a + \frac{b-2}b + \frac{c-2}c \right)\]
Which can be simplified to:
\[2 = \left( \frac{a-2}a + \frac{b-2}b + \frac{c-2}c \right)\]
Furthermore, we know that two of the polygons are congruent, thus WLOG $a=c$ . Our equation now becomes
\[2 = \left( 2\cdot\frac{a-2}a + \frac{b-2}b \right)\]
Multiply both sides by $ab$ and simplify to get $ab - 4b - 2a = 0$
Using the standard technique for Diophantine equations , we can add $8$ to both sides and rewrite the equation as $(a-4)(b-2)=8$
Remembering that $a,b\geq 3$ the only valid options for $(a-4,b-2)$ are: $(1,8)$ $(2,4)$ $(4,2)$ , and $(8,1)$
These correspond to the following pairs $(a,b)$ $(5,10)$ $(6,6)$ $(8,4)$ , and $(12,3)$
The perimeters of the resulting polygon for these four cases are $14$ $12$ $14$ , and $\boxed{21}$ | null | 21 |
83574b9d5ab8532dd9dfb1ea000af6e2 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_26 | Three non-overlapping regular plane polygons, at least two of which are congruent, all have sides of length $1$ . The polygons meet at a point $A$ in such a way that the sum of the three interior angles at $A$ is $360^{\circ}$ . Thus the three polygons form a new polygon with $A$ as an interior point. What is the largest possible perimeter that this polygon can have?
$\mathrm{(A) \ }12 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }21 \qquad \mathrm{(E) \ } 24$ | We want to maximize the number of sides of the two congruent polygons, so we need to make the third polygon have the fewest number of sides possible, i.e. a triangle. The interior angle measure of the two congruent polygons is therefore $\frac{360-60}{2}=150$ degrees, so they are dodecagons. Of all the $12 + 12 + 3 = 27$ sides, six of them are not part of the perimeter of the resulting polygon, so the resulting polygon has $\boxed{21}$ sides. | null | 21 |
83c88ce4d76cc4fbf728dd3f89285482 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_28 | Let $x_1, x_2, \ldots , x_n$ be a sequence of integers such that
(i) $-1 \le x_i \le 2$ for $i = 1,2, \ldots n$ (ii) $x_1 + \cdots + x_n = 19$ ; and
(iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99$ .
Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + \cdots + x_n^3$ , respectively. Then $\frac Mm =$
$\mathrm{(A) \ }3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ }7$ | Clearly, we can ignore the possibility that some $x_i$ are zero, as adding/removing such variables does not change the truth value of any condition, nor does it change the value of the sum of cubes. Thus we'll only consider $x_i\in\{-1,1,2\}$
Also, order of the $x_i$ does not matter, so we are only interested in the counts of the variables of each type. Let $a$ of the $x_i$ be equal to $-1$ $b$ equal to $1$ , and $c$ equal to $2$
The conditions (ii) and (iii) simplify to: (ii) $2c + b - a = 19$ (iii) $4c + a + b = 99$ and we want to find the maximum and minimum of $2^3c + 1^3b + (-1)^3a = 8c + b - a$ over all non-negative solutions of the above two equations.
Subtracting twice (ii) from (iii) we get $b = 3a-61$ . By entering that into one of the two equations and simplifying we get $c=40-a$
Thus all the solutions of our system of equations have the form $(a,3a-61,40-a)$
As all three variables must be non-negative integers, we have $3a-61 \geq 0 \Rightarrow a \geq 21$ and $40-a\geq 0 \Rightarrow a \leq 40$
For $(a,b,c)$ of the form $(a,3a-61,40-a)$ the expression we are maximizing/minimizing simplifies to $320-8a+3a-61-a = 259 - 6a$ . Clearly, the maximum is achieved for $a=21$ and the minimum for $a=40$ . Their values are $M=133$ and $m=19$ , and therefore $\frac Mm = \boxed{7}$ | null | 7 |
83c88ce4d76cc4fbf728dd3f89285482 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_28 | Let $x_1, x_2, \ldots , x_n$ be a sequence of integers such that
(i) $-1 \le x_i \le 2$ for $i = 1,2, \ldots n$ (ii) $x_1 + \cdots + x_n = 19$ ; and
(iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99$ .
Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + \cdots + x_n^3$ , respectively. Then $\frac Mm =$
$\mathrm{(A) \ }3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ }7$ | Let $a =$ number of $2$ s, $b =$ number of $1$ s, $c =$ number of $-1$
\[4a+b+c=99\] \[2a+b-c=19\]
Multiplying the second equation by $2$ gives $4a+2b-2c=38$ . By subtracting this equation from the first equation we get $3c-b=61$ $3c=61+b$ . As we need to minimize the value of $c$ $c = 21$ $b = 2$ $a = 19$
$x_1^3 + \cdots + x_n^3 = 8 \cdot 19 +2 - 21 = 133$
Therefore, $\frac Mm = \frac{133}{19}=\boxed{7}$ | E | 7 |
f37cf9aa0ce30c9ed032b065283fbca2 | https://artofproblemsolving.com/wiki/index.php/1999_AHSME_Problems/Problem_29 | A tetrahedron with four equilateral triangular faces has a sphere inscribed within it and a sphere circumscribed about it. For each of the four faces, there is a sphere tangent externally to the face at its center and to the circumscribed sphere. A point $P$ is selected at random inside the circumscribed sphere. The probability that $P$ lies inside one of the five small spheres is closest to
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }0.1 \qquad \mathrm{(C) \ }0.2 \qquad \mathrm{(D) \ }0.3 \qquad \mathrm{(E) \ }0.4$ | Let the side length of the tetrahedron be $1$
The altitude of the equilateral triangle base is $\frac{\sqrt{3}}{2}$ . Thus, the distance from the center of the equilateral triangle to its vertex is $\frac{\sqrt{3}}{2} \cdot \frac 23 = \frac{\sqrt{3}}{3}$ . Therefore, the altitude of the tetrahedron is $\sqrt{1^2-(\frac{\sqrt{3}}{3})^2} = \frac{\sqrt{6}}{3}$
Let the radius of the large sphere be $R$ \[(\frac{\sqrt{3}}{3})^2 + (\frac{\sqrt{6}}{3} - R)^2 = R^2\] \[R = \frac{\sqrt{6}}{4}\]
The distance from the center of the tetrahedron to the center of one of the bases is $\frac{\sqrt{6}}{3} - R = \frac{\sqrt{6}}{3} - \frac{\sqrt{6}}{4} = \frac{\sqrt{6}}{12}$
The distance from the center of the tetrahedron to the center of one of the bases is also the same as the radius of the small sphere in the center of the tetrahedron.
The radius of the smaller spheres tangent to the large sphere and the tetrahedron is $\frac{\frac{\sqrt{6}}{4}-\frac{\sqrt{6}}{12}}{2} = \frac{\sqrt{6}}{12}$
Therefore, the probability that $P$ lies inside one of the five small spheres is $\frac{5 \cdot (\frac{\sqrt{6}}{12})^3 \cdot \frac43 \cdot \pi }{(\frac{\sqrt{6}}{4})^3 \cdot \frac43 \cdot \pi} = \frac{5}{27} \approx \boxed{0.2}$ | C | 0.2 |
a7d995e81946d77f011abbf56c83e6ad | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_2 | Letters $A,B,C,$ and $D$ represent four different digits selected from $0,1,2,\ldots ,9.$ If $(A+B)/(C+D)$ is an integer that is as large as possible, what is the value of $A+B$
$\mathrm{(A) \ }13 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ }16 \qquad \mathrm{(E) \ } 17$ | If we want $\frac{A+B}{C+D}$ to be as large as possible, we want to try to maximize the numerator $A+B$ and minimize the denominator $C+D$ . Picking $A=9$ and $B=8$ will maximize the numerator, and picking $C=0$ and $D=1$ will minimize the denominator.
Checking to make sure the fraction is an integer, $\frac{A+B}{C+D} = \frac{17}{1} = 17$ , and so the values are correct, and $A+B = 17$ , giving the answer $\boxed{17}$ | E | 17 |
764a5391d32d56934290f8bb068c7792 | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_3 | If $\texttt{a,b,}$ and $\texttt{c}$ are digits for which
then $\texttt{a+b+c =}$
$\mathrm{(A) \ }14 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18$ | Working from right to left, we see that $2 - b = 3$ . Clearly if $b$ is a single digit integer, this cannot be possible. Therefore, there must be some borrowing from $a$ . Borrow $1$ from the digit $a$ , and you get $12 - b = 3$ , giving $b = 9$
Since $1$ was borrowed from $a$ , we have from the tens column $(a-1) - 8 = 7$ . Again for single digit integers this will not work. Again, borrow $1$ from $7$ , giving $10 + (a-1) - 8 = 7$ . Solving for $a$
$10 + a - 1 - 8 = 7$
$1 + a = 7$
$a = 6$
Finally, since $1$ was borrowed from the hundreds column, we have $7 - 1 - 4 = c$ , giving $c = 2$
As a check, the problem is $762 - 489 = 273$ , which is a true sentence.
The desired quantity is $a + b + c = 6 + 9 + 2 = 17$ , and the answer is $\boxed{17}$ | D | 17 |
790b1f0d20eb28f31dac2191fbad0dab | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_5 | If $2^{1998}-2^{1997}-2^{1996}+2^{1995} = k \cdot 2^{1995},$ what is the value of $k$
$\mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ } 5$ | Divide both sides of the original equation by $2^{1995}$ , giving:
$2^3 - 2^2 - 2^1 + 1 = k$
$k = 3$ , and the answer is $\boxed{3}$ | C | 3 |
65dbf9cb62b7ded94882f39cc5c29aed | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_6 | If $1998$ is written as a product of two positive integers whose difference is as small as possible, then the difference is
$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }17 \qquad \mathrm{(D) \ }47 \qquad \mathrm{(E) \ } 93$ | If we want the difference of the two factors to be as small as possible, then the two numbers must be as close to $\sqrt{1998}$ as possible.
Since $45^2 = 2025$ , the factors should be as close to $44$ or $45$ as possible.
Breaking down $1998$ into its prime factors gives $1998 = 2\cdot 3^3 \cdot 37$
$37$ is relatively close to $44$ , and no numbers between $38$ and $44$ are factors of $1998$ . Thus, the two factors are $37$ and $2\cdot 3^3 = 54$ , and the difference is $54 - 37 = 17$ , and the answer is $\boxed{17}$ | C | 17 |
0fda69661220838abe06bfe7854e0fba | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_8 | A square with sides of length $1$ is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find $x$ , the length of the longer parallel side of each trapezoid.
$\mathrm{(A) \ } \frac 35 \qquad \mathrm{(B) \ } \frac 23 \qquad \mathrm{(C) \ } \frac 34 \qquad \mathrm{(D) \ } \frac 56 \qquad \mathrm{(E) \ } \frac 78$ | The area of the trapezoid is $\frac{1}{3}$ , and the shorter base and height are both $\frac{1}{2}$ . Therefore, \[\frac{1}{3}=\frac{1}{2}\cdot \frac{1}{2}\cdot \left(\frac{1}{2}+x\right) \Rightarrow x=\frac{5}{6}\rightarrow \boxed{56}\] | D | 56 |
0fda69661220838abe06bfe7854e0fba | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_8 | A square with sides of length $1$ is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find $x$ , the length of the longer parallel side of each trapezoid.
$\mathrm{(A) \ } \frac 35 \qquad \mathrm{(B) \ } \frac 23 \qquad \mathrm{(C) \ } \frac 34 \qquad \mathrm{(D) \ } \frac 56 \qquad \mathrm{(E) \ } \frac 78$ | Divide the pentagon into 2 small congruent trapezoids by extending the common shorter base of the 2 larger trapezoids.
Since each of the smaller trapezoids has its area half each of the larger trapezoids, and each of them has a base $\frac{1}{2}$ , we have \[b_{large}+\frac{1}{2}=2(b_{small}+ \frac{1}{2})\] \[x+\frac{1}{2}=2((1-x)+\frac{1}{2})\] \[x=\frac{5}{6}\boxed{56}\] | D | 56 |
5de175a0135dc14d1ded277f3a81f9ad | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_9 | A speaker talked for sixty minutes to a full auditorium. Twenty percent of the audience heard the entire talk and ten percent slept through the entire talk. Half of the remainder heard one third of the talk and the other half heard two thirds of the talk. What was the average number of minutes of the talk heard by members of the audience?
$\mathrm{(A) \ } 24 \qquad \mathrm{(B) \ } 27\qquad \mathrm{(C) \ }30 \qquad \mathrm{(D) \ }33 \qquad \mathrm{(E) \ }36$ | Assume that there are $100$ people in the audience.
$20$ people heard $60$ minutes of the talk, for a total of $20\cdot 60 = 1200$ minutes heard.
$10$ people heard $0$ minutes.
$\frac{70}{2} = 35$ people heard $20$ minutes of the talk, for a total of $35\cdot 20 = 700$ minutes.
$35$ people heard $40$ minutes of the talk, for a total of $1400$ minutes.
Altogether, there were $1200 + 0 + 700 + 1400 = 3300$ minutes heard among $100$ people.
Thus, the average is $\frac{3300}{100} = 33$ minutes, and the answer is $\boxed{33}$ | D | 33 |
ecc0170b058b1067256080fd48bac97c | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_10 | A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimter of each of the congruent rectangles is $14$ . What is the area of the large square?
$\mathrm{(A) \ }49 \qquad \mathrm{(B) \ }64 \qquad \mathrm{(C) \ }100 \qquad \mathrm{(D) \ }121 \qquad \mathrm{(E) \ }196$ | Let the length of the longer side be $x$ , and the length of the shorter side be $y$ . We are given that $2x+2y=14\implies x+y=7$ . However, note that $x+y$ is also the length of a side of the larger square. Thus the area of the larger square is $(x+y)^2=7^2=\boxed{49}$ | A | 49 |
ecc0170b058b1067256080fd48bac97c | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_10 | A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimter of each of the congruent rectangles is $14$ . What is the area of the large square?
$\mathrm{(A) \ }49 \qquad \mathrm{(B) \ }64 \qquad \mathrm{(C) \ }100 \qquad \mathrm{(D) \ }121 \qquad \mathrm{(E) \ }196$ | Expand the small square so it basically equals the area of the large square. Two of the sides of the rectangles shrink to zero. The other two sides expand to equal the length of the large outer square, and have a length of $\frac{14}{2} = 7$ . Thus, the area of the larger square is $7^2=\boxed{49}$ | A | 49 |
7369bc0e59ff5c55d25ea14207b108c6 | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_11 | Let $R$ be a rectangle. How many circles in the plane of $R$ have a diameter both of whose endpoints are vertices of $R$
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6$ | There are $6$ pairs of vertices of $R$ . However, both diagonals determine the same circle, therefore the answer is $\boxed{5}$ | null | 5 |
e354eb947d30c88fb623c1e051b2c18f | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_15 | A regular hexagon and an equilateral triangle have equal areas. What is the ratio of the length of a side of the triangle to the length of a side of the hexagon?
$\mathrm{(A) \ }\sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }\sqrt{6} \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ }6$ | $A_{\triangle} = \frac{s_t^2\sqrt{3}}{4}$
$A_{hex} = \frac{6s_h^2\sqrt{3}}{4}$ since a regular hexagon is just six equilateral triangles.
Setting the areas equal, we get:
$s_t^2 = 6s_h^2$
$\left(\frac{s_t}{s_h}\right)^2 = 6$
$\frac{s_t}{s_h} = \sqrt{6}$ , and the answer is $\boxed{6}$ | C | 6 |
13eff65771947eb5884a951616736e29 | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_19 | How many triangles have area $10$ and vertices at $(-5,0),(5,0)$ and $(5\cos \theta, 5\sin \theta)$ for some angle $\theta$
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$ | The triangle can be seen as having the base on the $x$ axis and height $|5\sin\theta|$ . The length of the base is $10$ , thus the height must be $2$ . The equation $|\sin\theta| = \frac 25$ has $\boxed{4}$ solutions, one in each quadrant. | null | 4 |
13eff65771947eb5884a951616736e29 | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_19 | How many triangles have area $10$ and vertices at $(-5,0),(5,0)$ and $(5\cos \theta, 5\sin \theta)$ for some angle $\theta$
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$ | Alternatively, we use shoelace to get: \[\frac {1}{2}|(-5*0+5*5\sin (\theta)+5 \cos (\theta) *0)-(0*5+0*5 \cos (\theta)-5*5\sin (\theta))|=10 \implies \frac {1}{2}|50\sin (\theta)|=10\] This means $\sin (\theta)=\pm \frac {2}{5}$ . We see that if it equals $\frac {2}{5}$ , then $\cos (\theta)=\pm \frac {\sqrt {21}}{5}$ . Likewise, we see that if $\sin (\theta)=-\frac {2}{5}$ , then $\cos (\theta)$ has $2$ solutions. Thus, there are $\boxed{4}$ unique points such that the triangle has an area of $10$ , or $C$ | null | 4 |
67d669e904d58e2acacef12ea5705bcb | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_20 | Three cards, each with a positive integer written on it, are lying face-down on a table. Casey, Stacy, and Tracy are told that
First, Casey looks at the number on the leftmost card and says, "I don't have enough information to determine the other two numbers." Then Tracy looks at the number on the rightmost card and says, "I don't have enough information to determine the other two numbers." Finally, Stacy looks at the number on the middle card and says, "I don't have enough information to determine the other two numbers." Assume that each person knows that the other two reason perfectly and hears their comments. What number is on the middle card?
$\textrm{(A)}\ 2 \qquad \textrm{(B)}\ 3 \qquad \textrm{(C)}\ 4 \qquad \textrm{(D)}\ 5 \qquad \textrm{(E)}\ \text{There is not enough information to determine the number.}$ | Initially, there are the following possibilities for the numbers on the cards: $(1,2,10)$ $(1,3,9)$ $(1,4,8)$ $(1,5,7)$ $(2,3,8)$ $(2,4,7)$ $(2,5,6)$ , and $(3,4,6)$
If Casey saw the number $3$ , she would have known the other two numbers. As she does not, we eliminated the possibility $(3,4,6)$
At this moment, if the last card contained a $10$ $9$ , or a $6$ , Tracy would know the other two numbers. (Note that Tracy is aware of the fact that $(3,4,6)$ was eliminated. If she saw the number $6$ , she would know that the other two are $2$ and $5$ .) This eliminates three more possibilities.
Thus before Stacy took her look, we are left with four possible cases: $(2,4,7)$ $(1,4,8)$ $(1,5,7)$ , and $(2,3,8)$ . As Stacy could not find out the exact combination, the middle number must be $\boxed{4}$ | null | 4 |
b07655b9caf920e5d9d9ce5f85b313f2 | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_21 | In an $h$ -meter race, Sunny is exactly $d$ meters ahead of Windy when Sunny finishes the race. The next time they race, Sunny sportingly starts $d$ meters behind Windy, who is at the starting line. Both runners run at the same constant speed as they did in the first race. How many meters ahead is Sunny when Sunny finishes the second race?
$\mathrm{(A) \ } \frac dh \qquad \mathrm{(B) \ } 0 \qquad \mathrm{(C) \ } \frac {d^2}h \qquad \mathrm{(D) \ } \frac {h^2}d \qquad \mathrm{(E) \ } \frac{d^2}{h-d}$ | Let $s$ and $w$ be the speeds of Sunny and Windy. From the first race we know that $\frac sw = \frac h{h-d}$ . In the second race, Sunny's track length is $h+d$ . She will finish this track in $\frac{h+d}s$ . In this time, Windy will run the distance $w\cdot \frac{h+d}s = \frac{(h+d)(h-d)}h$ . This is less than $h$ , therefore Sunny is ahead. The exact distance between Windy and the finish is $h - \frac{(h+d)(h-d)}h = \frac{ h^2 - (h^2-d^2) }h = \boxed{2}$ | null | 2 |
74879b976aedf00c0701d771cacd8706 | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_23 | The graphs of $x^2 + y^2 = 4 + 12x + 6y$ and $x^2 + y^2 = k + 4x + 12y$ intersect when $k$ satisfies $a \le k \le b$ , and for no other values of $k$ . Find $b-a$
$\mathrm{(A) \ }5 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \ }104 \qquad \mathrm{(D) \ }140 \qquad \mathrm{(E) \ }144$ | Both sets of points are quite obviously circles. To show this, we can rewrite each of them in the form $(x-x_0)^2 + (y-y_0)^2 = r^2$
The first curve becomes $(x-6)^2 + (y-3)^2 = 7^2$ , which is a circle centered at $(6,3)$ with radius $7$
The second curve becomes $(x-2)^2 + (y-6)^2 = 40+k$ , which is a circle centered at $(2,6)$ with radius $r=\sqrt{40+k}$
The distance between the two centers is $5$ , and therefore the two circles intersect if $2\leq r \leq 12$
From $\sqrt{40+k} \geq 2$ we get that $k\geq -36$ . From $\sqrt{40+k}\leq 12$ we get $k\leq 104$
Therefore $b-a = 104 - (-36) = \boxed{140}$ | null | 140 |
34123d97c1525842aba2fd219af11733 | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_24 | Call a $7$ -digit telephone number $d_1d_2d_3-d_4d_5d_6d_7$ memorable if the prefix sequence $d_1d_2d_3$ is exactly the same as either of the sequences $d_4d_5d_6$ or $d_5d_6d_7$ (possibly both). Assuming that each $d_i$ can be any of the ten decimal digits $0,1,2, \ldots, 9$ , the number of different memorable telephone numbers is
$\mathrm{(A)}\ 19,810 \qquad\mathrm{(B)}\ 19,910 \qquad\mathrm{(C)}\ 19,990 \qquad\mathrm{(D)}\ 20,000 \qquad\mathrm{(E)}\ 20,100$ | In this problem, we only need to consider the digits $\overline{d_4d_5d_6d_7}$ . Each possibility of $\overline{d_4d_5d_6d_7}$ gives $2$ possibilities for $\overline{d_1d_2d_3}$ , which are $\overline{d_1d_2d_3}=\overline{d_4d_5d_6}$ and $\overline{d_1d_2d_3}=\overline{d_5d_6d_7}$ with the exception of the case of $d_4=d_5=d_6=d_7$ , which only gives one sequence. After accounting for overcounting, the answer is $(10 \times 10 \times 10 \times 10) \times 2 - 10=19990 \Rightarrow \boxed{19,990}$ | C | 19,990 |
c28e9768378573c529be87bd27a5240f | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_25 | A piece of graph paper is folded once so that $(0,2)$ is matched with $(4,0)$ , and $(7,3)$ is matched with $(m,n)$ . Find $m+n$
$\mathrm{(A) \ }6.7 \qquad \mathrm{(B) \ }6.8 \qquad \mathrm{(C) \ }6.9 \qquad \mathrm{(D) \ }7.0 \qquad \mathrm{(E) \ }8.0$ | The line of the fold is the perpendicular bisector of the segment that connects $(0,2)$ and $(4,0)$ .
The point $(m,n)$ is the image of the point $(7,3)$ according to this axis.
The situation looks as follows.
Now, we will compute the coordinates of the point $D$ , using the following facts:
As the triangles $SCT$ and $SDT$ are congruent, their areas are equal. The area of the triangle $SCT$ is $1/2$ of the size of the vector product $\overrightarrow{SC}\times\overrightarrow{ST}$ , and the area of $SDT$ is $1/2$ of the size of $\overrightarrow{ST}\times\overrightarrow{SD}$
We get that $\overrightarrow{SC}\times\overrightarrow{ST} = \overrightarrow{ST}\times\overrightarrow{SD}$
The equality remains valid if we multiply the vector $\overrightarrow{ST}$ by any constant. In other words, instead of $\overrightarrow{ST}$ we can use any vector with the same direction.
The axis of symmetry is perpendicular to $\overrightarrow{AB}$ . Thus its direction is $(2,4)=(1,2)$
We get that $\overrightarrow{SC}\times (1,2) = (1,2) \times\overrightarrow{SD}$
Substituting the coordinates $\overrightarrow{SC}=(5,2)$ and $\overrightarrow{SD}=(m-2,n-1)$ we get $5\cdot 2 - 2\cdot 1 = 1\cdot (n-1) - 2\cdot(m-2)$ . This simplifies to $n=2m+5$
We just discovered that the coordinates of $D$ are $(m,2m+5)$ . We will now use the second two facts mentioned above to find $m$
We have $|SD| = |SC|$ and therefore $|SD|^2 = |SC|^2$ . We know that $|SC|^2 = 5^2 + 2^2 = 29$ , and $|SD|^2 = (m-2)^2 + (2m+4)^2$ . Simplifying, we get the equation $5m^2 + 12m - 9 = 0$ . This has exactly one positive root $m=0.6$
It follows that $D=(0.6,6.2)$ , and that $m+n = 6.2 + 0.6 = \boxed{6.8}$ | null | 6.8 |
c28e9768378573c529be87bd27a5240f | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_25 | A piece of graph paper is folded once so that $(0,2)$ is matched with $(4,0)$ , and $(7,3)$ is matched with $(m,n)$ . Find $m+n$
$\mathrm{(A) \ }6.7 \qquad \mathrm{(B) \ }6.8 \qquad \mathrm{(C) \ }6.9 \qquad \mathrm{(D) \ }7.0 \qquad \mathrm{(E) \ }8.0$ | Note that the fold is the perpendicular bisector of $(0,2)$ and $(4,0)$ . Thus, the fold goes through the midpoint $(2,1)$
The fold also has a slope of $-\frac{1}{m}$ , where the $m$ is the slope of the line connecting these two points. We find $m = \frac{0 - 2}{4 - 0} = -\frac{1}{2}$ . Thus, the slope of the fold is $2$ and goes through $(2,1)$ , so the equation of the fold is $y = 2x - 3$
We want the line connecting $(7,3)$ and $(m,n)$ to have the same fold as a perpendicular bisector. The slope should be $-\frac{1}{2}$ , so we get $\frac{n - 3}{m - 7} = -\frac{1}{2}$ , which leads to $m = 13 -2n$
We also want $y = 2x - 3$ to bisect the segment from $(7,3)$ to $(m,n)$ . Thus, the midpoint $(x,y) = \left(\dfrac{7 + m}{2}, \frac{3 + n}{2}\right)$ must lie on the line. Plugging into the equation of the line, we find $\frac{3 + n}{2} = (7 + m) - 3$ , which simplifies to $n = 5 + 2m$
Solving the system of two equations in two variables $m = 13 - 2n$ and $n = 5 + 2m$ gives $m = \frac{3}{5}$ and $n = \frac{31}{5}$ , for a sum of $\boxed{6.8}$ | B | 6.8 |
6e718f6640e891a8a5d70690c06f372d | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_28 | In triangle $ABC$ , angle $C$ is a right angle and $CB > CA$ . Point $D$ is located on $\overline{BC}$ so that angle $CAD$ is twice angle $DAB$ . If $AC/AD = 2/3$ , then $CD/BD = m/n$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$
$\mathrm{(A) \ }10 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }18 \qquad \mathrm{(D) \ }22 \qquad \mathrm{(E) \ } 26$ | Let $AC = 2$ $AD = 3$ $\cos \angle CAD = \frac23$
By the pythagorean theorem $CD = \sqrt{3^2-2^2} = \sqrt{5}$
$\sin \angle BDA = \sin (180^{\circ} - \angle BDA) = \sin \angle CDA = \cos \angle (90^{\circ} - CDA) = \cos \angle CAD = \frac23$
$\sin \angle BAD = \sqrt{ \frac{1-cos (2\angle BAD)}{2} } = \sqrt{ \frac{1-\cos \angle CAD}{2} } = \sqrt{ \frac{1-\frac23}{2} } = \frac{\sqrt{6}}{6}$
By the Law of Sine, $\frac{ \sin \angle BDA }{AB} = \frac{ \sin \angle BAD }{BD}$
$\frac{ \frac23 }{ \sqrt{2^2 + ( \sqrt{5} + BD)^2} } = \frac{ \frac{\sqrt{6}}{6} }{BD}$
$8BD^2 = 3(9+ 2BD \sqrt{5} + BD^2)$
$5BD^2 - 6 BD \sqrt{5} -27=0$
As $BD>0$ $BD = \frac{6 \sqrt{5} + \sqrt{ (6 \sqrt{5})^2 - 4 \cdot 5 (-27) } }{10} = \frac{9\sqrt{5}}{5}$
$\frac{CD}{BD} = \frac{\sqrt{5}}{\frac{9\sqrt{5}}{5}} = \frac59$
$5+9=\boxed{14}$ | B | 14 |
83903a6085305e26e3fd6e6654a6ddec | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_29 | A point $(x,y)$ in the plane is called a lattice point if both $x$ and $y$ are integers. The area of the largest square that contains exactly three lattice points in its interior is closest to
$\mathrm{(A) \ } 4.0 \qquad \mathrm{(B) \ } 4.2 \qquad \mathrm{(C) \ } 4.5 \qquad \mathrm{(D) \ } 5.0 \qquad \mathrm{(E) \ } 5.6$ | [asy] real e = 0.1; dot((0,-1)); dot((1,-1)); dot((-1,0)); dot((0,0)); dot((1,0)); dot((2,0)); dot((-1,1)); dot((0,1)); dot((1,1)); dot((0,2)); dot((-1,-1)); dot((2,2)); dot((1,2)); dot((2,1)); dot((2,-1)); dot((-1,2)); draw((0.8, -1.4+e)--(1.8-e, 0.6)--(-0.2, 1.6-e)--(-1.2+e, -0.4)--cycle); [/asy] The best square's side length is slightly less than $\sqrt 5$ , yielding an answer of $\boxed{5.0}$ | D | 5.0 |
8f805f94fec91cd30bc8db090ef6b96f | https://artofproblemsolving.com/wiki/index.php/1998_AHSME_Problems/Problem_30 | For each positive integer $n$ , let
Let $k$ denote the smallest positive integer for which the rightmost nonzero digit of $a_k$ is odd. The rightmost nonzero digit of $a_k$ is
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ } 7 \qquad \mathrm{(E) \ } 9$ | We have $a_n = n(n+1)\dots (n+9)$
The value $a_n$ can be written as $2^{x_n} 5^{y_n} r_n$ , where $r_n$ is not divisible by 2 and 5. The number of trailing zeroes is $z_n = \min(x_n,y_n)$ . The last non-zero digit is the last digit of $2^{x_n-z_n} 5^{y_n-z_n} r_n$
Clearly, the last non-zero digit is even iff $x_n - z_n > 0$ iff $x_n > y_n$
Thus we are looking for the smallest $n$ such that the power of $5$ that divides $a_n$ is at least equal to the power of $2$ that divides $a_n$
The number $a_n$ is a product of $10$ consecutive integers. Out of these, $5$ are divisible by $2$ . Out of those $5$ , at least $2$ are divisible by $4$ , and out of those $2$ , one is divisible by $8$ . Therefore $x_n\geq 5+2+1=8$ for all $n$
On the other hand, exactly $2$ of our ten integers are divisible by $5$ , and at most one of them can be divisible by a higher power of $5$ . As we need $y_n\geq x_n\geq 8$ , one of the integers from $n$ to $n+9$ must be divisible by $5^7 = 78125$ . Therefore $n\geq 78116$
We can now take numbers starting with $78116$ , and write each of them in the form $2^x 5^y r$ . We are looking for 10 consecutive rows where the sum of $y$ s is at least equal to the sum of $x$ s.
At this point we can stop, as we just found out that $a_{78117}$ is of the form $2^8 5^8 r_{78117}$ . Therefore the $k$ we seek is $k=78117$
Now all we need to do is to compute the last non-zero digit. As the powers of $2$ and $5$ that divide $a_{78117}$ are equal, the last non-zero digit is simply the product of the last digits of the ten $z$ s. This is $7\cdot 9\cdot 9\cdot 3\cdot 1\cdot 1\cdot 3\cdot 1\cdot 1\cdot 3 \equiv\boxed{9}$ | null | 9 |
256fec896a3faa867059b798b80fdba9 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_1 | If $\texttt{a}$ and $\texttt{b}$ are digits for which
$\begin{array}{ccc}& 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{array}$
then $\texttt{a+b =}$
$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 9 \qquad \mathrm{(E) \ }12$ | From the units digit calculation, we see that the units digit of $a\times 3$ is $9$ . Since $0 \le a \le 9$ and $a$ is an integer, the only value of $a$ that works is is $a=3$ . As a double-check, that does work, since $23 \times 3 = 69$ , which is the first line of the multiplication.
The second line of the multiplication can be found by doing the multiplication $23\times b = 92$ . Dividing both sides by $23$ gives $b=4$
Thus, $a + b = 3 + 4 = 7$ , and the answer is $\boxed{7}$ | C | 7 |
5ff01725a4f6d9224ca9df708f876c14 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_2 | The adjacent sides of the decagon shown meet at right angles. What is its perimeter?
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; dot(origin);dot((12,0));dot((12,1));dot((9,1));dot((9,7));dot((7,7));dot((7,10));dot((3,10));dot((3,8));dot((0,8)); draw(origin--(12,0)--(12,1)--(9,1)--(9,7)--(7,7)--(7,10)--(3,10)--(3,8)--(0,8)--cycle); label("$8$",midpoint(origin--(0,8)),W); label("$2$",midpoint((3,8)--(3,10)),W); label("$12$",midpoint(origin--(12,0)),S);[/asy]
$\mathrm{(A)\ } 22 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \ } 34 \qquad \mathrm{(D) \ } 44 \qquad \mathrm{(E) \ }50$ | The three unlabelled vertical sides have the same sum as the two labelled vertical sides, which is $10$
The four unlabelled horizontal sides have the same sum as the one large horizontal side, which is $12$
Thus, the perimeter is $2(12+10) = 44$ , which is option $\boxed{44}$ | D | 44 |
6c542695d72f944e74dcb09ad1c340d1 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_3 | If $x$ $y$ , and $z$ are real numbers such that
then $x + y + z =$
$\mathrm{(A)\ } -12 \qquad \mathrm{(B) \ }0 \qquad \mathrm{(C) \ } 8 \qquad \mathrm{(D) \ } 12 \qquad \mathrm{(E) \ }50$ | If the sum of three squared expressions is zero, then each expression itself must be zero, since $a^2 \ge 0$ with the equality iff $a=0$
In this case, $x-3=0$ $y-4=0$ , and $z-5=0$ . Adding the three equations and moving the constant to the right gives $x + y + z = 12$ , and the answer is $\boxed{12}$ | D | 12 |
a1e2915dcb44b3de2f94fdf9d3ca8bd6 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_4 | If $a$ is $50\%$ larger than $c$ , and $b$ is $25\%$ larger than $c$ , then $a$ is what percent larger than $b$
$\mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \ } 50\% \qquad \mathrm{(D) \ } 100\% \qquad \mathrm{(E) \ }200\%$ | Translating each sentence into an equation, $a = 1.5c$ and $b = 1.25c$
We want a relationship between $a$ and $b$ . Dividing the second equation into the first will cancel the $c$ , so we try that and get:
$\frac{a}{b} = \frac{1.5}{1.25}$
$\frac{a}{b} = \frac{150}{125}$
$\frac{a}{b} = \frac{6}{5}$
$a = 1.2b$
In this case, $a$ is $1.2 - 1 = 0.2 = 20\%$ bigger than $b$ , and the answer is $\boxed{20}$ | A | 20 |
a1e2915dcb44b3de2f94fdf9d3ca8bd6 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_4 | If $a$ is $50\%$ larger than $c$ , and $b$ is $25\%$ larger than $c$ , then $a$ is what percent larger than $b$
$\mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \ } 50\% \qquad \mathrm{(D) \ } 100\% \qquad \mathrm{(E) \ }200\%$ | Arbitrarily assign a value to one of the variables. Since $c$ is the smallest variable, let $c = 100$
If $a$ is $50\%$ larger than $c$ , then $a = 150$
If $b$ is $25\%$ larger than $c$ , then $b = 125$
We see that $\frac{a}{b} = \frac{150}{125} = 1.2$ So, $a$ is $20\%$ bigger than $b$ , and the answer is $\boxed{20}$ | A | 20 |
efd49b137a35ab2b64d45aae578c0a21 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_5 | A rectangle with perimeter $176$ is divided into five congruent rectangles as shown in the diagram. What is the perimeter of one of the five congruent rectangles? [asy] defaultpen(linewidth(.8pt)); draw(origin--(0,3)--(4,3)--(4,0)--cycle); draw((0,1)--(4,1)); draw((2,0)--midpoint((0,1)--(4,1))); real r = 4/3; draw((r,3)--foot((r,3),(0,1),(4,1))); draw((2r,3)--foot((2r,3),(0,1),(4,1)));[/asy]
$\mathrm{(A)\ } 35.2 \qquad \mathrm{(B) \ }76 \qquad \mathrm{(C) \ } 80 \qquad \mathrm{(D) \ } 84 \qquad \mathrm{(E) \ }86$ | Let $l$ represent the length of one of the smaller rectangles, and let $w$ represent the width of one of the smaller rectangles, with $w < l$
From the large rectangle, we see that the top has length $3w$ , the right has length $l + w$ , the bottom has length $2l$ , and the left has length $l + 2$
Since the perimeter of the large rectangle is $176$ , we know that $172 = 3w + l + w + 2l + l + w$ , or $172 = 5w + 4l$
From the top and bottom of the large rectangle, we know that $3w = 2l$ , or $l = 1.5w$
Plugging that into the first equation, we get $176 = 5w + 4(1.5)w$
$176 = 11w$
$w = 16$
$l = 1.5w = 24$
$P = 2l + 2w = 2(16 + 24) = 80$ , and the answer is $\boxed{80}$ | C | 80 |
ef1497a9a1b097a777a0ed3b0ae3f95e | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_6 | Consider the sequence
$1,-2,3,-4,5,-6,\ldots,$
whose $n$ th term is $(-1)^{n+1}\cdot n$ . What is the average of the first $200$ terms of the sequence?
$\textbf{(A)}-\!1\qquad\textbf{(B)}-\!0.5\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 0.5\qquad\textbf{(E)}\ 1$ | The average of a list is the sum of all numbers divided by the size of the list.
The sum of the list can be found by adding the numbers in pairs: $(1 + -2) + (3 + -4) + ... + (199 + -200)$
The sum of each pair is $-1$ , and there are $100$ pairs, so the total sum is $-100$
There are $200$ numbers on the list, so the average is $\frac{-100}{200} = -0.5$ , and the answer is $\boxed{0.5}$ | B | 0.5 |
9b786fe16a682fa614f0327dde52fcf5 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_7 | The sum of seven integers is $-1$ . What is the maximum number of the seven integers that can be larger than $13$
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$ | If the first six integers are $14$ , the last number can be $(-14\cdot 6) - 1 = -85$ . The sum of all seven integers will be $-1$
However, if all seven integers are over $13$ , the smallest possible sum is $14\cdot 7 = 98$
Thus, the answer is $6$ , which is option $\boxed{6}$ | D | 6 |
e8413102a6cd81739a6f535a8a41c849 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_8 | Mientka Publishing Company prices its bestseller Where's Walter? as follows:
$C(n) =\left\{\begin{matrix}12n, &\text{if }1\le n\le 24\\ 11n, &\text{if }25\le n\le 48\\ 10n, &\text{if }49\le n\end{matrix}\right.$
where $n$ is the number of books ordered, and $C(n)$ is the cost in dollars of $n$ books. Notice that $25$ books cost less than $24$ books. For how many values of $n$ is it cheaper to buy more than $n$ books than to buy exactly $n$ books?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 8$ | Clearly, the areas of concern are where the piecewise function shifts value.
Since $C(25) = 11\cdot 25 = 275$ , we want to find the least value of $n$ for which $C(n) > 275$
If $n \le 24$ , then $C(n) = 12n$ , so for $C(n) > 275$ $12n > 275$ , which is equivalent to $n > 22.91$ . Thus, both $n=23$ and $n=24$ will be more expensive than $n=25$
Since $C(49) = 10\cdot 49 = 490$ , we want to find the least value of $n$ for which $C(n) > 490$
If $25 \le n \le 48$ , then $C(n) = 11n$ , so for $C(n) > 490$ , we have $11n > 490$ , leading to $n > 44.5$ . Thus, $n=45, 46, 47, 48$ will be more expensive than $n=49$
Thus, there are $2 + 4 = \boxed{6}$ | D | 6 |
b46ca9381a706a1cab75fb611fd5e287 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_11 | In the sixth, seventh, eighth, and ninth basketball games of the season, a player scored $23$ $14$ $11$ , and $20$ points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than $18$ , what is the least number of points she could have scored in the tenth game?
$\textbf{(A)}\ 26\qquad\textbf{(B)}\ 27\qquad\textbf{(C)}\ 28\qquad\textbf{(D)}\ 29\qquad\textbf{(E)}\ 30$ | The sum of the scores for games $6$ through $9$ is $68$ . The average in these four games is $\frac{68}{4} = 17$
The total points in all ten games is greater than $10\cdot 18 = 180$ . Thus, it must be at least $181$
There are at least $181 - 68 = 113$ points in the other six games: games $1-5$ and game $10$
Games $1-5$ must have an average of less than $17$ . Thus we cannot put more than $16 + 17 + 17 + 17 + 17 = 84$ points in those five games.
Thus, the tenth game must have at least $113 - 84 = 29$ points, and the answer is $\boxed{29}$ | D | 29 |
deb5be614dd6b6c17c2a7547e8ae0141 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_13 | How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of is a perfect square?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Let $N = 10t + u$ , where $t$ is the tens digit and $u$ is the units digit.
The condition of the problem is that $10t + u + 10u + t$ is a perfect square.
Simplifying and factoring, we want $11(t+u)$ to be a perfect square.
Thus, $t+u$ must at least be a multiple of $11$ , and since $t$ and $u$ are digits, the only multiple of $11$ that works is $11$ itself.
Thus, $(t,u) = (2,9)$ is the first solution, and $(t,u) = (9,2)$ is the last solution. There are $8$ solutions in total, leading to answer $\boxed{8}$ | E | 8 |
46a3476d54c2ccd83533d2b5bcf8bf09 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_14 | The number of geese in a flock increases so that the difference between the populations in year $n+2$ and year $n$ is directly proportional to the population in year $n+1$ . If the populations in the years $1994$ $1995$ , and $1997$ were $39$ $60$ , and $123$ , respectively, then the population in $1996$ was
$\textbf{(A)}\ 81\qquad\textbf{(B)}\ 84\qquad\textbf{(C)}\ 87\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 102$ | Let $x$ be the population in $1996$ , and let $k$ be the constant of proportionality.
If $n=1994$ , then the difference in population between $1996$ and $1994$ is directly proportional to the population in $1995$
Translating this sentence, $(x - 39) = k(60)$
Similarly, letting $n=1995$ gives the sentence $(123 - 60) = kx$
Since $kx = 63$ , we have $k = \frac{63}{x}$
Plugging this into the first equation, we have:
$(x - 39) = \frac{60\cdot 63}{x}$
$x - 39 = \frac{3780}{x}$
$x^2 - 39x - 3780 = 0$
$(x - 84)(x + 45) = 0$
Since $x>0$ , we must have $x=84$ , and the answer is $\boxed{84}$ | B | 84 |
e0846b797540ca5bc45aedfc39d36449 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_15 | Medians $BD$ and $CE$ of triangle $ABC$ are perpendicular, $BD=8$ , and $CE=12$ . The area of triangle $ABC$ is
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(rightanglemark(C,G,D,3));[/asy]
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96$ | [asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair F = midpoint(B--C); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G);dot(F); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE);label("$F$",F,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(A--F); draw(rightanglemark(B,G,E,3));[/asy]
One median divides a triangle into $2$ equal areas, so all three medians will divide a triangle into $6$ equal areas.
The median $CE$ is divided into a $2:1$ ratio at centroid $G$ , so $GE = \frac{1}{3}\cdot CE = \frac{1}{3}\cdot 12 = 4$
Similarly, $BG = \frac{2}{3}\cdot 8 = \frac{16}{3}$
The area of the right triangle $\triangle BEG$ is $\frac{1}{2}\cdot\frac{16}{3}\cdot 4$
The area of the whole figure is $6\cdot \frac{1}{2}\cdot\frac{16}{3}\cdot 4 = 64$ , and the correct answer is $\boxed{64}$ | D | 64 |
e0846b797540ca5bc45aedfc39d36449 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_15 | Medians $BD$ and $CE$ of triangle $ABC$ are perpendicular, $BD=8$ , and $CE=12$ . The area of triangle $ABC$ is
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(rightanglemark(C,G,D,3));[/asy]
$\textbf{(A)}\ 24\qquad\textbf{(B)}\ 32\qquad\textbf{(C)}\ 48\qquad\textbf{(D)}\ 64\qquad\textbf{(E)}\ 96$ | Notice that if you were to draw in line ED, you would get an orthodiagonal quadrilateral with diagonals 8 and 12. The area is going to be equal to 48. Now we need to examine the triangle AED. If the area we are trying to find is denoted as A, we can tell that the area of AEC is A/2. The area of AED is going to be half of that since AD = DC so it would be A/4. This means that 48 is 3/4 of A, so naturally A is going to be 64. Giving $\boxed{64}$ | D | 64 |
a67556f1f4d4658f72965fd9ca11ef73 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_16 | The three row sums and the three column sums of the array
\[\left[\begin{matrix}4 & 9 & 2\\ 8 & 1 & 6\\ 3 & 5 & 7\end{matrix}\right]\]
are the same. What is the least number of entries that must be altered to make all six sums different from one another?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$ | If you change $3$ numbers, then you either change one number in each column and row (ie sudoku-style):
\[\left[\begin{matrix}* & 9 & 2\\ 8 & * & 6\\ 3 & 5 & *\end{matrix}\right]\]
Or you leave at least one row and one column unchanged:
\[\left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & 5 & 7\end{matrix}\right]\]
In the first case, you are changing just one common number in two sums, so you wind up with three pairs of sums. (In the example given, the sum in row $x$ is the same as in column $x$ .)
In the second case, since two of the sums are unchanged, and the sums started out equal, they must remain equal. (In the second example given, row $3$ and column $3$ are untouched.)
Either way, $3$ changes is not enough. However, building on the second example, if you change either the untouched column or the untouched row, you will get a possible answer:
\[\left[\begin{matrix}* & 9 & 2\\ * & * & 6\\ 3 & * & 7\end{matrix}\right]\]
Letting the $*$ be a zero does indeed give $6$ different sums, so the answer is $4$ , which is option $\boxed{4}$ | D | 4 |
c59d439acdc6efdfdf0e4d5f6b6f56d0 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_17 | A line $x=k$ intersects the graph of $y=\log_5 x$ and the graph of $y=\log_5 (x + 4)$ . The distance between the points of intersection is $0.5$ . Given that $k = a + \sqrt{b}$ , where $a$ and $b$ are integers, what is $a+b$
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | Since the line $x=k$ is vertical, we are only concerned with vertical distance.
In other words, we want to find the value of $k$ for which the distance $|\log_5 x - \log_5 (x+4)| = \frac{1}{2}$
Since $\log_5 x$ is a strictly increasing function, we have:
$\log_5 (x + 4) - \log_5 x = \frac{1}{2}$
$\log_5 (\frac{x+4}{x}) = \frac{1}{2}$
$\frac{x+4}{x} = 5^\frac{1}{2}$
$x + 4 = x\sqrt{5}$
$x\sqrt{5} - x = 4$
$x = \frac{4}{\sqrt{5} - 1}$
$x = \frac{4(\sqrt{5} + 1)}{5 - 1^2}$
$x = 1 + \sqrt{5}$
The desired quantity is $1 + 5 = 6$ , and the answer is $\boxed{6}$ | A | 6 |
a0b74d71074408bfed8085ba3e8dadc1 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_18 | A list of integers has mode $32$ and mean $22$ . The smallest number in the list is $10$ . The median $m$ of the list is a member of the list. If the list member $m$ were replaced by $m+10$ , the mean and median of the new list would be $24$ and $m+10$ , respectively. If were $m$ instead replaced by $m-8$ , the median of the new list would be $m-4$ . What is $m$
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 17\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 19\qquad\textbf{(E)}\ 20$ | Let there be $n$ integers on the list. The list of $n$ integers has mean $22$ , so the sum of the integers is $22n$
Replacing $m$ with $m+10$ will increase the sum of the list from $22n$ to $22n + 10$
The new mean of the list is $24$ , so the new sum of the list is also $24n$
Thus, we get $22n + 10 = 24n$ , leading to $n=5$ numbers on the list.
If there are $5$ numbers on the list with mode $32$ and smallest number $10$ , then the list is $\{10, x, m, 32, 32\}$
Since replacing $m$ with $m-8$ gives a new median of $m-4$ , and $m-4$ must be on the list of $5$ integers since $5$ is odd, $x = m-4$ , and the list is now $\{10, m-4, m, 32, 32\}$
The sum of the numbers on this list is $22n = 22\cdot 5 = 110$ , so we get:
$10 + m - 4 + m + 32 + 32 = 110$
$70 + 2m = 110$
$m = 20$ , giving answer $\boxed{20}$ | E | 20 |
bbc6f2a4adcbad685908614142195b85 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_19 | A circle with center $O$ is tangent to the coordinate axes and to the hypotenuse of the $30^\circ$ $60^\circ$ $90^\circ$ triangle $ABC$ as shown, where $AB=1$ . To the nearest hundredth, what is the radius of the circle?
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$\textbf{(A)}\ 2.18\qquad\textbf{(B)}\ 2.24\qquad\textbf{(C)}\ 2.31\qquad\textbf{(D)}\ 2.37\qquad\textbf{(E)}\ 2.41$ | [asy] defaultpen(linewidth(.8pt)); dotfactor=3; pair A = origin; pair B = (1,0); pair C = (0,sqrt(3)); pair O = (2.33,2.33); pair D = (2.33,0); pair E = (0, 2.33); pair F = (0.35, 1.1); dot(A);dot(B);dot(C);dot(O);dot(D);dot(E);dot(F); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,W);label("$O$",O,NW);label("$D$",D,SW);label("$E$",E,SW);label("$F$", F, W); label("$1$",midpoint(A--B),S);label("$60^\circ$",B,2W + N); draw((3,0)--A--(0,3)); draw(B--C);draw(O--E);draw(O--D); draw(Arc(O,2.33,163,288.5));[/asy]
Draw radii $OE$ and $OD$ to the axes, and label the point of tangency to triangle $ABC$ point $F$ . Let the radius of the circle $O$ be $r$ . Square $OEAD$ has side length $r$
Because $BD$ and $BF$ are tangents from a common point $B$ $BD = BF$
$AD = AB + BD$
$r = 1 + BD$
$r = 1 + BF$
Similarly, $CF = CE$ , and we can write:
$AE = AC + CE$
$r = \sqrt{3} + CF$
Equating the radii lengths, we have $1 + BF = \sqrt{3} + CF$
This means $BF - CF = \sqrt{3} - 1$
$BF + CF = 2$ by the 30-60-90 triangle.
Therefore, $2BF = 2 + \sqrt{3} - 1$ , and we get $BF = \frac{1}{2} + \frac{\sqrt{3}}{2}$
The radius of the circle is $AD$ , which is $BF + 1 = \frac{3}{2} + \frac{\sqrt{3}}{2}$
Using decimal approximations, $r \approx 1.5 + \frac{1.73^+}{2} \approx 2.37$ , and the answer is $\boxed{2.37}$ | D | 2.37 |
233c41df098df9abda4d9f188a7c8112 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_22 | Ashley, Betty, Carlos, Dick, and Elgin went shopping. Each had a whole number of dollars to spend, and together they had $56$ dollars. The absolute difference between the amounts Ashley and Betty had to spend was $19$ dollars. The absolute difference between the amounts Betty and Carlos had was $7$ dollars, between Carlos and Dick was $5$ dollars, between Dick and Elgin was $4$ dollars, and between Elgin and Ashley was $11$ dollars. How many dollars did Elgin have?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$ | Working backwards, if $6 \le E \le 10$ , then $6 \pm 11 \le A \le 10 \pm 11$ . Since $A$ is a positive integer, $17 \le A \le 21$
Since $17 \le A \le 21$ , we know that $17 \pm 19 \le B \le 21 \pm 19$ . But if $B=36$ , which is the smallest possible "plus" value, then $E + A + B = 6 + 17 + 36 = 59$ , which is too much money.
Hence, $17 - 19 \le B \le 21 - 19$ . But since $B$ must be a positive integer, that leaves only two possibilities: $B = 1$ or $B=2$ , which correspond with $E = 9$ and $E = 10$
Concentrating only on $E=9$ , we have $E=9$ leading to $A = 9 + 11 = 20$ , which leads to $B = 20 - 19 = 1$ , which leads to $C = 1 + 7 = 8$ . Thus far we have given out $9 + 20 + 1 + 8 = 38$ dollars. This means that Dick must have $56 - 38 = 18$ dollars. However, the difference between Carlos and Dick is not $5$ dollars.
Thus, the right answer must be $\boxed{10}$ . Verifying, if $E = 10$ , then $A = 10 + 11 = 21$ $B = 21 - 19 = 2$ , which leads to $C = 2 + 7 = 9$ . Thus far, we have given out $10 + 21 + 2 + 9 = 42$ dollars, leaving $56 - 42 = 14$ dollars for Dick. Dick does indeed have $5$ dollars more than Carlos, and $4$ dollars more than Elgin. | E | 10 |
b3c46640b2abb4414385382f11ccc51a | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_24 | A rising number, such as $34689$ , is a positive integer each digit of which is larger than each of the digits to its left. There are $\binom{9}{5} = 126$ five-digit rising numbers. When these numbers are arranged from smallest to largest, the $97^{\text{th}}$ number in the list does not contain the digit
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | The list starts with $12345$ . There are $\binom{8}{4} = 70$ four-digit rising numbers that do not begin with $1$ , and thus also $70$ five digit rising numbers that do begin with $1$ that are formed by simply putting a $1$ before the four digit number.
Thus, the $71^{\text{st}}$ number is $23456$ . There are $\binom{6}{3} = 20$ three-digit rising numbers that do not begin with a $1,2$ or $3$ , and thus $20$ five digit rising numbers that begin with a $23$
Thus, the $91^{\text{st}}$ number is $24567$ . Counting up, $24568, 24569, 24578, 24579, 24589, 24678$ is the $97^{\text{th}}$ number, which does not contain the digit $\boxed{5}$ | B | 5 |
eaa6c15c986d60087c42ebff6d71592e | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_25 | Let $ABCD$ be a parallelogram and let $\overrightarrow{AA^\prime}$ $\overrightarrow{BB^\prime}$ $\overrightarrow{CC^\prime}$ , and $\overrightarrow{DD^\prime}$ be parallel rays in space on the same side of the plane determined by $ABCD$ . If $AA^{\prime} = 10$ $BB^{\prime}= 8$ $CC^\prime = 18$ , and $DD^\prime = 22$ and $M$ and $N$ are the midpoints of $A^{\prime} C^{\prime}$ and $B^{\prime}D^{\prime}$ , respectively, then $MN =$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 4$ | Let $ABCD$ be a unit square with $A(0,0,0)$ $B(0,1,0)$ $C(1,1,0)$ , and $D(1,0,0)$ . Assume that the rays go in the +z direction. In this case, $A^\prime(0,0,10)$ $B^\prime(0,1,8)$ $C^\prime(1,1,18)$ , and $D^\prime(1,0,22)$ . Finding the midpoints of $A^\prime C^\prime$ and $B^\prime D^\prime$ gives $M(\frac{1}{2}, \frac{1}{2}, 14)$ and $N(\frac{1}{2}, \frac{1}{2}, 15)$ . The distance $MN$ is $15 - 14 = 1$ , and the answer is $\boxed{1}$ | B | 1 |
2b2907bdc051d9810fec5bc87dc15f0b | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_26 | Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$ , angle $APB$ is twice angle $ACB$ , and $\overline{AC}$ intersects $\overline{BP}$ at point $D$ . If $PB = 3$ and $PD= 2$ , then $AD\cdot CD =$
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3,1); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); dot(A);dot(B);dot(C);dot(P);dot(D); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N); draw(A--B--P--cycle); draw(A--C--B--cycle);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | The product of two lengths with a common point brings to mind the Power of a Point Theorem
Since $PA = PB$ , we can make a circle with radius $PA$ that is centered on $P$ , and both $A$ and $B$ will be on that circle. Since $\angle APB = \widehat {AB} = 2 \angle ACB$ , we can see that point $C$ will also lie on the circle, since the measure of arc $\widehat {AB}$ is twice the measure of inscribed angle $\angle ACB$ , which is true for all inscribed angles.
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3.06,0.9); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); pair E = (0,4.5); dot(A);dot(B);dot(C);dot(P);dot(D);dot(E); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,NW);label("$D$",D,NE + N);label("$P$",P,N);label("$E$", E, NW); draw(A--B--P--cycle); draw(A--C--B--cycle); draw(circle(P, 2.46)); draw(P--E);[/asy]
Since $PDB$ is a line, we have $PD + DB = PB$ , which gives $3 = DB + 2$ , or $DB = 1$
We now extend radius $PB = 3$ to diameter $EB = 6$ . Since $EDB$ is a line, we have $ED + DB = EB$ , which gives $ED + 1 = 6$ , or $ED = 5$
Finally, we apply the power of a point theorem to point $D$ . This states that $AD \cdot DC = DB \cdot DE$ . Since $DB = 1$ and $DE = 5$ , the desired product is $5$ , which is $\boxed{5}$ | A | 5 |
2b2907bdc051d9810fec5bc87dc15f0b | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_26 | Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$ , angle $APB$ is twice angle $ACB$ , and $\overline{AC}$ intersects $\overline{BP}$ at point $D$ . If $PB = 3$ and $PD= 2$ , then $AD\cdot CD =$
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3,1); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); dot(A);dot(B);dot(C);dot(P);dot(D); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N); draw(A--B--P--cycle); draw(A--C--B--cycle);[/asy]
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 9$ | Construct the angle bisector of $\angle APD,$ and let it intersect $AD$ at $E.$ From the angle bisector theorem, we have $AE=3a$ and $DE=2a$ for some $a.$ Then, note that $\angle EPD = \angle BCD = x,$ so $EPCB$ is cyclic. Then, $\frac{PD}{ED} = \frac{CD}{BD}$ or $\frac{2}{2x} = \frac{CD}{1}.$ Thus, $AD \cdot DC = 5x \cdot DC = 5,$ or $\boxed{5}.$ | A | 5 |
60f82a4f625b02a1a7457402c532250b | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_27 | Consider those functions $f$ that satisfy $f(x+4)+f(x-4) = f(x)$ for all real $x$ . Any such function is periodic, and there is a least common positive period $p$ for all of them. Find $p$
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 32$ | Recall that $p$ is the fundamental period of function $f$ iff $p$ is the smallest positive $p$ such that $f(x) = f(x + p)$ for all $x$
In this case, we know that $f(x+ 4) + f(x - 4) = f(x)$ . Plugging in $x+4$ in for $x$ to get the next equation in the recursion, we also get $f(x + 8) + f(x) = f(x + 4)$ . Adding those two equations gives $f(x + 8) + f(x - 4) = 0$ after cancelling out common terms.
Again plugging in $x + 4$ in for $x$ in that last equation (in order to get $f(x)$ ), we find that $f(x) = -f(x + 12)$ . Now, plugging in $x+12$ for $x$ , we get $f(x + 12) = -f(x + 24)$ . This proves that $f(x) = f(x + 24)$ , so there is a period of $24$ , which gives answer $\boxed{24}$ . We now eliminate answers $A$ through $C$ | D | 24 |
73122aa8b12408873d2a86cbf98c30ed | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_28 | How many ordered triples of integers $(a,b,c)$ satisfy $|a+b|+c = 19$ and $ab+|c| = 97$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$ | WLOG , let $a \ge 0$ , and let $a \ge b$ . We can say this because if we have one solution $(a,b) = (a_0, b_0)$ with $a_0 \ge 0$ and $a_0 > b_0$ , we really have the four solutions $(a_0, b_0), (-a_0, -b_0), (b_0, a_0), (-b_0, -a_0)$ by the symmetry of the original problem.
Furthermore, we assert that these four solutions are distinct. We can say that $a > b$ , since if $a=b$ , we have $c = 19 - 2a$ for the first equation and either $c = 97 - a^2$ or $c = a^2 - 97$ for the second equation. Equating $19 - 2a = 97 - a^2$ gives no integer solution, while equating $19 - 2a = a^2 - 97$ also gives no integer solution.
Thus, we can now assume WLOG that $a \ge 0$ and $a > b$ , and each pair of $(a_0,b_0)$ that we get will generate four unique solutions: $(a_0, b_0), (b_0, a_0), (-a_0, -b_0), (-b_0, -a_0)$
We now divide the problem into $c \ge 0$ and $c < 0$
If $c \ge 0$ , we have $|a + b| + c = 19$ and $ab + c = 97$
Solving both equations for $c$ and equating them, we get that $ab - |a + b| = 78$ . Splitting these up, we find that either $ab - a - b = 78$ or $ab + a+ b = 78$ . Factoring both with SFFT gives $(a-1)(b-1) = 79$ or $(a +1)(b+1) = 79$ . We factor with the restrctions that $a \ge 0$ and $a > b$ . Since $79$ is prime, we have:
$a - 1 = 79$ and $b - 1 = 1$ , which leads to $(80,2)$
$a + 1 = 79$ and $b + 1 = 1$ , which leads to $(78, 0)$
Each of those solutions could generate $3$ more solutions, giving a total of $8$ potential solutions. However, in each the first set of four solutions, we have $|a + b| = 82$ , which from the original first equation $|a + b| + c = 19$ gives $c = 19 - 82$ , which contradicts our initial assumption that $c\ge 0$ . Similarly, for the second set of four solutions, we have $|a + b| = 78$ , which leads to $c = 19 - 78$ , also contradicting $c \ge 0$
If $c < 0$ , we have $|a + b| + c = 19$ and $ab - c = 97$ . We note that $19 - c$ must be positive whenever $c$ is negative, and thus $|a + b| = a + b$
Solving both equations for $c$ and using SFFT as above gives $(a+1)(b+1) = 117$ . Since $117 = 3^2\cdot 13$ , we factor with the restriction that $a\ge 0$ and $a > b$ . Thus, we can let $a+1 \in \{117, 39, 13\}$ , which means $a \in \{116, 38, 12\}$ . These give corresponding $b+1\in \{1, 3, 9\}$ , which leads to corresponding $b \in \{0, 2, 8\}$ . Combining the solutions, we have $(a,b) = (116, 0), (38, 2), (12, 8)$
Each of these three solutions permutes, negates, and permute-negates into $4$ solutions as described in the start of the solution, for a total of $12$ solutions.
Checking our solutions to ensure $c < 0$ , we find in the first set of four solutions, $|a + b| = 116$ , and thus $c = 19 - 116$ , which is indeed negative.
In the second set of four solutions, $|a + b| = 40$ , which leads to $c = 19 - 40$ , which is also negative.
Finally, in the third set of four solutoins, $|a + b| = 20$ , which leads to $c = 19 - 20$ , which is negative.
Thus, there are $12$ ordered triples, and the answer is $\boxed{12}$ | E | 12 |
f04e3cb9f98c0115f6fe43dd5075692e | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_29 | Call a positive real number special if it has a decimal representation that consists entirely of digits $0$ and $7$ . For example, $\frac{700}{99}= 7.\overline{07}= 7.070707\cdots$ and $77.007$ are special numbers. What is the smallest $n$ such that $1$ can be written as a sum of $n$ special numbers?
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 10\qquad\\ \textbf{(E)}\ \text{The number 1 cannot be represented as a sum of finitely many special numbers.}$ | Define a super-special number to be a number whose decimal expansion only consists of $0$ 's and $1$ 's. The problem is equivalent to finding the number of super-special numbers necessary to add up to $\frac{1}{7}=0.142857142857\hdots$ . This can be done in $8$ numbers if we take \[0.111111\hdots, 0.011111\hdots, 0.010111\hdots, 0.010111\hdots, 0.000111\hdots, 0.000101\hdots, 0.000101\hdots, 0.000100\hdots\] Now assume for sake of contradiction that we can do this with strictly less than $8$ super-special numbers (in particular, less than $10$ .) Then the result of the addition won't have any carry over, so each digit is simply the number of super-special numbers which had a $1$ in that place. This means that in order to obtain the $8$ in $0.1428\hdots$ , there must be $8$ super-special numbers, so the answer is $\boxed{8}$ | B | 8 |
22cef8226f118fa8dfa2bb9113054ea0 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_30 | For positive integers $n$ , denote $D(n)$ by the number of pairs of different adjacent digits in the binary (base two) representation of $n$ . For example, $D(3) = D(11_{2}) = 0$ $D(21) = D(10101_{2}) = 4$ , and $D(97) = D(1100001_{2}) = 2$ . For how many positive integers less than or equal $97$ to does $D(n) = 2$
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$ | If $D(n)$ is even, then the binary expansion of $n$ will both begin and end with a $1$ , because all positive binary numbers begin with a $1$ , and if you switch digits twice, you will have a $1$ at the end. Thus, we are only concerned with the $49$ odd numbers between $1$ and $98$ inclusive.
All of these odd numbers will have an even $D(n)$ $D(n) = 0$ will be given by the numbers $1, 11, 111, 1111, 11111, 111111$ , which is a total of $6$ numbers.
We skip $D(n) = 2$ for now, and move to $D(n) = 4$ , which is easier to count. The smallest $D(n) = 4$ happens when $n = 10101$ . To get another number such that $D(n) = 4$ , we may extend any of the five blocks of zeros or ones by one digit. This will form $110101, 100101, 101101, 101001, 101011$ , all of which are odd numbers that have $D(n) = 4$ . To find seven digit numbers that have $D(n) = 4$ , we can again extend any block by one, so long as it remains less than $1100001$ or under. There are five cases.
1) Extending $110101$ is impossible without going over $1100001$
2) Extending $100101$ by putting a $1$ at the beginning will go over $1100001$ , but the other four extensions work, giving $1000101, 1001101, 1001001, 1001011$
3) Extending $101101$ by putting a $1$ at the beginning will go over $1100001$ , but the other four extensions give $1001101, 1011101, 1011001, 1011011$ . However, $1001101$ already appeared in #2, giving only three new numbers.
4) Extending $101001$ at the first group is impossible. The other four extensions are $1001001, 1011001, 1010001, 1010011$ , but the first two are repeats. Thus, there are only two new numbers.
5) Extending $101011$ at the first group is impossible. The other four extensions give $1001011, 1011011, 1010011, 1010111$ , but only the last number is new.
Thus, there is $1$ five digit number, $5$ six digit numbers, and $4 + 3 + 2 + 1 = 10$ seven digit numbers under $1100001$ for which $D(n) = 4$ . That gives a total of $16$ numbers.
There smallest number for which $D(n) = 6$ is $1010101$ , which is under $98$ . Further extensions, as well as cases where $D(n) > 6$ , are not possible.
Thus, we know that there are $6$ odd numbers that have $D(n) = 0$ , and $16$ odd numbers that have $D(n) = 4$ , and $1$ number that has $D(n) = 6$ . The remaining odd numbers must have $D(n) = 2$ . This means there are $49 - 6 - 16 - 1 = 26$ numbers that have $D(n) = 2$ , which is option $\boxed{26}$ | C | 26 |
22cef8226f118fa8dfa2bb9113054ea0 | https://artofproblemsolving.com/wiki/index.php/1997_AHSME_Problems/Problem_30 | For positive integers $n$ , denote $D(n)$ by the number of pairs of different adjacent digits in the binary (base two) representation of $n$ . For example, $D(3) = D(11_{2}) = 0$ $D(21) = D(10101_{2}) = 4$ , and $D(97) = D(1100001_{2}) = 2$ . For how many positive integers less than or equal $97$ to does $D(n) = 2$
$\textbf{(A)}\ 16\qquad\textbf{(B)}\ 20\qquad\textbf{(C)}\ 26\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 35$ | For $D(n)$ to be $2$ $n$ must be in the form $1...10...01...1$ . Note that $n$ must be at least $3$ digits for $D(n)$ to be $2$
Case $1$ $n$ has $3$ digits
$n$ $101_2$ , there is only $1$ possible value for $n$ when $n$ has $3$ digits.
Case $2$ $n$ has $4$ digits
$n = 1$ _ _ $1_2$ , there could be $1$ zero or $2$ zeros between the first digit and the last digit. If there is only $1$ zero, the zero has $2$ possible places. If there are $2$ zeros, there is only $1$ possible placement. Therefore, there are $2+1=3$ possible values for $n$ when $n$ has $4$ digits.
Case $3$ $n$ has $5$ digits
$n = 1$ _ _ _ $1_2$ , there could be $1$ $2$ or $3$ zeros between the first digit and the last digit. If there is only $1$ zero, the zero has $3$ possible places. If there are $2$ zeros, there are $2$ possible placements. If there are $3$ zeros, there is only $1$ possible placement. Therefore, there are $3+2+1=6$ possible values for $n$ when $n$ has $5$ digits.
Case $4$ $n$ has $6$ digits
$n = 1$ _ _ _ _ $1_2$ , there could be $1$ $2$ $3$ or $4$ zeros between the first digit and the last digit. If there is only $1$ zero, the zero has $4$ possible places. If there are $2$ zeros, there are $3$ possible placements. If there are $3$ zeros, there are $2$ possible placements. If there are $4$ zeros, there is only $1$ possible placement. Therefore, there are $4+3+2+1=10$ possible values for $n$ when $n$ has $6$ digits.
Case $4$ $n$ has $7$ digits
Given by the problem, $n \le 1100001_2$ . We could first count the number of $n$ that is in the form of $10$ _ _ _ _ $1_2$ , where the digits between the second and last digit are in the form of $0...01...1$
There could be $0$ $1$ $2$ $3$ or $4$ zeros. Therefore, there are $5$ possible values for $n$ when $n=10$ _ _ _ _ $1_2$
There is also the case were $n = 1100001_2$ . Therefore, there are $5+1=6$ possible values for $n$ when $n$ has $7$ digits.
Therefore, the answer is $1+3+6+10+6=\boxed{26}$ | C | 26 |
6a37d35f20e264bc16acae7e32d1045e | https://artofproblemsolving.com/wiki/index.php/1996_AHSME_Problems/Problem_1 | The addition below is incorrect. What is the largest digit that can be changed to make the addition correct?
$\begin{tabular}{rr}&\ \texttt{6 4 1}\\ &\texttt{8 5 2}\\ &+\texttt{9 7 3}\\ \hline &\texttt{2 4 5 6}\end{tabular}$
$\text{(A)}\ 4\qquad\text{(B)}\ 5\qquad\text{(C)}\ 6\qquad\text{(D)}\ 7\qquad\text{(E)}\ 8$ | Doing the addition as is, we get $641 + 852 + 973 = 2466$ . This number is $10$ larger than the desired sum of $2456$ . Therefore, we must make one of the three numbers $10$ smaller.
We may either change $641 \rightarrow 631$ $852 \rightarrow 842$ , or $973 \rightarrow 963$ . Either change results in a valid sum. The largest digit that could be changed is thus the $7$ in the number $973$ , and the answer is $\boxed{7}$ | D | 7 |
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?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$ | If Walter had done his chores for $10$ days without doing any of them well, he would have earned $3 \cdot 10 = 30$ dollars. He got $6$ dollars more than this.
He gets a $5 - 3 = 2$ dollar bonus every day he does his chores well. Thus, he did his chores exceptionally well $\frac{6}{2} = 3$ days, and the answer is $\boxed{3}$ | A | 3 |
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?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\qquad\text{(E)}\ 7$ | If Walter had done his chores for $10$ days exceptionally well, he would have earned $5 \cdot 10 = 50$ dollars. He got $50 - 36 = 14$ dollars less than this.
He gets $2$ dollars docked from his pay if he doesn't do his chores well. Therefore, he didn't do his chores well on $\frac{14}{2} = 7$ days. The other $10 - 7 = 3$ days, he did them exceptionally well. Therefore, the answer is $\boxed{3}$ | A | 3 |