id int64 1 7.14k | link stringlengths 75 84 | no int64 1 14 | problem stringlengths 14 5.33k | solution stringlengths 21 6.43k | answer int64 0 999 |
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1,401 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_4 | 1 | The number of solutions to $\{1,~2\}\subseteq~X~\subseteq~\{1,~2,~3,~4,~5\}$ , where $X$ is a subset of $\{1,~2,~3,~4,~5\}$ is
$\textbf{(A) }2\qquad \textbf{(B) }4\qquad \textbf{(C) }6\qquad \textbf{(D) }8\qquad \textbf{(E) }\text{None of these}$ | $X$ has to contain $\{1,~2\}$ , so only $\{3,~4,~5\}$ matters. There are two choices for the elements; the element is either in $X$ or outside of $X$ . With this combinatorics in mind, the answer is simply $2^3=\boxed{8}.$ ~lopkiloinm | 8 |
1,402 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_24 | 1 | problem_id
4ca2bf599f1693b695997dd52d9e4774 A man walked a certain distance at a constant ...
4ca2bf599f1693b695997dd52d9e4774 A man walked a certain distance at a constant ...
Name: Text, dtype: object | We can make three equations out of the information, and since the distances are the same, we can equate these equations.
\[\frac{4t}{5}(x+\frac{1}{2})=xt=(t+\frac{5}{2})(x-\frac{1}{2})\] where $x$ is the man's rate and $t$ is the time it takes him.
Looking at the first two parts of the equations,
\[\frac{4t}{5}(x+\frac... | 15 |
1,403 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_25 | 1 | Inscribed in a circle is a quadrilateral having sides of lengths $25,~39,~52$ , and $60$ taken consecutively. The diameter of this circle has length
$\textbf{(A) }62\qquad \textbf{(B) }63\qquad \textbf{(C) }65\qquad \textbf{(D) }66\qquad \textbf{(E) }69$ | We note that $25^2+60^2=65^2$ and $39^2+52^2=65^2$ so our answer is $\boxed{65}$ | 65 |
1,404 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_28 | 1 | A circular disc with diameter $D$ is placed on an $8\times 8$ checkerboard with width $D$ so that the centers coincide. The number of checkerboard squares which are completely covered by the disc is
$\textbf{(A) }48\qquad \textbf{(B) }44\qquad \textbf{(C) }40\qquad \textbf{(D) }36\qquad \textbf{(E) }32$ | Consider the upper right half of the grid, which consists of a $4\times4$ section of the checkerboard and a quarter-circle of radius $4$ . We can draw this as a coordinate grid and shade in the complete squares. There are $8$ squares in the upper right corner, so there are $8 \cdot 4 = \boxed{32}$ whole squares in tota... | 32 |
1,405 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_31 | 1 | When the number $2^{1000}$ is divided by $13$ , the remainder in the division is
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }7\qquad \textbf{(E) }11$ | By Fermat's Little Theorem , we know that $2^{100} \equiv 2^{1000 \pmod{12}}\pmod{13}$ . However, we find that $1000 \equiv 4 \pmod{12}$ , so $2^{1000} \equiv 2^4 = 16 \equiv 3 \pmod{13}$ , so the answer is $\boxed{3}$ | 3 |
1,406 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_34 | 1 | Three times Dick's age plus Tom's age equals twice Harry's age.
Double the cube of Harry's age is equal to three times the cube of Dick's age added to the cube of Tom's age.
Their respective ages are relatively prime to each other. The sum of the squares of their ages is
$\textbf{(A) }42\qquad \textbf{(B) }46\qquad \... | \[t=2h-3d\] \[3d^3+t^3=2h^3\]
First, substitute in t into the second equation and get $3d^3+8h^3-36h^2d+54hd^2-27d^3=2h^3$ . That turns into $h^3-6h^2d+9hd^2-4d^3=0$ which is factored into $(h-4d)(h-d)^2 =0.$ WLOG, $d=1$ and consequently $h=4$ . Then $t=8-3=5$ . Everything appears to be relatively prime already. The an... | 42 |
1,407 | https://artofproblemsolving.com/wiki/index.php/1971_AHSME_Problems/Problem_2 | 1 | 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 ... | 2 |
1,408 | https://artofproblemsolving.com/wiki/index.php/1971_AHSME_Problems/Problem_27 | 1 | 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... | 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 $19*3=\boxed{57}$ | 57 |
1,409 | https://artofproblemsolving.com/wiki/index.php/1971_AHSME_Problems/Problem_29 | 1 | 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... | 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 eq... | 11 |
1,410 | https://artofproblemsolving.com/wiki/index.php/1970_AHSME_Problems/Problem_19 | 1 | 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} \\ \... | 5 |
1,411 | https://artofproblemsolving.com/wiki/index.php/1970_AHSME_Problems/Problem_33 | 1 | 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 fir... | 1 |
1,412 | https://artofproblemsolving.com/wiki/index.php/1970_AHSME_Problems/Problem_33 | 2 | 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 $4000\left( \frac{9 \cdot 10}{2} \right)+1=4000(45)+1=\boxed{180001}.$ | 1 |
1,413 | https://artofproblemsolving.com/wiki/index.php/1970_AHSME_Problems/Problem_33 | 3 | 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,... | 1 |
1,414 | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_8 | 1 | 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{(... | [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 s... | 61 |
1,415 | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_10 | 1 | 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 ... | 3 |
1,416 | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_18 | 1 | 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... | 4 |
1,417 | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_25 | 1 | 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 |
1,418 | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_26 | 1 | [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 $5$ i... | 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}{2... | 15 |
1,419 | https://artofproblemsolving.com/wiki/index.php/1969_AHSME_Problems/Problem_32 | 1 | 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 co... | 5 |
1,420 | https://artofproblemsolving.com/wiki/index.php/1968_AHSME_Problems/Problem_20 | 1 | 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 arit... | 9 |
1,421 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_1 | 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 a... | 6 |
1,422 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_4 | 1 | 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 $... | 2 |
1,423 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_5 | 1 | 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}... | 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, $\frac{P}{K} = \boxed{2}$ | 2 |
1,424 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_11 | 1 | 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 | 50 |
1,425 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_32 | 1 | 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 \cd... | 166 |
1,426 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_32 | 2 | 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; /* ... | 166 |
1,427 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_32 | 3 | 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 $\triangle AOD$ , wh... | 166 |
1,428 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_40 | 1 | 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)... | 79 |
1,429 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_40 | 2 | 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 $\frac{\sqrt{3}}{4}s^2$ ... | 79 |
1,430 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_40 | 3 | 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 t... | 79 |
1,431 | https://artofproblemsolving.com/wiki/index.php/1967_AHSME_Problems/Problem_40 | 4 | 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{\df... | 79 |
1,432 | https://artofproblemsolving.com/wiki/index.php/1966_AHSME_Problems/Problem_12 | 1 | 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 $\boxed{3}$ | 3 |
1,433 | https://artofproblemsolving.com/wiki/index.php/1966_AHSME_Problems/Problem_25 | 1 | 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 $2+50=\boxed{52}.$ | 52 |
1,434 | https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_1 | 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. | 2 |
1,435 | https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_1 | 2 | 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 $\boxed{1}$ | 1 |
1,436 | https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_3 | 1 | 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 $81^{\frac{1}{4}}=\sqrt[4]{81}=\boxed{3}$ | 3 |
1,437 | https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_5 | 1 | 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 $\boxed{15}$ | 15 |
1,438 | https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_9 | 1 | 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 ... | 16 |
1,439 | https://artofproblemsolving.com/wiki/index.php/1965_AHSME_Problems/Problem_40 | 1 | 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 $(x^2+3x+1)=x^4+6x^3+11x^2+6x+1$ \[... | 1 |
1,440 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_10 | 1 | 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... | 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 t... | 2 |
1,441 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_11 | 1 | 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:
$2y = (3y + 3) - 9$
$2y = 3y - 6$
$0 = y - 6$
$... | 27 |
1,442 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_14 | 1 | 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 \te... | 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... | 7 |
1,443 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_16 | 1 | 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 ... | 17 |
1,444 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_16 | 2 | 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... | 17 |
1,445 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_20 | 1 | 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,... | 1 |
1,446 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_23 | 1 | 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 $1=-7$ whi... | 48 |
1,447 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_28 | 1 | 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 $153$ . T... | 5 |
1,448 | https://artofproblemsolving.com/wiki/index.php/1964_AHSME_Problems/Problem_38 | 1 | 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, \[7=\sqrt{2(16)+2(49)-QR^2}\] \[49=32+98-QR^2\] \[QR^2=81\] \[QR=9=\boxed{9}\] | 9 |
1,449 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_2 | 1 | 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 $\boxed{14}$ | 14 |
1,450 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_9 | 1 | 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}}... | 21 |
1,451 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_12 | 1 | 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(r... | 9 |
1,452 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_13 | 1 | 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 w... | 0 |
1,453 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_16 | 1 | 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 +... | 12 |
1,454 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_19 | 1 | 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)}\ 2... | 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... | 210 |
1,455 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_20 | 1 | 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 s... | 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 s... | 4 |
1,456 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_23 | 1 | 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)}\ ... | 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... | 26 |
1,457 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_23 | 2 | 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)}\ ... | 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. | 26 |
1,458 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_24 | 1 | 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 $b=2$ , ... | 19 |
1,459 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_25 | 1 | 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... | 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, $... | 12 |
1,460 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_26 | 1 | 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... | 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 pr... | 4 |
1,461 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_27 | 1 | 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... | 22 |
1,462 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_27 | 2 | 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 $\boxed{22}$ | 22 |
1,463 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_27 | 3 | 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 $\boxed{22}$ | 22 |
1,464 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_29 | 1 | 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 $\boxed{400}$ | 400 |
1,465 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_31 | 1 | 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 $\boxed{127}$ | 127 |
1,466 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_31 | 2 | 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 S... | 127 |
1,467 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_32 | 1 | 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\qq... | 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 \... | 0 |
1,468 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_34 | 1 | 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... | 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 towa... | 120 |
1,469 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_35 | 1 | 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 prod... | 10 |
1,470 | https://artofproblemsolving.com/wiki/index.php/1963_AHSME_Problems/Problem_38 | 1 | 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 =... | 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 ... | 16 |
1,471 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_3 | 1 | 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 $\boxed{0}$ | 0 |
1,472 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_9 | 1 | 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 $\boxed{5}$ | 5 |
1,473 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_12 | 1 | 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 $\binom{6}{2}-\binom{6}{1}+\binom{6}{0}=15-6+1=\boxed{10}$ | 10 |
1,474 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_13 | 1 | $R$ 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
\[S=\frac12\sqrt{48\cdot75}=30\,.\] $\boxed{30}$ | 30 |
1,475 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_19 | 1 | 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 ... | 0 |
1,476 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_21 | 1 | 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 $s = 2(3+2i)(3-2i) = \boxed{26}$ | 26 |
1,477 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_24 | 1 | 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)}\ ... | 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}... | 23 |
1,478 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_25 | 1 | 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}=\s... | 5 |
1,479 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_31 | 1 | 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 $\frac{180 - \frac{360}{r}}{180 - \frac{360}{k}} = \frac{(r-2) \cdot k}{(k-2) \cdot r} = \frac{3}{2}$... | 3 |
1,480 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_35 | 1 | 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... | 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 \[180-\frac{11n}2=\pm110\]... | 40 |
1,481 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_36 | 1 | 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 $\boxed{2}$ soluti... | 2 |
1,482 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_38 | 1 | 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\... | 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... | 7 |
1,483 | https://artofproblemsolving.com/wiki/index.php/1962_AHSME_Problems/Problem_38 | 2 | 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\... | 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 ... | 7 |
1,484 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_2 | 1 | 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 $\boxed{10}$ | 10 |
1,485 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_10 | 1 | 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:
$\textbf{(A)}\ \sqrt{18} \qquad \textbf{(B)}\ \sqrt{28} \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ \sqrt{63} \qquad \textbf{(E)}\ \sq... | [asy] draw((0,0)--(50,0)--(25,43.301)--cycle); draw((25,43.301)--(25,0)); dot((0,0)); label("$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... | 63 |
1,486 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_11 | 1 | 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 \qq... | 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... | 40 |
1,487 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_17 | 1 | 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. T... | 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 alw... | 8 |
1,488 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_18 | 1 | 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 \te... | 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 ori... | 12 |
1,489 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_23 | 1 | 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)}\ 8... | [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 = ... | 70 |
1,490 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_28 | 1 | 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 ... | 7 |
1,491 | https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_32 | 1 | 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 ... | 12 |
1,492 | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_1 | 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 $\boxed{9}$ | 9 |
1,493 | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_3 | 1 | 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 $10,000\cdot0.64\cdot0.96=\textdollar{6,144}$
Taki... | 144 |
1,494 | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_20 | 1 | 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)^... | 14 |
1,495 | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_25 | 1 | 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 $a-b$ i... | 8 |
1,496 | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_27 | 1 | 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 \... | 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 simplif... | 700 |
1,497 | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_33 | 1 | 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.... | 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 ... | 0 |
1,498 | https://artofproblemsolving.com/wiki/index.php/1960_AHSME_Problems/Problem_34 | 1 | 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... | 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((... | 20 |
1,499 | https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_1 | 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 $... | 125 |
1,500 | https://artofproblemsolving.com/wiki/index.php/1959_AHSME_Problems/Problem_5 | 1 | 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}}... | 4 |
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