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5715c6643dc8dd837dd6e674d235ce63 | https://artofproblemsolving.com/wiki/index.php/1971_AHSME_Problems/Problem_2 | If $b$ men take $c$ days to lay $f$ bricks, then the number of days it will take $c$ men working at the same rate to lay $b$ bricks, is
$\textbf{(A) }fb^2\qquad \textbf{(B) }b/f^2\qquad \textbf{(C) }f^2/b\qquad \textbf{(D) }b^2/f\qquad \textbf{(E) }f/b^2$ | We can use a modified version of the equation $\text{Distance} = \text{Rate} \times {\text{Time}}$ , which is $\text{Work Done} = \text{Rate of Work} \times{ \text{Time Worked}}$ . In this case, the work done is the number of bricks laid, the rate of work is the number of men working, and the time worked is the number of days. With these definitions, we see from the first equation that $f = b \cdot c$ . If we let $X$ be the number of days it will take $c$ men working at the same rate to $b$ bricks, then we have the equation $b = c \cdot x$ . So, $x = \frac{b}{c}$ . The first equation says that $\frac{f}{c^2} = \frac{b}{c}$ , which leads to $x = \frac{f}{c^2}$ . This doesn't match any of our answer choices though, so we have to fiddle around a bit before we realize that $c = \frac{f}{b}$ , a substitution we can make to see that $x = \frac{b^2}{f}$ . Thus, the answer is $\boxed{2}$ | D | 2 |
c07df90b3690e435d8a32b2e8ecea7b8 | https://artofproblemsolving.com/wiki/index.php/1971_AHSME_Problems/Problem_27 | A box contains chips, each of which is red, white, or blue. The number of blue chips is at least half the number of white chips, and at most one third the number of red chips. The number which are white or blue is at least $55$ . The minimum number of red chips is
$\textbf{(A) }24\qquad \textbf{(B) }33\qquad \textbf{(C) }45\qquad \textbf{(D) }54\qquad \textbf{(E) }57$ | Let the number of white be $2x$ . The number of blue is then $x-y$ for some constant $y$ . So we want $2x+x-y=55\rightarrow 3x-y=55$ . We take mod 3 to find y. $55=1\pmod{3}$ , so blue is 1 more than half white. The number of whites is then 36, and the number of blues is 19. So $19*3=\boxed{57}$ | null | 57 |
b617cd9c64a578cf329a6c28c123b811 | https://artofproblemsolving.com/wiki/index.php/1971_AHSME_Problems/Problem_29 | Given the progression $10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}$ .
The least positive integer $n$ such that the product of the first $n$ terms of the progression exceeds $100,000$ is
$\textbf{(A) }7\qquad \textbf{(B) }8\qquad \textbf{(C) }9\qquad \textbf{(D) }10\qquad \textbf{(E) }11$ | The product of the sequence $10^{\dfrac{1}{11}}, 10^{\dfrac{2}{11}}, 10^{\dfrac{3}{11}}, 10^{\dfrac{4}{11}},\dots , 10^{\dfrac{n}{11}}$ is equal to $10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}$ since we are looking for the smallest value $n$ that will create $100,000$ , or $10^5$ . From there, we can set up the equation $10^{\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}}=10^5$ , which simplified to $\dfrac{1}{11}+\frac{2}{11}\dots\frac{n}{11}=5$ , or $1+2+3\dots n=55$ This can be converted to $\frac{n(1+n)}{2}=55$ This simplified to the quadratic $n^2+n-110=0$ Or $(n+11)(n-10)=0$ So $n=-11$ or $10$ Since only positive values of $n$ work, $n=10$ makes the expression equal $100000$ . However, we have to exceed $100000$ , so our answer is $\boxed{11}.$ | E | 11 |
9fdfd29526ba5de617c0d7be3102f919 | https://artofproblemsolving.com/wiki/index.php/1970_AHSME_Problems/Problem_19 | The sum of an infinite geometric series with common ratio $r$ such that $|r|<1$ is $15$ , and the sum of the squares of the terms of this series is $45$ . The first term of the series is
$\textbf{(A) } 12\quad \textbf{(B) } 10\quad \textbf{(C) } 5\quad \textbf{(D) } 3\quad \textbf{(E) 2}$ | We know that the formula for the sum of an infinite geometric series is $S = \frac{a}{1-r}$
So we can apply this to the conditions given by the problem.
We have two equations:
\begin{align*} 15 &= \frac{a}{1-r} \\ 45 &= \frac{a^{2}}{1-r^{2}} \end{align*}
We get
\begin{align*} a &= 15 - 15r \\ a^{2} &= 45 - 45r^{2} \\ \\ (15 - 15r)^{2} &= 45 - 45r^{2} \\ 270r^{2} - 450r + 180 &= 0 \\ 3r^{2} - 5r + 2 &= 0 \\ (3r - 2)(r - 1) &= 0 \end{align*}
Since $|r| < 1$ $r = \frac{2}{3}$ , so plug this into the equation above and we get $a = 15 - 15r = 15 - 10 = \boxed{5}$ | C | 5 |
4952444fe300b1eed94efd70557681cb | https://artofproblemsolving.com/wiki/index.php/1970_AHSME_Problems/Problem_33 | Find the sum of digits of all the numbers in the sequence $1,2,3,4,\cdots ,10000$
$\text{(A) } 180001\quad \text{(B) } 154756\quad \text{(C) } 45001\quad \text{(D) } 154755\quad \text{(E) } 270001$ | We can find the sum using the following method. We break it down into cases. The first case is the numbers $1$ to $9$ . The second case is the numbers $10$ to $99$ . The third case is the numbers $100$ to $999$ . The fourth case is the numbers $1,000$ to $9,999$ . And lastly, the sum of the digits in $10,000$ . The first case is just the sum of the numbers $1$ to $9$ which is, using $\frac{n(n+1)}{2}$ $45$ . In the second case, every number $1$ to $9$ is used $19$ times. $10$ times in the tens place, and $9$ times in the ones place. So the sum is just $19(45)$ . Similarly, in the third case, every number $1$ to $9$ is used $100$ times in the hundreds place, $90$ times in the tens place, and $90$ times in the ones place, for a total sum of $280(45)$ . By the same method, every number $1$ to $9$ is used $1,000$ times in the thousands place, $900$ times in the hundreds place, $900$ times in the tens place, and $900$ times in the ones place, for a total of $3700(45)$ . Thus, our final sum is $45+19(45)+280(45)+3700(45)+1=4000(45)+1=\boxed{180001}.$ | A | 180001 |
4952444fe300b1eed94efd70557681cb | https://artofproblemsolving.com/wiki/index.php/1970_AHSME_Problems/Problem_33 | Find the sum of digits of all the numbers in the sequence $1,2,3,4,\cdots ,10000$
$\text{(A) } 180001\quad \text{(B) } 154756\quad \text{(C) } 45001\quad \text{(D) } 154755\quad \text{(E) } 270001$ | Consider the numbers from $0000-9999$ . We have $40000$ digits and each has equal an probability of being $0,1,2....9$ .
Thus, our desired sum is $4000\left( \frac{9 \cdot 10}{2} \right)+1=4000(45)+1=\boxed{180001}.$ | A | 180001 |
4952444fe300b1eed94efd70557681cb | https://artofproblemsolving.com/wiki/index.php/1970_AHSME_Problems/Problem_33 | Find the sum of digits of all the numbers in the sequence $1,2,3,4,\cdots ,10000$
$\text{(A) } 180001\quad \text{(B) } 154756\quad \text{(C) } 45001\quad \text{(D) } 154755\quad \text{(E) } 270001$ | As in Solution 2, we consider the four digit numbers from $0000-9999.$ We see that we have $10000\times4=40000$ digits with each digit appearing equally.
Thus, the digit sum will be the average of the digits multiplied by $40000.$ The digit average comes out to be $\frac{0+9}{2},$ since all digits are consecutive.
So, our answer will be $\frac{9}{2} \times 40000 = 180000.$ However, since we purposely did not include $10000,$ we add one to get our final answer as $\boxed{180001}.$ | A | 180001 |
4f8eaefceb628aa32dc40045de6c9670 | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_8 | Triangle $ABC$ is inscribed in a circle. The measure of the non-overlapping minor arcs $AB$ $BC$ and $CA$ are, respectively, $x+75^{\circ} , 2x+25^{\circ},3x-22^{\circ}$ . Then one interior angle of the triangle is:
$\text{(A) } 57\tfrac{1}{2}^{\circ}\quad \text{(B) } 59^{\circ}\quad \text{(C) } 60^{\circ}\quad \text{(D) } 61^{\circ}\quad \text{(E) } 122^{\circ}$ | [asy] draw(circle((0,0),65)); draw((25,60)--(39,-52)--(-52,-39)--(25,60)); dot((25,60)); dot((39,-52)); dot((-52,-39)); dot((0,0)); draw((0,0)--(-52,-39)); draw((0,0)--(39,-52)); draw((0,0)--(25,60)); label("A",(-52,-39),SW); label("B",(25,60),NE); label("C",(39,-52),SE); [/asy] Because the triangle is inscribed, the sum of the minor arcs equals $360^\circ$ . Thus, \[x+75+2x+25+3x-22=360\] \[6x+78=360\] Solving this yields $x = 47$ , so the inscribed angles are $122^\circ$ $99^\circ$ , and $119^\circ$ . Noting that an angle of $\triangle ABC$ is half of its corresponding inscribed angle, so the angles of $\triangle ABC$ are $59.5^\circ$ $49.5^\circ$ , and $\boxed{61}$ | D | 61 |
006f10571390d748bcd0a1536ffb49aa | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_10 | The number of points equidistant from a circle and two parallel tangents to the circle is:
$\text{(A) } 0\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } \infty$ | [asy] draw(circle((0,0),100)); draw((-300,100)--(300,100),Arrows); draw((-300,-100)--(300,-100),Arrows); draw((-300,0)--(300,0),dotted,Arrows); dot((-200,0)); dot((0,0)); dot((200,0)); draw((-200,100)--(-200,-100),dotted); draw((200,100)--(200,-100),dotted); [/asy] The distance between the two parallel tangents is the length of the circle's diameter, so the distance from a point that satisfies the conditions and the two tangents is the length of the circle's radius. From the diagram, there are $\boxed{3}$ points that satisfies the conditions. | C | 3 |
b15d25bbc8780fe098f1b621249a1771 | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_18 | The number of points common to the graphs of $(x-y+2)(3x+y-4)=0 \text{ and } (x+y-2)(2x-5y+7)=0$ is:
$\text{(A) } 2\quad \text{(B) } 4\quad \text{(C) } 6\quad \text{(D) } 16\quad \text{(E) } \infty$ | By the Zero Product Property $x-y+2=0$ or $3x+y-4=0$ in the first equation and $x+y-2=0$ or $2x-5y+7=0$ in the second equation. Thus, from the first equation, $y = x+2$ or $y =-3x+4$ , and from the second equation, $y=-x+2$ or $y = \frac{2}{5}x + \frac{7}{5}$
If a point is common to the two graphs, then the point must be in one of the lines in the first equation as well as one of the lines in the second equation. Since the slopes of the lines are different, none of the lines are parallel. Thus, there are $2 \cdot 2 = \boxed{4}$ points of intersection in the two graphs. | B | 4 |
ae3602f29da4c3292630d53ca457547f | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_22 | Let $K$ be the measure of the area bounded by the $x$ -axis, the line $x=8$ , and the curve defined by
\[f={(x,y)\quad |\quad y=x \text{ when } 0 \le x \le 5, y=2x-5 \text{ when } 5 \le x \le 8}.\]
Then $K$ is:
$\text{(A) } 21.5\quad \text{(B) } 36.4\quad \text{(C) } 36.5\quad \text{(D) } 44\quad \text{(E) less than 44 but arbitrarily close to it}$ | [asy] import graph; size(9.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=9.2,ymin=-5.2,ymax=13.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); draw((0,0)--(5,5)--(8,11)--(8,0)--(0,0)); dot((0,0)); dot((5,5)); dot((8,11)); dot((8,0)); [/asy]
The shape can be divided into a triangle and a trapezoid. For the triangle, the base is $5$ and the height is $5$ , so the area is $\frac{5 \cdot 5}{2} = \frac{25}{2}$ . For the trapezoid, the two bases are $5$ and $11$ and the height is $3$ , so the area is $\frac{3(5+11)}{2} = 24$ . Thus, the total area is $\frac{25}{2} + 24 = \frac{73}{2} = \boxed{36.5}$ | C | 36.5 |
28026254e4a079847a2e6ec89cc365bc | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_25 | If it is known that $\log_2(a)+\log_2(b) \ge 6$ , then the least value that can be taken on by $a+b$ is:
$\text{(A) } 2\sqrt{6}\quad \text{(B) } 6\quad \text{(C) } 8\sqrt{2}\quad \text{(D) } 16\quad \text{(E) none of these}$ | We use the logarithm property of addition: \begin{align*} \log_2(a)+\log_2(b) \ge 6 &= \log_2(ab) \ge 6\\ &\Rightarrow 2^{log_2(ab)} \ge 2^6\\ &= ab \ge 64 \end{align*} Due to the Quadratic Optimization or the AM-GM Inequality , the least value is obtained when $a = b$ .
Therefore, $a = b = 8 \Rightarrow a + b = \boxed{16}$ | D | 16 |
19becd1d83c241ce6243d3261eaa6a02 | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_26 | [asy] draw(arc((0,-1),2,30,150),dashed+linewidth(.75)); draw((-1.7,0)--(0,0)--(1.7,0),dot); draw((0,0)--(0,.98),dot); MP("A",(-1.7,0),W);MP("B",(1.7,0),E);MP("M",(0,0),S);MP("C",(0,1),N); [/asy]
A parabolic arch has a height of $16$ inches and a span of $40$ inches. The height, in inches, of the arch at the point $5$ inches from the center $M$ is:
$\text{(A) } 1\quad \text{(B) } 15\quad \text{(C) } 15\tfrac{1}{3}\quad \text{(D) } 15\tfrac{1}{2}\quad \text{(E) } 15\tfrac{3}{4}$ | Because the arch has a height of $16$ inches, an equation that models the arch is $y = ax^2 + 16$ , where $x$ is the horizontal distance from the center and $y$ is the height. The arch has a span of $40$ inches, so the arch meets the ground $20$ inches from the center. That means $0 = 400a + 16$ , so $a = -\frac{1}{25}$
Thus, the equation that models height based on distance from the center $y = -\frac{1}{25}x^2 + 16$ , so the height of the arch $5$ inches from the center is $-\frac{1}{25} \cdot 5^2 + 16 = \boxed{15}$ inches. | B | 15 |
31f11e3f16a77b7221fb95b15ac74fc3 | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_32 | Let a sequence $\{u_n\}$ be defined by $u_1=5$ and the relationship $u_{n+1}-u_n=3+4(n-1), n=1,2,3\cdots.$ If $u_n$ is expressed as a polynomial in $n$ , the algebraic sum of its coefficients is:
$\text{(A) 3} \quad \text{(B) 4} \quad \text{(C) 5} \quad \text{(D) 6} \quad \text{(E) 11}$ | Note that the first differences create a linear function, so the sequence ${u_n}$ is quadratic.
The first three terms of the sequence are $5$ $8$ , and $15$ . From there, a system of equations can be written. \[a+b+c=5\] \[4a+2b+c=8\] \[9a+3b+c=15\] Solve the system to get $a=2$ $b=-3$ , and $c=6$ . The sum of the coefficients is $\boxed{5}$ | C | 5 |
a01f700286fd2da97a42897d3f9b43e3 | https://artofproblemsolving.com/wiki/index.php/1968_AHSME_Problems/Problem_20 | The measures of the interior angles of a convex polygon of $n$ sides are in arithmetic progression. If the common difference is $5^{\circ}$ and the largest angle is $160^{\circ}$ , then $n$ equals:
$\text{(A) } 9\quad \text{(B) } 10\quad \text{(C) } 12\quad \text{(D) } 16\quad \text{(E) } 32$ | The formula for the sum of the angles in any polygon is $180(n-2)$ . Because this particular polygon is convex and has its angles in an arithmetic sequence with its largest angle being $160$ , we can find the sum of the angles.
$a_{n}=160$
$a_{1}=160-5(n-1)$
Plugging this into the formula for finding the sum of an arithmetic sequence...
$n(\frac{160+160-5(n-1)}{2})=180(n-2)$
Simplifying, we get $n^2+7n-144$
Since we want the positive solution to the quadratic, we can easily factor and find the answer is $n=\boxed{9}$ | null | 9 |
7e25c36078ec81d559edac4902071586 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_1 | The three-digit number $2a3$ is added to the number $326$ to give the three-digit number $5b9$ . If $5b9$ is divisible by 9, then $a+b$ equals
$\text{(A)}\ 2\qquad\text{(B)}\ 4\qquad\text{(C)}\ 6\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$ | If $5b9$ is divisible by $9$ , this must mean that $5 + b + 9$ is a multiple of $9$ . So, \[5 + b + 9 = 9, 18, 27, 36...\]
Because $5 + 9 = 14$ and $b$ is in between 0 and 9,
\[5 + b + 9 = 18\] \[b = 4\]
The question states that \[2a3 + 326 = 549\] so \[2a3 = 549 - 326\] \[2a3 = 223\] \[a = 2\]
\[a + b = 6\] which is answer choice $\boxed{6}$ | C | 6 |
5a4dfff6353b3a0ce1e0ad4815a3712c | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_4 | Given $\frac{\log{a}}{p}=\frac{\log{b}}{q}=\frac{\log{c}}{r}=\log{x}$ , all logarithms to the same base and $x \not= 1$ . If $\frac{b^2}{ac}=x^y$ , then $y$ is:
$\text{(A)}\ \frac{q^2}{p+r}\qquad\text{(B)}\ \frac{p+r}{2q}\qquad\text{(C)}\ 2q-p-r\qquad\text{(D)}\ 2q-pr\qquad\text{(E)}\ q^2-pr$ | We are given: \[\frac{b^2}{ac} = x^y\]
Taking the logarithm on both sides: \[\log{\left(\frac{b^2}{ac}\right)} = \log{x^y}\]
Using the properties of logarithms: \[2\log{b} - \log{a} - \log{c} = y \log{x}\]
Substituting the values given in the problem statement: \[2q \log{x} - p \log{x} - r \log{x} = y \log{x}\]
Since $x \neq 1$ , dividing each side by $\log{x}$ we get: \[y = \boxed{2}\] | C | 2 |
d63690a8f604498417b3956cf8e623b1 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_5 | A triangle is circumscribed about a circle of radius $r$ inches. If the perimeter of the triangle is $P$ inches and the area is $K$ square inches, then $\frac{P}{K}$ is:
$\text{(A)}\text{ independent of the value of} \; r\qquad\text{(B)}\ \frac{\sqrt{2}}{r}\qquad\text{(C)}\ \frac{2}{\sqrt{r}}\qquad\text{(D)}\ \frac{2}{r}\qquad\text{(E)}\ \frac{r}{2}$ | The area $K$ of the triangle can be expressed in terms of its inradius $r$ and its semiperimeter $s$ as: \[K = r \times s = r \times \frac{P}{2}\]
So, $\frac{P}{K} = \boxed{2}$ | C | 2 |
73242762882170910d6d8cd0050a9c27 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_11 | If the perimeter of rectangle $ABCD$ is $20$ inches, the least value of diagonal $\overline{AC}$ , in inches, is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \sqrt{50}\qquad \textbf{(C)}\ 10\qquad \textbf{(D)}\ \sqrt{200}\qquad \textbf{(E)}\ \text{none of these}$ | For rectangle $ABCD$ with perimeter 20, the diagonal $AC$ is given by: \[AC = \sqrt{l^2 + w^2}\] To minimize $AC$ $l$ and $w$ should be equal (i.e., the rectangle is a square). Thus, $l = w = 5$ .
So, the minimum $AC$ is: \[AC = \sqrt{5^2 + 5^2} = \boxed{50}\] ~ proloto | B | 50 |
c9da80067ca3683e69f0a11d8799c592 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_32 | In quadrilateral $ABCD$ with diagonals $AC$ and $BD$ , intersecting at $O$ $BO=4$ $OD = 6$ $AO=8$ $OC=3$ , and $AB=6$ . The length of $AD$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$ | After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of $AB, BD(BD = BO + OD)$ , but we want to find the value of $AD$ . We can apply stewart's theorem now, letting $m = 4, n = 6, AD = X, AB = 6$ , and we have $10 \cdot 6 \cdot 4 + 8 \cdot 8 \cdot 10 = x^2 + 36 \cdot 6$ , and we see that $x = \boxed{166}$ | E | 166 |
c9da80067ca3683e69f0a11d8799c592 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_32 | In quadrilateral $ABCD$ with diagonals $AC$ and $BD$ , intersecting at $O$ $BO=4$ $OD = 6$ $AO=8$ $OC=3$ , and $AB=6$ . The length of $AD$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$ | [asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(5cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -9.78, xmax = 9.78, ymin = -5.72, ymax = 5.72; /* image dimensions */ draw((-1,4)--(-4.08,3.78)--(-3.1,-3.42)--(1.56,-0.22)--cycle, linewidth(2)); /* draw figures */ draw((-1,4)--(-4.08,3.78), linewidth(2)); draw((-4.08,3.78)--(-3.1,-3.42), linewidth(2)); draw((-3.1,-3.42)--(1.56,-0.22), linewidth(2)); draw((1.56,-0.22)--(-1,4), linewidth(2)); draw(circle((-2.287661623108666,0.35726272352132055), 3.863626188061437), linewidth(2)); draw((-1,4)--(-3.1,-3.42), linewidth(2)); draw((-4.08,3.78)--(1.56,-0.22), linewidth(2)); /* dots and labels */ dot((-1,4),dotstyle); label("$A$", (-0.92,4.2), NE * labelscalefactor); dot((-4.08,3.78),dotstyle); label("$B$", (-4.42,4), NE * labelscalefactor); dot((-3.1,-3.42),dotstyle); label("$C$", (-3.4,-3.94), NE * labelscalefactor); dot((1.56,-0.22),dotstyle); label("$D$", (1.8,-0.4), NE * labelscalefactor); dot((-1.566733533935139,1.9975415134291763),linewidth(4pt) + dotstyle); label("$O$", (-1.34,1.98), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy] (Diagram not to scale)
Since $AO \cdot OC = BO \cdot OD$ $ABCD$ is cyclic through power of a point. From the given information, we see that $\triangle{AOB}\sim \triangle{DOC}$ and $\triangle{BOC} \sim \triangle{AOD}$ . Hence, we can find $CD=\frac{9}{2}$ and $AD=2 \cdot BC$ . Letting $BC$ be $x$ , we can use Ptolemy's to get \[6 \cdot \frac{9}{2} + 2x^2=10 \cdot 11 \implies x=\sqrt{\frac{83}{2}}\] Since we are solving for $AD=2x=2\cdot\sqrt{\frac{83}{2}}=\sqrt{4\cdot\frac{83}{2}} = \boxed{166}$ | E | 166 |
c9da80067ca3683e69f0a11d8799c592 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_32 | In quadrilateral $ABCD$ with diagonals $AC$ and $BD$ , intersecting at $O$ $BO=4$ $OD = 6$ $AO=8$ $OC=3$ , and $AB=6$ . The length of $AD$ is: $\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$ | The solution says it all. Since $\angle AOD$ is supplementary to $\angle AOB$ $cos(\angle AOD) = cos(180^{\circ} - \angle AOB)=-cos(\angle AOB)$ . The law of cosines on $\triangle AOB$ gives us $cos(\angle AOB)=\frac {8^2+4^2-6^2}{(2)(8)(4)}=\frac {11}{16}$ . Again, we can use the law of cosines on $\triangle AOD$ , which gives us \[AD=\sqrt {8^2+6^2-2(8)(6)cos(\angle AOD)}\] \[=\sqrt {8^2+6^2-(2)(8)(6)cos(\angle 180^{\circ} - AOB)}\] \[=\sqrt {8^2+6^2+(2)(8)(6)cos(\angle AOB)}\] \[=\sqrt {8^2+6^2+(2)(8)(6)(\frac {11}{16})}\] \[=\sqrt{166}\] which gives us $\boxed{166}$ | E | 166 |
51800ec2b52ef308827d818cdebf20bb | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_40 | Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$ $PB=6$ , and $PC=10$ . To the nearest integer the area of triangle $ABC$ is:
$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$ | [asy] draw((0,10)--(8.66,-5)--(-8.66,-5)--cycle); label("$A$",(0,10),N); label("$B$",(-9.5,-5.2),N); label("$C$",(9.5,-5.2),N); dot((-3,0)); label("$P$",(-3,-2),N); draw((-3,0)--(0,10)); draw((-3,0)--(-8.66,-5)); draw((-3,0)--(8.66,-5)); dot((-9,7.5)); label("$P'$",(-9.2,7.5),N); draw((-9,7.5)--(0,10)); draw((-9,7.5)--(-8.66,-5)); draw((-9,7.5)--(-3,0)); [/asy]
Notice that $6^2+8^2=10^2.$ That makes us want to construct a right triangle.
Rotate $\triangle APC$ $60^{\circ}$ about A. Note that $\triangle PAC \cong \triangle P'AB$ , so \[\angle P'AP = \angle PAB + \angle P'AB = \angle PAB + \angle PAC = 60^{\circ}.\]
Therefore, $\triangle APP'$ is equilateral, so $P'P=8$ , which means $\angle P'PB = 90^{\circ}.$
Let $\angle BP'P = \alpha .$ Notice that $\cos\alpha = \frac{8}{10}=\frac{4}{5},$ and $\sin\alpha = \frac{3}{5}.$
Applying the Law of Cosines to $\triangle APC$ (remembering $\angle APC = \angle AP'B$ ): \begin{align*} AC^2 &= 10^2+8^2-2\cdot10\cdot8\cdot \cos(60^{\circ}+\alpha)\\&= 164-160(\cos60\cos\alpha-\sin60\sin\alpha)\\&= 164-160(\frac{2}{5}-\frac{3\sqrt3}{10}) \\&= 164-16(4-3\sqrt3) \\ &= 100+48\sqrt3.\end{align*}
We want to find the area of $\triangle ABC$ , which is \[AC^2\cdot\frac{\sqrt3}{4}=25\sqrt3+36\approx\boxed{79}.\] | D | 79 |
51800ec2b52ef308827d818cdebf20bb | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_40 | Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$ $PB=6$ , and $PC=10$ . To the nearest integer the area of triangle $ABC$ is:
$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$ | Fun formula: Given a point whose distances from the vertices of an equilateral triangle are $a$ $b$ , and $c$ , the side length of the triangle is:
\[s=\sqrt{\frac{1}{2}\left(a^2+b^2+c^2\pm\sqrt{6(a^2b^2+b^2c^2+c^2a^2)-3(a^4+b^4+c^4)}\right)}\]
Given that the area of an equilateral triangle is $\frac{\sqrt{3}}{4}s^2$ , the answer is:
\begin{align*} [ABC] &= \frac{\sqrt{3}}{4}\cdot\frac{1}{2}\left(a^2+b^2+c^2\pm\sqrt{6(a^2b^2+b^2c^2+c^2a^2)-3(a^4+b^4+c^4)}\right)\\ &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{6\cdot 16(3^2 4^2+4^2 5^2+5^2 3^2)-3\cdot16(3^4+4^4+5^4)}\right)\\ &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96(144+400+225)-48(81+256+625)}\right)\\ &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot769-48\cdot962}\right) = \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot769-96\cdot481}\right)\\ &= \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot288}\right) = \frac{\sqrt{3}}{8}\left(200\pm\sqrt{96\cdot96\cdot3}\right)\\ &= 25\sqrt{3}\pm36 \approx \{6.5, \text{or } 78.5\} \end{align*}
$6.5$ is not a choice, therefore the answer is $\boxed{79}$ | D | 79 |
51800ec2b52ef308827d818cdebf20bb | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_40 | Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$ $PB=6$ , and $PC=10$ . To the nearest integer the area of triangle $ABC$ is:
$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$ | Rotate $P$ and $B$ $60^{\circ}$ CCW around $A$ , becoming $X$ and $C$ . Rotate $P$ and $C$ $60^{\circ}$ CCW around $B$ , becoming $Y$ and $A$ . Rotate $P$ and $A$ $60^{\circ}$ CCW around $C$ , becoming $Z$ and $B$
[asy] import graph; import geometry; size(12cm); pair A, B, C, P, X, Y, Z; // Define the equilateral triangle ABC real a = sqrt(100+48*sqrt(3)); A = (0, 0); B = rotate(60)*A + (a, 0); C = rotate(120)*B + (a, 0); // Define point P using given distances pair[] P_candidates = intersectionpoints(Circle(A,8), Circle(B,6)); for (pair candidate : P_candidates) { if (length(candidate - C) < 10.1 && length(candidate - C) > 9.9) { P = candidate; break; } } // Rotate C and P about A through 60 degrees to get B and X X = rotate(60,A)*P; // Rotate A and P about B through 60 degrees to get C and Y Y = rotate(60,B)*P; // Rotate B and P about C through 60 degrees to get A and Z Z = rotate(60,C)*P; // Draw the triangle and the segments draw(A--B--C--cycle); draw(A--P); draw(B--P); draw(C--P); // Connect X, Y, Z to P and to the vertices of the triangle draw(X--P, dashed); draw(Y--P, dashed); draw(Z--P, dashed); draw(X--A, dashed); draw(X--C, dashed); draw(Y--A, dashed); draw(Y--B, dashed); draw(Z--B, dashed); draw(Z--C, dashed); // Label the points label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$P$", P, NNE*2); label("$X$", X, NW); label("$Y$", Y, S); label("$Z$", Z, E); // Add the distances label("$8$", (A+P)/2, NW); label("$6$", (B+P)/2, NE); label("$10$", (C+P)/2, N); // Add right angle marks draw(rightanglemark(C,X,P,15)); draw(rightanglemark(P,B,Z,15)); draw(rightanglemark(A,P,Y,15)); [/asy]
Notice that since $\triangle AXC\cong\triangle APB$ $\triangle BYA\cong\triangle BPC$ , and $\triangle CZB\cong\triangle CPA$ , then
\[[AYBZCX]=2\cdot[ABC]\]
Now the area of the big hexagon is easy to compute since it's comprised of 3 equilateral triangle and 3 right triangles:
\begin{align*}[ABC] &= \frac{1}{2}[AYBZCX] = \frac{1}{2}\left(\underbrace{3\cdot\frac12\cdot6\cdot8}_{\text{3 right triangles}}+\underbrace{\frac{\sqrt{3}}{4}\left(6^2+8^2+10^2\right)}_{\text{3 equilateral triangles}}\right)\\ &= \frac{1}{2}\left(72+\frac{\sqrt{3}}{4}\cdot200)\right) = 36+25\sqrt{3}\\ &\approx \boxed{79} | D | 79 |
51800ec2b52ef308827d818cdebf20bb | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_40 | Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$ $PB=6$ , and $PC=10$ . To the nearest integer the area of triangle $ABC$ is:
$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$ | Let $s$ be the side length of $ABC.$ Notice that $s\le 14$ by the triangle inequality. This means that \[[ABC]\le\dfrac{14^2\sqrt{3}}{2}\approx 84.87.\] This automatically rules out choices $A, B,$ and $C.$ Now, we will look at if the area is $50$ . By the equilateral triangle area formula, $s$ would equal $10\sqrt{\dfrac{2}{\sqrt{3}}}\approx 10.75.$ This is very close to $10.$ If $s=10,$ $\angle APB=90$ and $\angle APC, \angle BPC<90$ by the Pythagorean theorem and Pythagorean inequalities. Thus, \[\angle APB+\angle APC+\angle BPC<270.\] $\angle APB+\angle APC+\angle BPC$ needs to be $360,$ and it probably cannot increase by more than $90$ by just adding $0.75$ to $s$ (more rigorous proof below) Thus, the only viable answer choice is $\boxed{79}.$ | null | 79 |
8eb39dfc4034fb5401144172411d1829 | https://artofproblemsolving.com/wiki/index.php/1966_AHSME_Problems/Problem_12 | The number of real values of $x$ that satisfy the equation \[(2^{6x+3})(4^{3x+6})=8^{4x+5}\] is:
$\text{(A) zero} \qquad \text{(B) one} \qquad \text{(C) two} \qquad \text{(D) three} \qquad \text{(E) greater than 3}$ | We know that $2^{6x+3}\cdot4^{3x+6}=2^{6x+3}\cdot(2^2)^{3x+6}=2^{6x+3}\cdot2^{6x+12}=2^{12x+15}$ .
We also know that $8^{4x+5}=(2^3)^{4x+5}=2^{12x+15}$ .
There are infinite solutions to the equation $2^{12x+15}=2^{12x+15}$ , so the answer is $\boxed{3}$ | E | 3 |
548986a99b68a15e9e521a2f3a34f859 | https://artofproblemsolving.com/wiki/index.php/1966_AHSME_Problems/Problem_25 | If $F(n+1)=\frac{2F(n)+1}{2}$ for $n=1,2,\cdots$ and $F(1)=2$ , then $F(101)$ equals:
$\text{(A) } 49 \quad \text{(B) } 50 \quad \text{(C) } 51 \quad \text{(D) } 52 \quad \text{(E) } 53$ | Notice that $\frac{2F(n)+1}{2}=F(n)+\frac{1}{2}.$
This means that for every single increment $n$ goes up from $1$ $F(n)$ will increase by $\frac{1}{2}.$ Since $101$ is $100$ increments from $1$ $F(n)$ will increase $\frac{1}{2}\times100=50.$
Since $F(1)=2,$ $F(101)$ will equal $2+50=\boxed{52}.$ | D | 52 |
212e80b4f9896d94a6c128f3d1726d38 | https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_1 | The number of real values of $x$ satisfying the equation $2^{2x^2 - 7x + 5} = 1$ is:
$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 3 \qquad \textbf{(E) }\ \text{more than 4}$ | Solution by e_power_pi_times_i
Take the logarithm with a base of $2$ to both sides, resulting in the equation $2x^2-7x+5 = 0$ . Factoring results in $(2x-5)(x-1) = 0$ , so there are $\boxed{2}$ real solutions. | C | 2 |
212e80b4f9896d94a6c128f3d1726d38 | https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_1 | The number of real values of $x$ satisfying the equation $2^{2x^2 - 7x + 5} = 1$ is:
$\textbf{(A)}\ 0 \qquad \textbf{(B) }\ 1 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 3 \qquad \textbf{(E) }\ \text{more than 4}$ | Notice that $a^0=1, a>0$ . So $2^0=1$ . So $2x^2-7x+5=0$ . Evaluating the discriminant, we see that it is equal to $7^2-4*2*5=9$ . So this means that the equation has two real solutions. Therefore, select $\boxed{1}$ | B | 1 |
acf84c5b62ee15e8bea40363393584fa | https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_3 | The expression $(81)^{-2^{-2}}$ has the same value as:
$\textbf{(A)}\ \frac {1}{81} \qquad \textbf{(B) }\ \frac {1}{3} \qquad \textbf{(C) }\ 3 \qquad \textbf{(D) }\ 81\qquad \textbf{(E) }\ 81^4$ | Let us recall $\text{PEMDAS}$ . We calculate the exponent first. $(-2)^{-2}=\frac{1}{(-2)^2}=\frac{1}{4}$ When we substitute, we get $81^{\frac{1}{4}}=\sqrt[4]{81}=\boxed{3}$ | C | 3 |
a8dcdd30722d86f6ca8474c9f556b9ae | https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_5 | When the repeating decimal $0.363636\ldots$ is written in simplest fractional form, the sum of the numerator and denominator is:
$\textbf{(A)}\ 15 \qquad \textbf{(B) }\ 45 \qquad \textbf{(C) }\ 114 \qquad \textbf{(D) }\ 135 \qquad \textbf{(E) }\ 150$ | We let $x=0.\overline{36}$ . Thus, $100x=36.\overline{36}$ . We find that $100x-x=99x=36.\overline{36}-0.\overline{36}=36$ , or $x=\frac{36}{99}=\frac{4}{11}$ . Since $4+11=15$ , the answer is $\boxed{15}$ | A | 15 |
e9e2e710fdb30acc9bb0ef011f7425b9 | https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_9 | The vertex of the parabola $y = x^2 - 8x + c$ will be a point on the $x$ -axis if the value of $c$ is:
$\textbf{(A)}\ - 16 \qquad \textbf{(B) }\ - 4 \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 8 \qquad \textbf{(E) }\ 16$ | Notice that if the vertex of a parabola is on the x-axis, then the x-coordinate of the vertex must be a solution to the quadratic. Since the quadratic is strictly increasing on either side of the vertex, the solution must have double multiplicity, or the quadratic is a perfect square trinomial. This means that for the vertex of $y = x^2 - 8x + c$ to be on the x-axis, the trinomial must be a perfect square, and have discriminant of zero. So,
\begin{align*} 0 &= b^2-4ac\\ 0 &= (-8)^2-4c\\ c &= 64\\ c &= 16\\ \end{align*}
Therefore $c=16$ , and our answer is $\boxed{16}$ | E | 16 |
d69a68eb75828c83b10a93059963b84a | https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_40 | Let $n$ be the number of integer values of $x$ such that $P = x^4 + 6x^3 + 11x^2 + 3x + 31$ is the square of an integer. Then $n$ is:
$\textbf{(A)}\ 4 \qquad \textbf{(B) }\ 3 \qquad \textbf{(C) }\ 2 \qquad \textbf{(D) }\ 1 \qquad \textbf{(E) }\ 0$ | First, we wish to factor $P$ into a more manageable form.
From the beginning of $P$ , we notice $x^4+6x^3$ , which gives us the idea to use $(x^2+3x)^2=x^4+6x^3+9x^2$
This gives us \[P=(x^2+3x)^2+2x^2+3x+31\] This is not useful, but it gives us a place to start from.
We can then try $(x^2+3x+1)=x^4+6x^3+11x^2+6x+1$ \[P=(x^2+3x+1)^2-3(x-10)\] This is much more useful, as it moves all non-linear terms inside of a squared expression.
We can then say $P=(x^2+3x+1)^2-3(x-10)=a^2$ , where $a^2$ is the square of an integer mentioned on the problem. Right from here, we can set $x=10$ , which cancels out the $3(x-10)$ , giving $(100+30+1)^2=a^2$ . This gives us one solution, $x=10$ $a=131$
We can then rearrange the expression, giving us $(x^2+3x+1)^2-a^2=3(x-10)$ . Factoring using difference of squares, we obtain \[(x^2+3x+1+|a|)(x^2+3x+1-|a|)=3(x-10)\]
We can then state that when $x$ is greater than $3$ and less than $-9$ $x^2+3x+1$ will be greater than $|3x-10|$ . This is obtained by setting $x^2+3x+1>|3x-10|$ and then solving the inequality. We can then conclude that $x^2+3x+1+|a|>|3x-10|$
Next, we claim that $x^2+3x+1-|a| \geq 1$ or $x^2+3x+1-|a| \leq -1$ when $x \neq 10$ . We can prove this by first noting that since $x$ and $a$ are integers, $x^2+3x+1-|a|$ is an integer. Next, we shall assume that $x^2+3x+1-|a|=0$ . Solving this and plugging back into the original equation, we obtain $(2|a|)(0)=3(x-10)$ . Solving we obtain $x=10$ , which is a contraction to $x \neq 10$ . Therefore, $x^2+3x+1-|a| \neq 0$ and $x^2+3x+1-|a| \geq 1$ or $x^2+3x+1-|a| \leq -1$
Finally, we can go back to the equation \[(x^2+3x+1+|a|)(x^2+3x+1-|a|)=3(x-10)\] We note that since $(x^2+3x+1+|a|)$ is larger than $3(x-10)$ , in order for there to be solutions, $(x^2+3x+1-|a|)$ must be in the range $(-1,1)$ . However, this contradicts what was proven earlier, so when $x \neq 10$ and $x < -9$ or $x > 3$ , there are no solutions for $x$
Now, all that remains to be checked are values of $x$ between $-9$ and $3$ . Using brute force and checking each value individually, we can assert that there are no such solutions for $x$ , leaving us with only $1$ solution, $x=10$ . Therefore, the answer is $\boxed{1}$ | D | 1 |
5361b49d41ae23c53e3a81fa3477e7f7 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_10 | Given a square side of length $s$ . On a diagonal as base a triangle with three unequal sides is constructed so that its area
equals that of the square. The length of the altitude drawn to the base is:
$\textbf{(A)}\ s\sqrt{2} \qquad \textbf{(B)}\ s/\sqrt{2} \qquad \textbf{(C)}\ 2s \qquad \textbf{(D)}\ 2\sqrt{s} \qquad \textbf{(E)}\ 2/\sqrt{s}$ | The area of the square is $s^2$ . The diagonal of a square with side $s$ bisects the square into two $45-45-90$ right triangles, so the diagonal has length $s\sqrt{2}$
The area of the triangle is $\frac{1}{2}bh$ . The base $b$ of the triangle is the diagonal of the square, which is $b = s\sqrt{2}$ . If the area of the triangle is equal to the area of the square, we have:
$s^2 = \frac{1}{2}bh$
$s^2 = \frac{1}{2}s\sqrt{2}\cdot h$
$s = \frac{\sqrt{2}}{2}h$
$h = \frac{2}{\sqrt{2}}s$
$h = s\sqrt{2}$
This is option $\boxed{2}$ | A | 2 |
58c36628c635bdf5ead2bc24829f6417 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_11 | Given $2^x=8^{y+1}$ and $9^y=3^{x-9}$ , find the value of $x+y$
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 21 \qquad \textbf{(C)}\ 24 \qquad \textbf{(D)}\ 27 \qquad \textbf{(E)}\ 30$ | Since $8^{y + 1} = 2^{3(y+1)}$ and $9^y = 3^{2y}$ , we have:
$2^x = 2^{3(y+1)}$ and $3^{2y} = 3^{x - 9}$
Note that if $a^b = a^c$ , then $b=c$ . Setting the exponents equal gives $x = 3y + 3$ and $2y = x - 9$ . Plugging the first equation into the second equation gives:
$2y = (3y + 3) - 9$
$2y = 3y - 6$
$0 = y - 6$
$y = 6$
Plugging that back in to $x = 3y + 3$ gives $x = 3(6) + 3$ , or $x = 21$ . Thus, $x+y = 21 + 6$ , or $x+y=27$ , which is option $\boxed{27}$ | D | 27 |
643c1bcdfeea3264d30daf5a8c7ef0b8 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_14 | A farmer bought $749$ sheep. He sold $700$ of them for the price paid for the $749$ sheep.
The remaining $49$ sheep were sold at the same price per head as the other $700$ .
Based on the cost, the percent gain on the entire transaction is:
$\textbf{(A)}\ 6.5 \qquad \textbf{(B)}\ 6.75 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 7.5 \qquad \textbf{(E)}\ 8$ | Let us say each sheep cost $x$ dollars. The farmer paid $749x$ for the sheep. He sold $700$ of them for $749x$ , so each sheep sold for $\frac{749}{700} = 1.07x$
Since every sheep sold for the same price per head, and since every sheep cost $x$ and sold for $1.07x$ , there is an increase of $\frac{1.07x - 1x}{1x} = 0.07$ , or $7\%$ , which is option $\boxed{7}$ | C | 7 |
d3d373e38c15d44478e948a861e67def | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_16 | Let $f(x)=x^2+3x+2$ and let $S$ be the set of integers $\{0, 1, 2, \dots , 25 \}$ .
The number of members $s$ of $S$ such that $f(s)$ has remainder zero when divided by $6$ is:
$\textbf{(A)}\ 25\qquad \textbf{(B)}\ 22\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 17$ | Note that for all polynomials $f(x)$ $f(x + 6) \equiv f(x) \pmod 6$
Proof:
If $f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ , then $f(x+6) = a_n(x+6)^n + a_{n-1}(x+6)^{n-1} +...+ a_0$ . In the second equation, we can use the binomial expansion to expand every term, and then subtract off all terms that have a factor of $6^1$ or higher, since subtracting a multiple of $6$ will not change congruence $\pmod 6$ . This leaves $a_nx^n + a_{n-1}x^{n-1} + ... + a_0$ , which is $f(x)$ , so $f(x+6) \equiv f(x) \pmod 6$
So, we only need to test when $f(x)$ has a remainder of $0$ for $0, 1, 2, 3, 4, 5$ . The set of numbers $6, 7, 8, 9, 10, 11$ will repeat remainders, as will all other sets. The remainders are $2, 0, 0, 2, 0, 0$
This means for $s=1, 2, 4, 5$ $f(s)$ is divisible by $6$ . Since $f(1)$ is divisible, so is $f(s)$ for $s=7, 13, 19, 25$ , which is $5$ values of $s$ that work. Since $f(2)$ is divisible, so is $f(s)$ for $s=8, 14, 20$ , which is $4$ more values of $s$ that work. The values of $s=4, 5$ will also generate $4$ solutions each, just like $f(2)$ . This is a total of $17$ values of $s$ , for an answer of $\boxed{17}$ | E | 17 |
d3d373e38c15d44478e948a861e67def | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_16 | Let $f(x)=x^2+3x+2$ and let $S$ be the set of integers $\{0, 1, 2, \dots , 25 \}$ .
The number of members $s$ of $S$ such that $f(s)$ has remainder zero when divided by $6$ is:
$\textbf{(A)}\ 25\qquad \textbf{(B)}\ 22\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 17$ | $f(x)$ $x^2$ $3x$ $2$ is $0$ congruent modulo 6 that implies $x+1$ or/and $x+2$ is $0$ congruent modulo $6$ .The numbers are of the form $6k+5$ and $6k+4$ for some integer $k$ and due to restrictions of number of elements in the set $S$ we get the inequality $k<4$ .Then we calculate the possible combinations to get $17$ as answer i.e. option $\boxed{17}$ | E | 17 |
2b12149748c3c473480531601e492283 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_20 | The sum of the numerical coefficients of all the terms in the expansion of $(x-2y)^{18}$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 19\qquad \textbf{(D)}\ -1\qquad \textbf{(E)}\ -19$ | For any polynomial, even a polynomial with more than one variable, the sum of all the coefficients (including the constant, which is the coefficient of $x^0y^0$ ) is found by setting all variables equal to $1$ . Note that we don't have to worry about whether a constant is a coefficient of an "invisible" $x^0y^0$ term, because there is no such term here.
Setting $x=y=1$ gives $(-1)^{18}$ , which is equal to $1$ , which is answer $\boxed{1}$ | B | 1 |
cd5f51b4b0c817f9a14769532e599fd5 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_23 | Two numbers are such that their difference, their sum, and their product are to one another as $1:7:24$ . The product of the two numbers is:
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 12\qquad \textbf{(C)}\ 24\qquad \textbf{(D)}\ 48\qquad \textbf{(E)}\ 96$ | Set the two numbers as $x$ and $y$ . Therefore, $x+y=7(x-y), xy=24(x-y)$ , and $24(x+y)=7xy$ . Simplifying the first equation gives $y=\frac{3}{4}x$ . Substituting for $y$ in the second equation gives $\frac{3}{4}x^2=6x.$ Solving yields $x=8$ or $x=0$ . Substituting $x=0$ back into the first equation yields $1=-7$ which is false, so $x=0$ is not valid and $x=8$ . Substituting into $y=\frac{3}{4}x$ gives $y=6$ and $xy=\boxed{48}$ | D | 48 |
0092d2f502a0a817269e060dfa0cf4c4 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_28 | The sum of $n$ terms of an arithmetic progression is $153$ , and the common difference is $2$ .
If the first term is an integer, and $n>1$ , then the number of possible values for $n$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5\qquad \textbf{(E)}\ 6$ | Let the progression start at $a$ , have common difference $2$ , and end at $a + 2(n-1)$
The average term is $\frac{a + (a + 2(n-1))}{2}$ , or $a + n - 1$ . Since the number of terms is $n$ , and the sum of the terms is $153$ , we have:
$n(a+n-1) = 153$
Since $n$ is a positive integer, it must be a factor of $153$ . This means $n = 1, 3, 9, 17, 51, 153$ are the only possibilities. We are given $n>1$ , leaving the other five factors.
We now must check if $a$ is an integer. We have $a = \frac{153}{n} + 1 - n$ . If $n$ is a factor of $153$ , then $\frac{153}{n}$ will be an integer. Adding $1-n$ wil keep it an integer.
Thus, there are $5$ possible values for $n$ , which is answer $\boxed{5}$ | D | 5 |
4e80d5a36c3a970e475337dbc224cbfb | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_38 | The sides $PQ$ and $PR$ of triangle $PQR$ are respectively of lengths $4$ inches, and $7$ inches. The median $PM$ is $3\frac{1}{2}$ inches. Then $QR$ , in inches, is:
$\textbf{(A) }6\qquad\textbf{(B) }7\qquad\textbf{(C) }8\qquad\textbf{(D) }9\qquad \textbf{(E) }10$ | By the Median Formula $PM = \frac12\sqrt{2PQ^2+2PR^2-QR^2}$
Plugging in the numbers given in the problem, we get \[\frac72=\frac12\sqrt{2\cdot4^2+2\cdot7^2-QR^2}\]
Solving, \[7=\sqrt{2(16)+2(49)-QR^2}\] \[49=32+98-QR^2\] \[QR^2=81\] \[QR=9=\boxed{9}\] | D | 9 |
b535b3e0164ec912b10a6d68d2d482c6 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_2 | let $n=x-y^{x-y}$ . Find $n$ when $x=2$ and $y=-2$
$\textbf{(A)}\ -14 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 256$ | Substitute the variables to determine the value of $n$ \[n = 2 - (-2)^{2-(-2)}\] \[n = 2 - (-2)^4\] \[n = 2 - 16\] \[n = -14\] The answer is $\boxed{14}$ | A | 14 |
725baca28b4c124879cbe3c29e565e5c | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_9 | In the expansion of $\left(a-\dfrac{1}{\sqrt{a}}\right)^7$ the coefficient of $a^{-\dfrac{1}{2}}$ is:
$\textbf{(A)}\ -7 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ -21 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 35$ | By the Binomial Theorem , each term of the expansion is $\binom{7}{n}(a)^{7-n}(\frac{-1}{\sqrt{a}})^n$
We want the exponent of $a$ to be $-\frac{1}{2}$ , so \[(7-n)-\frac{1}{2}n=-\frac{1}{2}\] \[-\frac{3}{2}n = -\frac{15}{2}\] \[n = 5\]
If $n=5$ , then the corresponding term is \[\binom{7}{5}(a)^{2}(\frac{-1}{\sqrt{a}})^5\] \[-21a^{-\frac{1}{2}}\]
The answer is $\boxed{21}$ | C | 21 |
7c2327d72e4a4f67b6766e25e98a9c1c | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_11 | The arithmetic mean of a set of $50$ numbers is $38$ . If two numbers of the set, namely $45$ and $55$ , are discarded,
the arithmetic mean of the remaining set of numbers is:
$\textbf{(A)}\ 38.5 \qquad \textbf{(B)}\ 37.5 \qquad \textbf{(C)}\ 37 \qquad \textbf{(D)}\ 36.5 \qquad \textbf{(E)}\ 36$ | If the arithmetic mean of a set of $50$ numbers is $38$ , then the sum of the $50$ numbers equals $1900$ . Since $45$ and $55$ are being removed, subtract $100$ to get the sum of the remaining $48$ numbers, which is $1800$ . Therefore, the new mean is $37.5$ , which is answer choice $\boxed{37.5}$ | B | 37.5 |
e0b68fbe2e57407eac60daeecaabd47d | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_12 | Three vertices of parallelogram $PQRS$ are $P(-3,-2), Q(1,-5), R(9,1)$ with $P$ and $R$ diagonally opposite.
The sum of the coordinates of vertex $S$ is:
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 9$ | [asy] import graph; size(7.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.2,xmax=10.2,ymin=-6.2,ymax=5.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype("2 2"); real gx=1,gy=1; for(real i=ceil(xmin/gx)*gx;i<=floor(xmax/gx)*gx;i+=gx) draw((i,ymin)--(i,ymax),gs); for(real i=ceil(ymin/gy)*gy;i<=floor(ymax/gy)*gy;i+=gy) draw((xmin,i)--(xmax,i),gs); Label laxis; laxis.p=fontsize(10); xaxis(xmin,xmax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); yaxis(ymin,ymax,defaultpen+black,Ticks(laxis,Step=1.0,Size=2,NoZero),Arrows(6),above=true); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); dot((-3,-2)); label("P",(-3,-2),NW); dot((1,-5)); label("Q",(1,-5),NW); dot((9,1)); label("R",(9,1),NW); [/asy] Graph the three points on the coordinate grid. Noting that the opposite sides of a parallelogram are congruent and parallel, count boxes to find out that point $S$ is on $(5,4)$ . The sum of the x-coordinates and y-coordinates is $9$ , so the answer is $\boxed{9}$ | E | 9 |
855e533a3649b5978ce30d5a11229e5d | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_13 | If $2^a+2^b=3^c+3^d$ , the number of integers $a,b,c,d$ which can possibly be negative, is, at most:
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 0$ | Assume $c,d \ge 0$ , and WLOG, assume $a<0$ and $a \le b$ . This also takes into account when $b$ is negative. That means \[\frac{1}{2^{-a}} + 2^b = 3^c + 3^d\] Multiply both sides by $2^{-a}$ to get \[1 + 2^{-a+b} = 2^{-a} (3^c + 3^d)\] Note that both sides are integers. If $a \ne b$ , then the right side is even while the left side is odd, so equality can not happen. If $a = b$ , then $2^{-a} (3^c + 3^d) = 2$ , and since $a<0$ $a = -1$ and $3^c + 3^d = 1$ . No nonnegative value of $c$ and $d$ works, so equality can not happen. Thus, $a$ and $b$ can not be negative when $c,d \ge 0$
Assume $a,b \ge 0$ , and WLOG, assume $c < 0$ and $c \le d$ . This also takes into account when $d$ is negative. That means \[2^a + 2^b = \frac{1}{3^{-c}} + 3^d\] Multiply both sides by $3^{-c}$ to get \[3^{-c} (2^a + 2^b) = 1 + 3^{d-c}\] That makes both sides integers. The left side is congruent to $0$ modulo $3$ while the right side is congruent to $1$ or $2$ modulo $3$ , so equality can not happen. Thus, $c$ and $d$ can not be negative when $a,b \ge 0$
Assume $a,c < 0$ , and WLOG, let $a \le b$ and $c \le d$ . This also takes into account when $b$ or $d$ is negative. That means \[\frac{1}{2^{-a}} + 2^b = \frac{1}{3^{-c}} + 3^d\] Multiply both sides by $2^{-a} \cdot 3^{-c}$ to get \[3^{-c} (1 + 2^{b-a}) = 2^{-a} (1 + 3^{d-c})\] That makes both sides integers. The left side is congruent to $0$ modulo $3$ while the right side is congruent to $1$ or $2$ modulo $3$ , so equality can not happen. Thus, $a$ and $c$ can not be negative.
Putting all the information together, none of $a,b,c,d$ can be negative, so the answer is $\boxed{0}$ | E | 0 |
bf4ce210cd84a0a3ea1ab5431bed3554 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_16 | Three numbers $a,b,c$ , none zero, form an arithmetic progression. Increasing $a$ by $1$ or increasing $c$ by $2$ results
in a geometric progression. Then $b$ equals:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$ | Let $d$ be the common difference of the arithmetic sequence , so $a = b-d$ and $c = b+d$
Since increasing $a$ by $1$ or $c$ by $2$ results in a geometric sequence \[\frac{b}{b-d+1} = \frac{b+d}{b}\] \[\frac{b}{b-d} = \frac{b+d+2}{b}\] Cross-multiply in both equations to get a system of equations \[b^2 = b^2 - d^2 + b + d\] \[b^2 = b^2 - d^2 + 2b - 2d\] Rearranging terms results in \[d^2 = b+d\] \[d^2 = 2b-2d\] Substitute and solve for $d$ \[b+d = 2b-2d\] \[d = \frac{b}{3}\] Finally, substitute $d$ and solve for $b$ . Since $b \ne 0$ , dividing by $b$ is allowed. \[(\frac{b}{3})^2 = b + \frac{b}{3}\] \[\frac{b^2}{9} = \frac{4b}{3}\] \[\frac{b}{9} = \frac{4}{3}\] \[b = 12\] The answer is $\boxed{12}$ | C | 12 |
3be3edcb70f15975b810265c50650053 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_19 | In counting $n$ colored balls, some red and some black, it was found that $49$ of the first $50$ counted were red.
Thereafter, $7$ out of every $8$ counted were red. If, in all, $90$ % or more of the balls counted were red, the maximum value of $n$ is:
$\textbf{(A)}\ 225 \qquad \textbf{(B)}\ 210 \qquad \textbf{(C)}\ 200 \qquad \textbf{(D)}\ 180 \qquad \textbf{(E)}\ 175$ | The desired percentage of red balls is more than $90$ percent, so write an inequality
\[\frac{49+7x}{50+8x} \ge 0.9\]
Since $x >0$ , the sign does not need to be swapped after multiplying both sides by $50+8x$
\[49+7x \ge 45+7.2x\] \[4 \ge 0.2x\] \[20 \ge x\]
Thus, up to $20$ batches of balls can be used, so a total of $20 \cdot 8 + 50 = 210$ balls can be counted while satisfying the requirements. The answer is $\boxed{210}$ | B | 210 |
d3093a566a85ea3baee4fc8354535d29 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_20 | Two men at points $R$ and $S$ $76$ miles apart, set out at the same time to walk towards each other.
The man at $R$ walks uniformly at the rate of $4\tfrac{1}{2}$ miles per hour; the man at $S$ walks at the constant
rate of $3\tfrac{1}{4}$ miles per hour for the first hour, at $3\tfrac{3}{4}$ miles per hour for the second hour,
and so on, in arithmetic progression. If the men meet $x$ miles nearer $R$ than $S$ in an integral number of hours, then $x$ is:
$\textbf{(A)}\ 10 \qquad \textbf{(B)}\ 8 \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 2$ | First, find the number of hours it takes for the two to meet together. After $h$ hours, the person at $R$ walks $4.5h$ miles. In the same amount of time, the person at $S$ has been walking at $3.25+0.5(h-1)$ mph for the past hour, so the person walks $\frac{h(6.5+0.5(h-1))}{2}$ miles.
In order for both to meet, the sum of both of the distances walked must total $76$ miles, so \[4.5h + \frac{h(6+0.5h)}{2} = 76\] \[4.5h + 3h + 0.25h^2 = 76\] \[0.25h^2 + 7.5h - 76 = 0\] \[h^2 + 30h - 304 = 0\] \[(h + 38)(h - 8) = 0\] Since $h$ must be positive, $h = 8$ . Because it takes $8$ hours to meet, the person from $R$ traveled $36$ miles while the person from $S$ traveled $40$ miles. Thus, they are $4$ miles closer to $R$ than $S$ , so the answer is $\boxed{4}$ | D | 4 |
ccec92e370d78617c5a23276bc7d8bc2 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_23 | A gives $B$ as many cents as $B$ has and $C$ as many cents as $C$ has. Similarly, $B$ then gives $A$ and $C$ as many cents as each then has. $C$ , similarly, then gives $A$ and $B$ as many cents as each then has. If each finally has $16$ cents, with how many cents does $A$ start?
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 26\qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 32$ | Let $a$ be number of cents $A$ originally had, $b$ be number of cents $B$ originally had, and $c$ be number of cents $C$ originally had.
After $A$ gave his money away, $A$ has $a-b-c$ cents, $B$ has $2b$ cents, and $C$ has $2c$ cents.
After $B$ gave his money away, $A$ has $2a-2b-2c$ cents, $B$ has $-a+3b-c$ cents, and $C$ has $4c$ cents.
After $C$ gave his money away, $A$ has $4a-4b-4c$ cents, $B$ has $-2a+6b-2c$ cents, and $C$ has $-a-b+7c$ cents.
Since all of them have $16$ cents in the end, we can write a system of equations \[4a-4b-4c=16\] \[-2a+6b-2c=16\] \[-a-b+7c=16\] Note that adding the three equation yields $a+b+c=48$ , so $4a+4b+4c=192$ . Therefore, $8a=208$ , so $a = 26$ . Solving for $a$ can also be done traditionally.
Thus, $A$ started out with $26$ cents, which is answer choice $\boxed{26}$ | B | 26 |
ccec92e370d78617c5a23276bc7d8bc2 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_23 | A gives $B$ as many cents as $B$ has and $C$ as many cents as $C$ has. Similarly, $B$ then gives $A$ and $C$ as many cents as each then has. $C$ , similarly, then gives $A$ and $B$ as many cents as each then has. If each finally has $16$ cents, with how many cents does $A$ start?
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 26\qquad \textbf{(C)}\ 28 \qquad \textbf{(D)}\ 30 \qquad \textbf{(E)}\ 32$ | We know that people not giving away money on the previous turn now have twice what they previously had. Using the fact that the sum of their money is $48$ cents, we can work backward \[16,16,16\] \[8,8,32\] \[4,28,16\] \[26,14,8\] Thus at the beginning $A$ has $26\boxed{26}$ cents. | B | 26 |
985ab119b03fbbafb731dfbebadab273 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_24 | Consider equations of the form $x^2 + bx + c = 0$ . How many such equations have real roots and have coefficients $b$ and $c$ selected
from the set of integers $\{1,2,3, 4, 5,6\}$
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 19 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 16$ | The discriminant of the quadratic is $b^2 - 4c$ . Since the quadratic has real roots, \[b^2 - 4c \ge 0\] \[b^2 \ge 4c\] If $b = 6$ , then $c$ can be from $1$ to $6$ . If $b = 5$ , then $c$ can also be from $1$ to $6$ . If $b=4$ , then $c$ can be from $1$ to $4$ . If $b=3$ , then $c$ can be $1$ or $2$ . If $b=2$ , then $c$ can only be $1$ . If $b = 1$ , no values of $c$ in the set would work.
Thus, there are a total of $19$ equations that work. The answer is $\boxed{19}$ | B | 19 |
fcb55b06b1816b55c027dd035c2683fe | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_25 | Point $F$ is taken in side $AD$ of square $ABCD$ . At $C$ a perpendicular is drawn to $CF$ , meeting $AB$ extended at $E$ .
The area of $ABCD$ is $256$ square inches and the area of $\triangle CEF$ is $200$ square inches. Then the number of inches in $BE$ is:
[asy] size(6cm); pair A = (0, 0), B = (1, 0), C = (1, 1), D = (0, 1), E = (1.3, 0), F = (0, 0.7); draw(A--B--C--D--cycle); draw(F--C--E--B); label("$A$", A, SW); label("$B$", B, S); label("$C$", C, N); label("$D$", D, NW); label("$E$", E, SE); label("$F$", F, W); //Credit to MSTang for the asymptote[/asy]
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 14 \qquad \textbf{(C)}\ 15 \qquad \textbf{(D)}\ 16 \qquad \textbf{(E)}\ 20$ | Because $ABCD$ is a square $DC = CB = 16$ $DC \perp DA$ , and $CB \perp BA$ . Also, because $\angle DCF + \angle FCB = \angle FCB + \angle BCE = 90^\circ$ $\angle DCF = \angle BCE$ . Thus, by ASA Congruency, $\triangle DCF \cong \triangle BCF$
From the congruency, $CF = CE$ . Using the area formula for a triangle, $CE = 20$ . Finally, by the Pythagorean Theorem $BE = \sqrt{20^2 - 16^2} = 12$ , which is answer choice $\boxed{12}$ | A | 12 |
73fa6ab7224b0884b408655b68263299 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_26 | Consider the statements:
$\textbf{(1)}\ p\text{ }\wedge\sim q\wedge r\qquad\textbf{(2)}\ \sim p\text{ }\wedge\sim q\wedge r\qquad\textbf{(3)}\ p\text{ }\wedge\sim q\text{ }\wedge\sim r\qquad\textbf{(4)}\ \sim p\text{ }\wedge q\text{ }\wedge r$
where $p,q$ , and $r$ are propositions. How many of these imply the truth of $(p\rightarrow q)\rightarrow r$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | Statement $1$ states that $p$ is true and $q$ is false. Therefore, $p \rightarrow q$ is false, because a premise being true and a conclusion being false is, itself, false. This means that $(p \rightarrow q) \rightarrow X$ , where $X$ is any logical statement (or series of logical statements) must be true - if your premise is false, then the implication is automatically true. So statement $1$ implies the truth of the given statement.
Statement $3$ similarly has $p$ as true and $q$ is false, so it also implies the truth of the given statement.
Statement $2$ states that $p$ and $q$ are both false. This in turn means that $p \rightarrow q$ is true. Since $r$ is also true from statement $2$ , this means that $(p \rightarrow q) \rightarrow r$ is true, since $T \rightarrow T$ is $T$ . Thus statement $2$ implies the truth of the given statement.
Statement $4$ states that $p$ is false and $q$ is true. In this case, $p \rightarrow q$ is true - your conclusion can be true even if your premise is false. And, since $r$ is also true from statement $4$ , this means $(p \rightarrow q) \rightarrow r$ is true. Thus, statement $4$ implies the truth of the given statement.
All four statements imply the truth of the given statement, so the answer is $\boxed{4}$ | E | 4 |
72ea8ffbf997ec780133410e0f59cbae | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_27 | Six straight lines are drawn in a plane with no two parallel and no three concurrent. The number of regions into which they divide the plane is:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$ | The first line divides the plane into two regions. The second line intersects one line, creating two regions. The third line intersects two lines, creating three regions. Similarly, the fourth line intersects three lines and creates four regions, the fifth line intersects four lines and creates five regions, and the sixth line intersects five lines and creates six regions.
Totaling the regions created results in $2 + 2 + 3 + 4 + 5 + 6 = 22$ regions, which is answer choice $\boxed{22}$ | C | 22 |
72ea8ffbf997ec780133410e0f59cbae | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_27 | Six straight lines are drawn in a plane with no two parallel and no three concurrent. The number of regions into which they divide the plane is:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$ | 1963 AHSME Problem 27.png
With careful drawing, one can draw all six lines and count the regions. There are $22$ regions in total, which is answer choice $\boxed{22}$ | C | 22 |
72ea8ffbf997ec780133410e0f59cbae | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_27 | Six straight lines are drawn in a plane with no two parallel and no three concurrent. The number of regions into which they divide the plane is:
$\textbf{(A)}\ 16 \qquad \textbf{(B)}\ 20\qquad \textbf{(C)}\ 22 \qquad \textbf{(D)}\ 24 \qquad \textbf{(E)}\ 26$ | We can use the fact that the number of regions that $n$ lines divide a plane is given by the equation $L_n = \frac{n^2 + n +2}{2}$ , and in this problems, $n=6$ , from which the answer is $\boxed{22}$ | null | 22 |
92bf52271840482cbfc663632c07aea3 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_29 | A particle projected vertically upward reaches, at the end of $t$ seconds, an elevation of $s$ feet where $s = 160 t - 16t^2$ . The highest elevation is:
$\textbf{(A)}\ 800 \qquad \textbf{(B)}\ 640\qquad \textbf{(C)}\ 400 \qquad \textbf{(D)}\ 320 \qquad \textbf{(E)}\ 160$ | The highest elevation a particle can reach is the vertex of the quadratic. The x-value that can get the maximum is $\frac{-160}{-2 \cdot 16} = 5$ , so the highest elevation is $160(5) - 16(5^2) = 400$ feet, which is answer choice $\boxed{400}$ | C | 400 |
7a99bc1192d4487d25440fde8998380b | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_31 | The number of solutions in positive integers of $2x+3y=763$ is:
$\textbf{(A)}\ 255 \qquad \textbf{(B)}\ 254\qquad \textbf{(C)}\ 128 \qquad \textbf{(D)}\ 127 \qquad \textbf{(E)}\ 0$ | Solving for $x$ in the equation yields $x =rfthe meaning of theta 0$ . Solving the inequality results in $y \le 254 \frac{1}{3}$ . From the two conditions, $y$ can be an odd number from $1$ to $253$ , so there are $127$ solutions where $x$ and $y$ are integers. The answer is $\boxed{127}$ | D | 127 |
7a99bc1192d4487d25440fde8998380b | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_31 | The number of solutions in positive integers of $2x+3y=763$ is:
$\textbf{(A)}\ 255 \qquad \textbf{(B)}\ 254\qquad \textbf{(C)}\ 128 \qquad \textbf{(D)}\ 127 \qquad \textbf{(E)}\ 0$ | We can solve for first solution by applying extended euclids division algorithm or we can apply hit and trial for the first solution to get $x_0$ $380$ and $y_0$ $1$ . then the general solution of the given diophanitine equation will be $x$ $x_0$ $3t$ and $y$ $y_0$ $2t$ . Since we need only positive integer solutions So we solve $380$ $3t$ $>$ $0$ and $1$ $2t$ $>$ $0$ to get $t$ $>$ $0$ (applying Greatest integer function) also we can clearly see that $t_{(min)}$ $=$ $0$ so,t $<$ $GIF$ $383$ $3$ ). That implies $t$ ranges from $0$ to $127$ . Hence,the correct answer is $127$ $\boxed{127}$ | D | 127 |
5bd11d14152ba5421f7ef2ce93f3345b | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_32 | The dimensions of a rectangle $R$ are $a$ and $b$ $a < b$ . It is required to obtain a rectangle with dimensions $x$ and $y$ $x < a, y < a$ ,
so that its perimeter is one-third that of $R$ , and its area is one-third that of $R$ . The number of such (different) rectangles is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ \infty$ | Using the perimeter and area formulas, \[2(x+y) = \frac{2}{3}(a+b)\] \[x+y = \frac{a+b}{3}\] \[xy = \frac{ab}{3}\] Dividing the second equation by the last equation results in \[\frac1y + \frac1x = \frac1b + \frac1a\] Since $x,y < a$ $\tfrac1a < \tfrac1x, \tfrac1y$ . Since $a < b$ $\tfrac1b < \tfrac1a$ . That means \[\tfrac1x + \tfrac1y > \tfrac1a + \tfrac1a > \tfrac1a + \tfrac1b\] This is a contradiction, so there are $\boxed{0}$ rectangles that satisfy the conditions. | A | 0 |
d2863fcb6a5e38a8981ce5afc0df14aa | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_34 | In $\triangle ABC$ , side $a = \sqrt{3}$ , side $b = \sqrt{3}$ , and side $c > 3$ . Let $x$ be the largest number such that the magnitude,
in degrees, of the angle opposite side $c$ exceeds $x$ . Then $x$ equals:
$\textbf{(A)}\ 150^{\circ} \qquad \textbf{(B)}\ 120^{\circ}\qquad \textbf{(C)}\ 105^{\circ} \qquad \textbf{(D)}\ 90^{\circ} \qquad \textbf{(E)}\ 60^{\circ}$ | Using the Law of Cosines \[\sqrt{3 + 3 - 2\cdot 3 \cdot \cos{x^\circ}}>3\]
Both sides are positive, so squaring both sides will not affect the inequality.
\[6 - 6 \cos{x^\circ}>9\] \[\cos{x^\circ} < -\frac{1}{2}\]
Note that $\cos{120^\circ} = -\frac{1}{2}$ . As $x$ gets closer to $180^{\circ}$ $\cos{x}$ decreases towards $-1$ . Thus, $x > 120$ , so the answer is $\boxed{120}$ | B | 120 |
32d3d72749fc3f382798eb638e296909 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_35 | The lengths of the sides of a triangle are integers, and its area is also an integer.
One side is $21$ and the perimeter is $48$ . The shortest side is:
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 16$ | Let $b$ and $c$ be the other two sides of the triangle. The perimeter of the triangle is $48$ units, so $c = 27-b$ and the semiperimeter equals $24$ units.
By Heron's Formula , the area of the triangle is $\sqrt{24 \cdot 3(24-b)(b-3)}$ . Plug in the answer choices for $b$ and write the prime factorization of the product to make sure it is a perfect square.
Testing $b = 8$ results in the area being $\sqrt{6 \cdot 4 \cdot 3 \cdot 16 \cdot 5} = \sqrt{2^7 \cdot 3^2 \cdot 5}$ , so $8$ does not work. However, testing $b = 10$ results in the area being $\sqrt{6 \cdot 4 \cdot 3 \cdot 14 \cdot 7} = \sqrt{2^4 \cdot 3^2 \cdot 7^2}$ , so $10$ works. The third side is $17$ , and the sides satisfy the Triangle Inequality , so the answer is $\boxed{10}$ | B | 10 |
93b83770742e3fda2e01323a6030eafe | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_38 | Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD$ $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G$ .
If $EF = 32$ and $GF = 24$ , then $BE$ equals:
[asy] size(7cm); pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5); pair G = intersectionpoints(B--F, D--C)[0]; pair E = intersectionpoints(A--C, B--F)[0]; draw(A--D--C--B--cycle); draw(A--C); draw(D--F--B); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, NE); label("$D$", D, NW); label("$F$", F, N); label("$G$", G, NE); label("$E$", E, SE); //Credit to MSTang for the asymptote[/asy]
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 10 \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 16$ | Let $BE = x$ and $BC = y$ . Since $AF \parallel BC$ , by AA Similarity, $\triangle AFE \sim \triangle CBE$ . That means $\frac{AF}{CB} = \frac{FE}{BE}$ . Substituting in values results in \[\frac{AF}{y} = \frac{32}{x}\] Thus, $AF = \frac{32y}{x}$ , so $FD = \frac{32y - xy}{x}$
In addition, $DC \parallel AB$ , so by AA Similarity, $\triangle FDG = \triangle FAB$ . That means \[\frac{\frac{32y-xy}{x}}{\frac{32y}{x}} = \frac{24}{x+32}\] Cross multiply to get \[\frac{y(32-x)}{x} (x+32) = \frac{32y}{x} \cdot 24\] Since $x \ne 0$ and $y \ne 0$ \[(32-x)(32+x) = 32 \cdot 24\] \[32^2 - x^2 = 32 \cdot 24\] \[32 \cdot 8 = x^2\] Thus, $x = 16$ , which is answer choice $\boxed{16}$ | E | 16 |
bcf9bf25c27b5accae9d1a9f55854011 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_3 | The first three terms of an arithmetic progression are $x - 1, x + 1, 2x + 3$ , in the order shown. The value of $x$ is:
$\textbf{(A)}\ -2\qquad\textbf{(B)}\ 0\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{undetermined}$ | Let $y$ represent the common difference between the terms. We have $(x+1)-y=(x-1)\implies y=2$
Substituting gives us $(2x+3)-2=(x+1)\implies 2x+1=x+1\implies x=0$
Therefore, our answer is $\boxed{0}$ | B | 0 |
ce0e9752f1efd1a5251a228c26ef9762 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_9 | When $x^9-x$ is factored as completely as possible into polynomials and monomials with integral coefficients, the number of factors is:
$\textbf{(A)}\ \text{more than 5}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ 2$ | Obviously, we can factor out an $x$ first to get $x(x^8-1)$ .
Next, we repeatedly factor differences of squares: \[x(x^4+1)(x^4-1)\] \[x(x^4+1)(x^2+1)(x^2-1)\] \[x(x^4+1)(x^2+1)(x+1)(x-1)\] None of these 5 factors can be factored further, so the answer is $\boxed{5}$ | B | 5 |
fadf28bd4298b88ffba29daf42254f61 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_12 | When $\left ( 1 - \frac{1}{a} \right ) ^6$ is expanded the sum of the last three coefficients is:
$\textbf{(A)}\ 22\qquad\textbf{(B)}\ 11\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ -10\qquad\textbf{(E)}\ -11$ | This is equivalent to $\frac{(a-1)^6}{a^6}.$ Its expansion has 7 terms, whose coefficients are the same as those of $(a-1)^6$ .
By the Binomial Theorem, the sum of the last three coefficients is $\binom{6}{2}-\binom{6}{1}+\binom{6}{0}=15-6+1=\boxed{10}$ | C | 10 |
e056fc8ef482145ffeef18182c29f6e2 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_13 | $R$ varies directly as $S$ and inversely as $T$ . When $R = \frac{4}{3}$ and $T = \frac {9}{14}$ $S = \frac37$ . Find $S$ when $R = \sqrt {48}$ and $T = \sqrt {75}$
$\textbf{(A)}\ 28\qquad\textbf{(B)}\ 30\qquad\textbf{(C)}\ 40\qquad\textbf{(D)}\ 42\qquad\textbf{(E)}\ 60$ | \[R=c\cdot\frac{S}T\]
for some constant $c$
You know that
\[\frac43=c\cdot\frac{3/7}{9/14}=c\cdot\frac37\cdot\frac{14}9=c\cdot\frac23\,,\]
so
\[c=\frac{4/3}{2/3}=2\,.\]
When $R=\sqrt{48}$ and $T=\sqrt{75}$ we have
\[\sqrt{48}=\frac{2S}{\sqrt{75}}\,,\]
so
\[S=\frac12\sqrt{48\cdot75}=30\,.\] $\boxed{30}$ | B | 30 |
60de9b60dbdbd1ce9c848a3f1f886d0e | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_14 | Let $s$ be the limiting sum of the geometric series $4- \frac83 + \frac{16}{9} - \dots$ , as the number of terms increases without bound. Then $s$ equals:
$\textbf{(A)}\ \text{a number between 0 and 1}\qquad\textbf{(B)}\ 2.4\qquad\textbf{(C)}\ 2.5\qquad\textbf{(D)}\ 3.6\qquad\textbf{(E)}\ 12$ | The infinite sum of a geometric series with first term $a$ and common ratio $r$ $-1<r<1$ ) is $\frac{a}{1-r}$ .
Now, in this geometric series, $a=4$ , and $r=-\frac23$ . Plugging these into the formula, we get $\frac4{1-(-\frac23)}$ , which simplifies to $\frac{12}5$ , or $\boxed{2.4}$ | B | 2.4 |
26b15bf2617138b9e854ec8a5b1f5bf7 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_19 | If the parabola $y = ax^2 + bx + c$ passes through the points $( - 1, 12)$ $(0, 5)$ , and $(2, - 3)$ , the value of $a + b + c$ is:
$\textbf{(A)}\ -4\qquad\textbf{(B)}\ -2\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 1\qquad\textbf{(E)}\ 2$ | Substituting in the $(x, y)$ pairs gives the following system of equations: \[a-b+c=12\] \[c=5\] \[4a+2b+c=-3\] We know $c=5$ , so plugging this in reduces the system to two variables: \[a-b=7\] \[4a+2b=-8\] Dividing the second equation by 2 gives $2a+b=-4$ , which can be added to the first equation to get $3a=3$ , or $a=1$ . So the solution set is $(1, -6, 5)$ , and the sum is $\boxed{0}$ | C | 0 |
e28b3c69e545683d8f6b2fb6e2792f72 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_20 | The angles of a pentagon are in arithmetic progression. One of the angles in degrees, must be:
$\textbf{(A)}\ 108\qquad\textbf{(B)}\ 90\qquad\textbf{(C)}\ 72\qquad\textbf{(D)}\ 54\qquad\textbf{(E)}\ 36$ | If the angles are in an arithmetic progression, they can be expressed as $a$ $a+n$ $a+2n$ $a+3n$ , and $a+4n$ for some real numbers $a$ and $n$ .
Now we know that the sum of the degree measures of the angles of a pentagon is $180(5-2)=540$ .
Adding our expressions for the five angles together, we get $5a+10n=540$ .
We now divide by 5 to get $a+2n=108$ . It so happens that $a+2n$ is one of the angles we defined earlier, so that angle must have a measure of $\boxed{108}$ .
(In fact, for any arithmetic progression with an odd number of terms,
the middle term is equal to the average of all the terms.) | A | 108 |
40c2a9b048c1b2919f3a8ae01650af0a | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_21 | It is given that one root of $2x^2 + rx + s = 0$ , with $r$ and $s$ real numbers, is $3+2i (i = \sqrt{-1})$ . The value of $s$ is:
$\textbf{(A)}\ \text{undetermined}\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ -13\qquad\textbf{(E)}\ 26$ | If a quadratic with real coefficients has two non-real roots, the two roots must be complex conjugates of one another.
This means the other root of the given quadratic is $\overline{3+2i}=3-2i$ .
Now Vieta's formulas say that $s/2$ is equal to the product of the two roots, so $s = 2(3+2i)(3-2i) = \boxed{26}$ | E | 26 |
ce88ae4bc3a7b036f78da9d7f4d381a4 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_24 | Three machines $\text{P, Q, and R,}$ working together, can do a job in $x$ hours. When working alone, $\text{P}$ needs an additional $6$ hours to do the job; $\text{Q}$ , one additional hour; and $R$ $x$ additional hours. The value of $x$ is:
$\textbf{(A)}\ \frac{2}3\qquad\textbf{(B)}\ \frac{11}{12}\qquad\textbf{(C)}\ \frac{3}2\qquad\textbf{(D)}\ 2\qquad\textbf{(E)}\ 3$ | Machine P takes $x+6$ hours, machine Q takes $x+1$ hours, and machine R takes $2x$ hours.
We also know that all three working together take $x$ hours.
Now the time it takes for all three machines to complete the job is the harmonic mean of the times the three machines take individually; that is, \[x=\frac1{\frac1{x+6}+\frac1{x+1}+\frac1{2x}}\] \[\frac{x}{x+6}+\frac{x}{x+1}+\frac{x}{2x}=1\] \[\frac{x}{x+6}+\frac{x}{x+1}=\frac12\] \[2x(x+1)+2x(x+6)=(x+1)(x+6)\] \[2x^2+2x+2x^2+12x=x^2+7x+6\] \[3x^2+7x-6=0\] \[(3x-2)(x+3)=0\] \[x\in\{\frac23, -3\}\] Obviously, the number of hours is positive, so the answer is $\boxed{23}$ | A | 23 |
993ea3600fa02254048d0ef4a0d8075d | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_25 | Given square $ABCD$ with side $8$ feet. A circle is drawn through vertices $A$ and $D$ and tangent to side $BC$ . The radius of the circle, in feet, is:
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 4\sqrt{2}\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5\sqrt{2}\qquad\textbf{(E)}\ 6$ | Let $O$ be the center of the circle and $E$ be the point of tangency of the circle and $BC$ and let $F$ be the point of intersection of lines $OE$ and $AD$ Because of the symmetry, $BE=EC=AF=FD=4$ feet. Let the length of $OF$ be $x$ . The length of $OE$ is $EF-OF=-x+8$ . By Pythagorean Theorem, $OA=OD=\sqrt{x^2+4^2}=\sqrt{x^2+16}$ . Because $OA$ $OD$ , and $OE$ are radii of the same circle, $-x+8=OE=OA=AD=\sqrt{x^2+16}$ . So, $\sqrt{x^2+16}=-x+8$ . Squaring both sides, we obtain $x^2+16=x^2-16x+64$ . Subtracting $x^2+16$ from both sides and adding $16x$ , our equation becomes $16x=80$ , so our answer is $x=\boxed{5}$ | C | 5 |
859abe90d629c8ce96af522d81e46877 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_31 | The ratio of the interior angles of two regular polygons with sides of unit length is $3: 2$ . How many such pairs are there?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ \text{infinitely many}$ | The formula for the measure of the interior angle of a regular polygon with $n$ -sides is $180 - \frac{360}{n}$ . Letting our two polygons have side length $r$ and $k$ , we have that the ratio of the interior angles is $\frac{180 - \frac{360}{r}}{180 - \frac{360}{k}} = \frac{(r-2) \cdot k}{(k-2) \cdot r} = \frac{3}{2}$ . Cross multiplying both sides, we have $2rk-4k = 3kr-6r \Rightarrow -rk-4k+6r = 0$ . Using Simon's Favorite Factoring Trick, we have $-k(r+4)+6r+24 = 24 \Rightarrow (r+4)(6-k)=24$ . Because $k$ and $r$ are both more than $2$ , we know that $6-k < r+4$ . Now, we just set these factors equal to the factors of 24. We can set $6-k$ to $1$ $2$ $3$ , or $4$ and $r+4$ to $24$ $12$ $8$ , or $6$ respectively to get the following pairs for $(k, r)$ $(5, 20)$ $(4, 8)$ $(3, 4)$ , and $(2, 2)$ . However, we have to take out the solution with $k = 2$ , because $k$ and $r$ are both more than $2$ , leaving us with $\boxed{3}$ as the correct answer. | C | 3 |
1396c741a259ac5a22c69ae247c2802c | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_35 | A man on his way to dinner short after $6: 00$ p.m. observes that the hands of his watch form an angle of $110^{\circ}$ . Returning before $7: 00$ p.m. he notices that again the hands of his watch form an angle of $110^{\circ}$ . The number of minutes that he has been away is:
$\textbf{(A)}\ 36\frac{2}3\qquad\textbf{(B)}\ 40\qquad\textbf{(C)}\ 42\qquad\textbf{(D)}\ 42.4\qquad\textbf{(E)}\ 45$ | Let $n$ be the number of minutes after 6:00. Let $h(n)=180+\frac{n}2$ be the angle, in degrees, of the hour hand (with $0^{\circ}$ at the top and increasing in the clockwise direction); similarly, let $m(n)=6n$ be the angle of the minute hand. We want $|h(n)-m(n)|=110$ . This is equivalent to \[180-\frac{11n}2=\pm110\] \[-\frac{11n}2\in\{-70,-290\}\] \[\frac{11n}2\in\{70,290\}\] \[11n\in\{140,580\}\] \[n\in\{\frac{140}{11},\frac{580}{11}\}\] The difference between the two values of $n$ is $\frac{440}{11}=\boxed{40}$ | B | 40 |
11d50bab727fe4a3d21b438a6f464ed0 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_36 | If both $x$ and $y$ are both integers, how many pairs of solutions are there of the equation $(x-8)(x-10) = 2^y$
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \text{more than 3}$ | The equality implies $x-8$ and $x-10$ are both powers of two; since they differ by two, it must be the case that $(x-8,x-10) = (4,2)$ or $(x-8,x-10) = (-2,-4)$ . (Note that $(1,-1)$ is not allowed because then the product is negative.) These yield $(x,y) = (12,3)$ or $(x,y) = (6,3)$ , for a total of $\boxed{2}$ solutions. | C | 2 |
21f540ec14c54018ed1046b15ed17d8e | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_38 | The population of Nosuch Junction at one time was a perfect square. Later, with an increase of $100$ , the population was one more than a perfect square. Now, with an additional increase of $100$ , the population is again a perfect square.
The original population is a multiple of:
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 17$ | Let $a^2$ $=$ original population count, $b^2+1$ $=$ the second population count, and $c^2$ $=$ the third population count
We first see that $a^2 + 100 = b^2 + 1$ or $99$ $=$ $b^2-a^2$ .
We then factor the right side getting $99$ $=$ $(b-a)(b+a)$ .
Since we can only have an nonnegative integral population, clearly $b+a$ $>$ $b-a$ and both factor $99$ .
We factor $99$ into $3^2 \cdot 11$ $=$ $(b-a)(b+a)$ There are a few cases to look at: $1)$ $b+a$ $=$ $11$ and $b-a$ $=$ $9$ .
Adding the two equations we get $2b$ $=$ $20$ or $b$ $=$ $10$ , which means $a$ $=$ $1$ .
But looking at the restriction that the second population + $100$ $=$ third population... $10^2$ $+$ $1$ $+$ $100$ $=$ $201$ $\neq$ a perfect square.
$2)$ $b+a$ $=$ $33$ and $b-a$ $=$ $3$ .
Adding the two equations we get $2b$ $=$ $36$ or $b$ $=$ $18$ , which means $a$ $=$ $15$ .
Looking at the same restriction, we get $18^2$ $1$ $100$ $=$ $324$ $101$ $=$ $425$ , which is NOT a perfect square.
Finally, $b+a$ $=$ $99$ and $b-a$ $=$ $1$ $2b$ $=$ $100$ or $b$ $=$ $50$ , which means $a$ $=$ $49$ .
Looking at the same restriction, we get $50^2$ $1$ $100$ $=$ $2500$ $101$ $=$ $2601$ $=$ $51^2$ . Thus we find that the original population is $a^2$ $=$ $49^2$ $=$ $7^4$ . Or $a^2$ is a multiple of $\boxed{7}$ | B | 7 |
21f540ec14c54018ed1046b15ed17d8e | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_38 | The population of Nosuch Junction at one time was a perfect square. Later, with an increase of $100$ , the population was one more than a perfect square. Now, with an additional increase of $100$ , the population is again a perfect square.
The original population is a multiple of:
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 11\qquad\textbf{(E)}\ 17$ | Let $P$ $=$ original population. Translating the word problem into a system of equations, we got: \begin{align} P &= x^2 \\ P + 100 &= y^2 + 1 \\ P + 200 &= z^2 \end{align} for some positive integers $x$ $y$ and $z$ .
Now, by subtracting $(2)$ from $(3)$ (i.e. $(3) - (2)$ ), we got: \begin{align*} 100 &= z^2 - y^2 - 1 \\ 101 &= z^2 - y^2 \\ 101 &= (z - y)(z + y) \end{align*} Since y and z are both positive integers and 101 is a prime, by factoring, the only working solution for us is $y = 50$ and $z = 51$ .
Plugging that back to $(2)$ and simplify, we got $P = 2401 = (49)^2 = x^2$ , a multiple of $7$ .
Therefore, the answer is $\boxed{7}$ . -nullptr07 | B | 7 |
7e06dbb3eb8e89f6f188648f5a362f1a | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_2 | An automobile travels $a/6$ feet in $r$ seconds. If this rate is maintained for $3$ minutes, how many yards does it travel in $3$ minutes?
$\textbf{(A)}\ \frac{a}{1080r}\qquad \textbf{(B)}\ \frac{30r}{a}\qquad \textbf{(C)}\ \frac{30a}{r}\qquad \textbf{(D)}\ \frac{10r}{a}\qquad \textbf{(E)}\ \frac{10a}{r}$ | Use dimensional analysis. \[\frac{a/6 \text{ feet}}{r \text{ seconds}} \cdot \frac{1 \text{ yard}}{3 \text{ feet}} \cdot \frac{60 \text{ seconds}}{1 \text{ minute}} \cdot 3 \text{ minutes}\] \[\frac{10a}{r} \text{ yards}\] The answer is $\boxed{10}$ | E | 10 |
8ecf04f2380b6248612d5a2005042e39 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_10 | Each side of $\triangle ABC$ is $12$ units. $D$ is the foot of the perpendicular dropped from $A$ on $BC$ ,
and $E$ is the midpoint of $AD$ . The length of $BE$ , in the same unit, is:
$\textbf{(A)}\ \sqrt{18} \qquad \textbf{(B)}\ \sqrt{28} \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ \sqrt{63} \qquad \textbf{(E)}\ \sqrt{98}$ | [asy] draw((0,0)--(50,0)--(25,43.301)--cycle); draw((25,43.301)--(25,0)); dot((0,0)); label("$B$",(0,0),SW); dot((50,0)); label("$C$",(50,0),SE); dot((25,43.301)); label("$A$",(25,43.301),N); dot((25,0)); label("$D$",(25,0),S); dot((25,21.651)); label("$E$",(25,21.651),E); draw((25,21.651)--(0,0)); label("$12$",(10,25)); label("$6$",(12.5,-5)); label("$6$",(37.5,-5)); label("$12$",(40,25)); draw((25,3)--(28,3)--(28,0)); label("$3\sqrt{3}$",(30.5,11)); [/asy] Note that $\triangle ABC$ is an equilateral triangle . From the Pythagorean Theorem (or by using 30-60-90 triangles), $AD = 6\sqrt{3}$ . That means $DE = 3\sqrt{3}$ . Using the Pythagorean Theorem again, $BE = \sqrt{63}$ , which is answer choice $\boxed{63}$ | D | 63 |
97b29d82e1cfa1c6a3c71b243680bcc2 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_11 | Two tangents are drawn to a circle from an exterior point $A$ ; they touch the circle at points $B$ and $C$ respectively.
A third tangent intersects segment $AB$ in $P$ and $AC$ in $R$ , and touches the circle at $Q$ . If $AB=20$ , then the perimeter of $\triangle APR$ is
$\textbf{(A)}\ 42\qquad \textbf{(B)}\ 40.5 \qquad \textbf{(C)}\ 40\qquad \textbf{(D)}\ 39\frac{7}{8} \qquad \textbf{(E)}\ \text{not determined by the given information}$ | Since $Q$ can be anywhere on the circle between $A$ and $B$ , it can basically be "on top" of $A$ . Then $R$ will be at the same point as $A$ , so $APR$ form a degenerate triable with side length $AB=20$ . So its perimeter will be $40$ . Since $BP=PQ$ and $QR=CR$ by power of a point, as $AP$ and $AR$ decrease in length, $PR=PQ+QR$ will "grow" to compensate, so the perimeter will stay constant with a value of $\boxed{40}$ | C | 40 |
d7566d38604832fc24309e0fd98801a1 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_17 | In the base ten number system the number $526$ means $5 \times 10^2+2 \times 10 + 6$ .
In the Land of Mathesis, however, numbers are written in the base $r$ .
Jones purchases an automobile there for $440$ monetary units (abbreviated m.u).
He gives the salesman a $1000$ m.u bill, and receives, in change, $340$ m.u. The base $r$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 12$ | If Jones received $340$ m.u. change after paying $1000$ m.u. for something that costs $440$ m.u., then \[440_r + 340_r = 1000_r\] This equation can be rewritten as \[4r^2 + 4r + 3r^2 + 4r = r^3\] Bring all of the terms to one side to get \[r^3 - 7r^2 - 8r = 0\] Factor to get \[r(r-8)(r+1)=0\] Since base numbers are always positive, base $r$ is $8$ , which is answer choice $\boxed{8}$ | D | 8 |
5836872ef6cfed1a833c0a0cc4abfa65 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_18 | The yearly changes in the population census of a town for four consecutive years are,
respectively, 25% increase, 25% increase, 25% decrease, 25% decrease.
The net change over the four years, to the nearest percent, is:
$\textbf{(A)}\ -12 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 12$ | A 25% increase means the new population is $\frac{5}{4}$ of the original population. A 25% decrease means the new population is $\frac{3}{4}$ of the original population.
Thus, after four years, the population is $1 \cdot \frac{5}{4} \cdot \frac{5}{4} \cdot \frac{3}{4} \cdot \frac{3}{4} = \frac{225}{256}$ times the original population.
Thus, the net change is -12%, so the answer is $\boxed{12}$ | A | 12 |
3fa1353535cd3b75d78600ad9ba9ee6b | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_23 | Points $P$ and $Q$ are both in the line segment $AB$ and on the same side of its midpoint. $P$ divides $AB$ in the ratio $2:3$ ,
and $Q$ divides $AB$ in the ratio $3:4$ . If $PQ=2$ , then the length of $AB$ is:
$\textbf{(A)}\ 60\qquad \textbf{(B)}\ 70\qquad \textbf{(C)}\ 75\qquad \textbf{(D)}\ 80\qquad \textbf{(E)}\ 85$ | [asy] draw((0,0)--(70,0)); dot((0,0)); dot((28,0)); dot((30,0)); dot((70,0)); label("2",(29,0),N); label("x",(14,0),N); label("y",(50,0),N); label("A",(0,0),S); label("P",(28,0),SW); label("Q",(30,0),SE); label("B",(70,0),S); [/asy]
Draw diagram as shown, where $P$ and $Q$ are on the same side. Let $AP = x$ and $QB = y$
Since $P$ divides $AB$ in the ratio $2:3$ $\frac{x}{y+2} = \frac{2}{3}$ . Since $Q$ divides $AB$ in the ratio $3:4$ $\frac{x+2}{y} = \frac{3}{4}$ . Cross multiply to get a system of equations \[3x = 2y+4\] \[4x+8=3y\] Solve the system to get $x = 28$ and $y = 40$ . Thus, $AB = 40 + 28 + 2 = 70$ , so the answer is $\boxed{70}$ | B | 70 |
7e69e65336f26c1820793ca63e531812 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_26 | For a given arithmetic series the sum of the first $50$ terms is $200$ , and the sum of the next $50$ terms is $2700$ .
The first term in the series is:
$\textbf{(A)}\ -1221 \qquad \textbf{(B)}\ -21.5 \qquad \textbf{(C)}\ -20.5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 3.5$ | Let the first term of the arithmetic sequence be $a$ and the common difference be $d$
The $50^{\text{th}}$ term of the sequence is $a+49d$ , so the sum of the first $50$ terms is $\frac{50(a + a + 49d)}{2}$
The $51^{\text{th}}$ term of the sequence is $a+50d$ and the $100^{\text{th}}$ term of the sequence is $a+99d$ , so the sum of the next $50$ terms is $\frac{50(a+50d+a+99d)}{2}$
Substituting in values results in this system of equations \[2a+49d=8\] \[2a+149d=108\] Solving for $a$ yields $a = \frac{-41}{2}$ . The first term is $-20.5$ , which is answer choice $\boxed{20.5}$ | C | 20.5 |
dc7ae85cc1bd1b16a8e56bac58896721 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_28 | If $2137^{753}$ is multiplied out, the units' digit in the final product is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$ | $7^1$ has a unit digit of $7$ $7^2$ has a unit digit of $9$ $7^3$ has a unit digit of $3$ $7^4$ has a unit digit of $1$ $7^5$ has a unit digit of $7$
Notice that the unit digit eventually cycles to itself when the exponent is increased by $4$ . It also does not matter what the other digits are in the base because the units digit is found by multiplying by only the units digit. Since $753$ leaves a remainder of $1$ after being divided by $4$ , the units digit of $2137^{753}$ is $7$ , which is answer choice $\boxed{7}$ | D | 7 |
e544ba0602a7f9b8f3c712c0be18fdfd | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_32 | A regular polygon of $n$ sides is inscribed in a circle of radius $R$ . The area of the polygon is $3R^2$ . Then $n$ equals:
$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 15\qquad \textbf{(E)}\ 18$ | Note that the distance from the center of the circle to each of the vertices of the inscribed regular polygon equals the radius $R$ . Since each side of a regular polygon is the same length, all the angles between the two lines from the center to the two vertices of a side is the same.
That means each of these angles between the two lines from the center to the two vertices of a side equals $\frac{360}{n}$ degrees. Thus, the area of the polygon is \[n \cdot \frac{1}{2}R^2\sin\left(\frac{360}{n}^{\circ}\right) = 3R^2\] Dividing both sides by $R^2$ yields \[\frac{n}{2}\sin\left(\frac{360}{n}^{\circ}\right) = 3\] Multiply both sides by $\frac{2}{n}$ to get \[\sin\left(\frac{360}{n}^{\circ}\right) = \frac{6}{n}\] At this point, use trial-and-error for each of the answer choices. When checking $n = 12$ , the equation results in $\sin(30^{\circ}) = \frac{1}{2}$ , which is correct. Thus, the answer is $\boxed{12}$ | C | 12 |
36f3c1611e8bef67fffe9b4262f1907d | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_35 | The number $695$ is to be written with a factorial base of numeration, that is, $695=a_1+a_2\times2!+a_3\times3!+ \ldots a_n \times n!$ where $a_1, a_2, a_3 ... a_n$ are integers such that $0 \le a_k \le k,$ and $n!$ means $n(n-1)(n-2)...2 \times 1$ . Find $a_4$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4$ | This problem can be approached similarly to other base number problems.
Since $120 < 695 < 720$ , divide $695$ by $120$ . The quotient is $5$ and the remainder is $95$ , so rewrite the number as \[695 = 5 \cdot 120 + 95\] Similarly, dividing $95$ by $24$ results in a quotient of $3$ and a remainder of $23$ , so the number can be rewritten as \[695 = 5 \cdot 120 + 3 \cdot 24 + 23\] Repeat the steps to get \[695 = 5 \cdot 120 + 3 \cdot 24 + 3 \cdot 6 + 2 \cdot 2 + 1\] The answer is $\boxed{3}$ . One can also stop at the second step by noting $23 < 24$ | D | 3 |
a5ead4687a44cf4408bc16b44594de5a | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_1 | If $2$ is a solution (root) of $x^3+hx+10=0$ , then $h$ equals:
$\textbf{(A) }10\qquad \textbf{(B) }9 \qquad \textbf{(C) }2\qquad \textbf{(D) }-2\qquad \textbf{(E) }-9$ | Substitute $2$ for $x$ . We are given that this equation is true. Solving for $h$ gives $h=-9$ . The answer is $\boxed{9}$ | E | 9 |
88a6fa32f261761fa21537b0eab428c0 | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_3 | Applied to a bill for $\textdollar{10,000}$ the difference between a discount of $40$ % and two successive discounts of $36$ % and $4$ %,
expressed in dollars, is:
$\textbf{(A)}0\qquad \textbf{(B)}144\qquad \textbf{(C)}256\qquad \textbf{(D)}400\qquad \textbf{(E)}416$ | Taking the discount of $40$ % means you're only paying $60$ % of the bill. That results in $10,000\cdot0.6=\textdollar{6,000}$
Likewise, taking two discounts of $36$ % and $4$ % means taking $64$ % of the original amount and then $96$ % of the result. That results in $10,000\cdot0.64\cdot0.96=\textdollar{6,144}$
Taking the difference results in $6,144-6,000=\textdollar{144}$ , or answer choice $\boxed{144}$ | B | 144 |
ef944140f9eac47972e8d294aa270443 | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_20 | The coefficient of $x^7$ in the expansion of $\left(\frac{x^2}{2}-\frac{2}{x}\right)^8$ is:
$\textbf{(A)}\ 56\qquad \textbf{(B)}\ -56\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ -14\qquad \textbf{(E)}\ 0$ | By the Binomial Theorem , each term of the expansion is $\binom{8}{n}\left(\frac{x^2}{2}\right)^{8-n}\left(\frac{-2}{x}\right)^n$
We want the exponent of $x$ to be $7$ , so \[2(8-n)-n=7\] \[16-3n=7\] \[n=3\]
If $n=3$ , then the corresponding term is \[\binom{8}{3}\left(\frac{x^2}{2}\right)^{5}\left(\frac{-2}{x}\right)^3\] \[56 \cdot \frac{x^{10}}{32} \cdot \frac{-8}{x^3}\] \[-14x^7\]
The answer is $\boxed{14}$ | D | 14 |
3e465bd78d4808f4dbd328bb082fa008 | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_25 | Let $m$ and $n$ be any two odd numbers, with $n$ less than $m$ .
The largest integer which divides all possible numbers of the form $m^2-n^2$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 16$ | First, factor the difference of squares \[(m+n)(m-n)\] Since $m$ and $n$ are odd numbers, let $m=2a+1$ and $n=2b+1$ , where $a$ and $b$ can be any integer. \[(2a+2b+2)(2a-2b)\] Factor the resulting expression. \[4(a+b+1)(a-b)\] If $a$ and $b$ are both even, then $a-b$ is even. If $a$ and $b$ are both odd, then $a-b$ is even as well. If $a$ is odd and $b$ is even (or vise versa), then $a+b+1$ is even. Therefore, in all cases, $8$ can be divided into all numbers with the form $m^2-n^2$
This can be confirmed by setting $m=3$ and $n=1$ , making $m^2-n^2=9-1=8$ . Since $8$ is not a multiple of $3$ and is less than $16$ , we can confirm that the answer is $\boxed{8}$ | D | 8 |
a35daf8aab94d48b126ad086f51e9a1d | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_27 | Let $S$ be the sum of the interior angles of a polygon $P$ for which each interior angle is $7\frac{1}{2}$ times the
exterior angle at the same vertex. Then
$\textbf{(A)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may be regular}\qquad \\ \textbf{(B)}\ S=2660^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad \\ \textbf{(C)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is regular}\qquad \\ \textbf{(D)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{is not regular}\qquad \\ \textbf{(E)}\ S=2700^{\circ} \text{ } \text{and} \text{ } P \text{ } \text{may or may not be regular}$ | Let $a_n$ be the interior angle of the nth vertex, and let $b_n$ be the exterior angle of the nth vertex. From the conditions in the problem, \[a_n = 7.5b_n\] That means \[a_1 + a_2 \cdots a_n = 7.5(b_1 + b_2 \cdots b_n)\] Since the sum of the exterior angles of a polygon is $360^{\circ}$ , the equation can be simplified as \[a_1 + a_2 \cdots a_n = 7.5 \cdot 360\] \[a_1 + a_2 \cdots a_n = 2700\] The sum of the interior angles is $2700$ degrees. However, there is no other constraint on what angle the interior angles can be, so we can not for sure claim that the polygon is regular or not. Thus, the answer is $\boxed{2700}$ | E | 2700 |
9a9f5798a88acaa7adf7afa105365e11 | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_33 | You are given a sequence of $58$ terms; each term has the form $P+n$ where $P$ stands for the product $2 \times 3 \times 5 \times\ldots \times 61$ of all prime numbers less than or equal to $61$ , and $n$ takes, successively, the values $2, 3, 4,\ldots, 59$ .
Let $N$ be the number of primes appearing in this sequence. Then $N$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 16\qquad \textbf{(C)}\ 17\qquad \textbf{(D)}\ 57\qquad \textbf{(E)}\ 58$ | First, note that $n$ does not have a prime number larger than $61$ as one of its factors. Also, note that $n$ does not equal $1$
Therefore, since the prime factorization of $n$ only has primes from $2$ to $59$ $n$ and $P$ share at least one common factor other than $1$ . Therefore $P+n$ is not prime for any $n$ , so the answer is $\Rightarrow{\boxed{0}$ | A | 0 |
50fd559e49d1069bf13c1fa4de08561d | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_34 | Two swimmers, at opposite ends of a $90$ -foot pool, start to swim the length of the pool,
one at the rate of $3$ feet per second, the other at $2$ feet per second.
They swim back and forth for $12$ minutes. Allowing no loss of times at the turns, find the number of times they pass each other.
$\textbf{(A)}\ 24\qquad \textbf{(B)}\ 21\qquad \textbf{(C)}\ 20\qquad \textbf{(D)}\ 19\qquad \textbf{(E)}\ 18$ | First, note that it will take $30$ seconds for the first swimmer to reach the other side and $45$ seconds for the second swimmer to reach the other side. Also, note that after $180$ seconds (or $3$ minutes), both swimmers will complete an even number of laps, essentially returning to their starting point.
[asy] draw((0,0)--(0,105),EndArrow); draw((0,0)--(105,0),EndArrow); for (int i=0; i<6;++i) { dot((0,18i)); } for (int j=0;j<7;++j) { dot((15j,0)); } label("0",(0,0),SW); label("90",(0,90),W); label("180",(90,0),S); draw((0,0)--(15,90)--(30,0)--(45,90)--(60,0)--(75,90)--(90,0),red); draw((0,90)--(22.5,0)--(45,90)--(67.5,0)--(90,90),blue); [/asy]
At this point, find the number of meeting points in the first $3$ minutes, then multiply by four to get the answer. From the graph (where the x-axis is the time in seconds and the y-axis is distance from one side of the pool), there are five meeting points, so the two swimmers will pass each other $20$ times, which is answer choice $\boxed{20}$ | C | 20 |
fd6e5d66f69f5ee0cbd185fefdd5226c | https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_1 | Each edge of a cube is increased by $50$ %. The percent of increase of the surface area of the cube is: $\textbf{(A)}\ 50 \qquad\textbf{(B)}\ 125\qquad\textbf{(C)}\ 150\qquad\textbf{(D)}\ 300\qquad\textbf{(E)}\ 750$ | Note that increasing the length of each edge by $50$ % with result in a cube that is similar to the original cube with scale factor $1.5$ . Therefore, the surface area will increase by a factor of $1.5^2$ , or $2.25$ . Converting this back into a percent, the percent increase will be $125$ %. Therefore, the answer is $\boxed{125}$ | B | 125 |
420408c06e5138eeb94b9b5d39a331e7 | https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_5 | The value of $\left(256\right)^{.16}\left(256\right)^{.09}$ is:
$\textbf{(A)}\ 4 \qquad \\ \textbf{(B)}\ 16\qquad \\ \textbf{(C)}\ 64\qquad \\ \textbf{(D)}\ 256.25\qquad \\ \textbf{(E)}\ -16$ | When we multiply numbers with exponents, we add the exponents together and leave the bases unchanged. We can apply this concept to computate $256^{0.16} \cdot 256^{0.09}$ \[256^{0.16} \cdot 256^{0.09} = 256^{0.16+0.09}=256^{0.25}.\] Now we can convert the decimal exponent to a fraction: \[256^{0.25} = 256^{\frac{1}{4}}.\] Now, let us convert the expression into radical form. Since $4$ is the denominator of the fractional exponent, it will be the index exponent: \[256^{\frac{1}{4}}=\sqrt[4]{256}.\] Since $256 =16^2=(4^2)^2=4^4$ , we can solve for the fourth root of $256$ \[\sqrt[4]{256}=\sqrt[4]{4^4}=4.\] Therefore, $(256)^{.16} \cdot (256)^{.09}=\boxed{4}.$ | A | 4 |