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906a9cb460bb0bada0c7b76fcbd0c37c | https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_17 | Given that $i^2=-1$ , for how many integers $n$ is $(n+i)^4$ an integer?
$\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4$ | $(n+i)^4=n^4+4in^3-6n^2-4in+1$ , and this has to be an integer, so the sum of the imaginary parts must be $0$ \[4in^3-4in=0\] \[4in^3=4in\] \[n^3=n\] Since $n^3=n$ , there are $\boxed{3}$ solutions for $n$ $0$ and $\pm1$ | null | 3 |
3703b99f33ccfd6b9e661437c68aaa49 | https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_24 | For some real number $r$ , the polynomial $8x^3-4x^2-42x+45$ is divisible by $(x-r)^2$ . Which of the following numbers is closest to $r$
$\text{(A)} \ 1.22 \qquad \text{(B)} \ 1.32 \qquad \text{(C)} \ 1.42 \qquad \text{(D)} \ 1.52 \qquad \text{(E)} \ 1.62$ | Solution by e_power_pi_times_i
Denote $s$ as the third solution. Then, by Vieta's, $2r+s = \dfrac{1}{2}$ $r^2+2rs = -\dfrac{21}{4}$ , and $r^2s = -\dfrac{45}{8}$ . Multiplying the top equation by $2r$ and eliminating, we have $3r^2 = r+\dfrac{21}{4}$ . Combined with the fact that $s = \dfrac{1}{2}-2r$ , the third equation can be written as $(\dfrac{r+\dfrac{21}{4}}{3})(\dfrac{1}{2}-2r) = -\dfrac{45}{8}$ , or $(4r+21)(4r-1) = 135$ . Solving, we get $r = \dfrac{3}{2}, -\dfrac{13}{2}$ . Plugging the solutions back in, we see that $-\dfrac{13}{2}$ is an extraneous solution, and thus the answer is $\boxed{1.52}$ | D | 1.52 |
6a072c0a614f889d57b52c1b8952e41e | https://artofproblemsolving.com/wiki/index.php/1980_AHSME_Problems/Problem_25 | In the non-decreasing sequence of odd integers $\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}$ each odd positive integer $k$ appears $k$ times. It is a fact that there are integers $b, c$ , and $d$ such that for all positive integers $n$ $a_n=b\lfloor \sqrt{n+c} \rfloor +d$ ,
where $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$ . The sum $b+c+d$ equals
$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4$ | Solution by e_power_pi_times_i
Because the set consists of odd numbers, and since $\lfloor{}\sqrt{n+c}\rfloor{}$ is an integer and can be odd or even, $b = 2$ and $|a| = 1$ . However, given that $\lfloor{}\sqrt{n+c}\rfloor{}$ can be $0$ $a = 1$ . Then, $a_1 = 1 = 2\lfloor{}\sqrt{1+c}\rfloor{}+1$ , and $\lfloor{}\sqrt{1+c}\rfloor{}$ = 0, and $c = -1$ because $c$ is an integer. $b+c+d = 2+(-1)+1 = \boxed{2}$ | C | 2 |
5089c17dab6071b312491c201325e8f6 | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_1 | [asy] draw((-2,1)--(2,1)--(2,-1)--(-2,-1)--cycle); draw((0,0)--(0,-1)--(-2,-1)--(-2,0)--cycle); label("$F$",(0,0),E); label("$A$",(-2,1),W); label("$B$",(2,1),E); label("$C$", (2,-1),E); label("$D$",(-2,-1),WSW); label("$E$",(-2,0),W); label("$G$",(0,-1),S); //Credit to TheMaskedMagician for the diagram [/asy]
If rectangle ABCD has area 72 square meters and E and G are the midpoints of sides AD and CD, respectively, then the area of rectangle DEFG in square meters is
$\textbf{(A) }8\qquad \textbf{(B) }9\qquad \textbf{(C) }12\qquad \textbf{(D) }18\qquad \textbf{(E) }24$ | Solution by e_power_pi_times_i
Since the dimensions of $DEFG$ are half of the dimensions of $ABCD$ , the area of $DEFG$ is $\dfrac{1}{2}\cdot\dfrac{1}{2}$ of $ABCD$ , so the area of $ABCD$ is $\dfrac{1}{4}\cdot72 = \boxed{18}$ | D | 18 |
b90fc9a521928719b59d8f00b2ef3bef | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_2 | For all non-zero real numbers $x$ and $y$ such that $x-y=xy, \frac{1}{x}-\frac{1}{y}$ equals
$\textbf{(A) }\frac{1}{xy}\qquad \textbf{(B) }\frac{1}{x-y}\qquad \textbf{(C) }0\qquad \textbf{(D) }-1\qquad \textbf{(E) }y-x$ | Moving all variables to one side of the equation, we can use Simon's Favorite Factoring Trick to factor the equation into \[(x+1)(y-1) = -1\] Plugging in $-1$ and $1$ as the $x$ and $y$ sides respectively, we get $x = -2$ and $y = 2$ . Plugging this in to $\frac{1}{x}-\frac{1}{y}$ gives us $\boxed{1}$ as our final answer. | null | 1 |
b90fc9a521928719b59d8f00b2ef3bef | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_2 | For all non-zero real numbers $x$ and $y$ such that $x-y=xy, \frac{1}{x}-\frac{1}{y}$ equals
$\textbf{(A) }\frac{1}{xy}\qquad \textbf{(B) }\frac{1}{x-y}\qquad \textbf{(C) }0\qquad \textbf{(D) }-1\qquad \textbf{(E) }y-x$ | Notice that we can do $\frac{x-y}{xy} = \frac{xy}{xy}$ . We are left with $\frac{1}{y} - \frac{1}{x} = 1$ . Multiply by $-1$ to achieve $\frac{1}{x} - \frac{1}{y} = \boxed{1}$ | null | 1 |
e3dbb094237e1af9d3aa764687b22a44 | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_3 | [asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD$ .
What is the measure of $\measuredangle AED$ in degrees?
$\textbf{(A) }10\qquad \textbf{(B) }12.5\qquad \textbf{(C) }15\qquad \textbf{(D) }20\qquad \textbf{(E) }25$ | Solution by e_power_pi_times_i
Notice that $\measuredangle DAE = 90^\circ+60^\circ = 150^\circ$ and that $AD = AE$ . Then triangle $ADE$ is isosceles, so $\measuredangle AED = \dfrac{180^\circ-150^\circ}{2} = \boxed{15}$ | C | 15 |
e3dbb094237e1af9d3aa764687b22a44 | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_3 | [asy] real s=sqrt(3)/2; draw(box((0,0),(1,1))); draw((1+s,0.5)--(1,1)); draw((1+s,0.5)--(1,0)); draw((0,1)--(1+s,0.5)); label("$A$",(1,1),N); label("$B$",(1,0),S); label("$C$",(0,0),W); label("$D$",(0,1),W); label("$E$",(1+s,0.5),E); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $ABCD$ is a square, $ABE$ is an equilateral triangle and point $E$ is outside square $ABCD$ .
What is the measure of $\measuredangle AED$ in degrees?
$\textbf{(A) }10\qquad \textbf{(B) }12.5\qquad \textbf{(C) }15\qquad \textbf{(D) }20\qquad \textbf{(E) }25$ | WLOG, let the side length of the square and the equilateral triangle be $1$ $\angle{DAE}=90^\circ+60^\circ=150^\circ$ . Apply the law of cosines then the law of sines, we find that $\angle{AED}=15^\circ$ . Select $\boxed{15}$ | C | 15 |
c948c562cff2b61bb646514403c6f121 | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_5 | Find the sum of the digits of the largest even three digit number (in base ten representation)
which is not changed when its units and hundreds digits are interchanged.
$\textbf{(A) }22\qquad \textbf{(B) }23\qquad \textbf{(C) }24\qquad \textbf{(D) }25\qquad \textbf{(E) }26$ | Solution by e_power_pi_times_i
Since the number doesn't change when the units and hundreds digits are switched, the number must be of the form $aba$ . We want to create the largest even $3$ -digit number, so $a = 8$ and $b = 9$ . The sum of the digits is $8+9+8 = \boxed{25}$ | D | 25 |
5a6f941f7786ada85122a0632c6f7870 | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_11 | Find a positive integral solution to the equation $\frac{1+3+5+\dots+(2n-1)}{2+4+6+\dots+2n}=\frac{115}{116}$
$\textbf{(A) }110\qquad \textbf{(B) }115\qquad \textbf{(C) }116\qquad \textbf{(D) }231\qquad\\ \textbf{(E) }\text{The equation has no positive integral solutions.}$ | Solution by e_power_pi_times_i
Notice that the numerator and denominator are the sum of the first $n$ odd and even numbers, respectively. Then the numerator is $n^2$ , and the denominator is $n(n+1)$ . Then $\frac{n}{n+1} = \frac{115}{116}$ , so $n = \boxed{115}$ | B | 115 |
26a60b36e9b66b14919bbb1075de9841 | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_12 | [asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,S);label("$D$",D,E);label("$E$",EE,NE);label("$O$",O,S);label("$45^\circ$",(0.25,0.1),fontsize(10pt)); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $CD$ is the diameter of a semi-circle with center $O$ . Point $A$ lies on the extension of $DC$ past $C$ ;
point $E$ lies on the semi-circle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semi-circle.
If length $AB$ equals length $OD$ , and the measure of $\measuredangle EOD$ is $45^\circ$ , then the
measure of $\measuredangle BAO$ is
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }25^\circ\qquad \textbf{(E) }30^\circ$ | Solution by e_power_pi_times_i
Because $AB = OD$ , triangles $ABO$ and $BOE$ are isosceles. Denote $\measuredangle BAO = \measuredangle AOB = \theta$ . Then $\measuredangle ABO = 180^\circ-2\theta$ , and $\measuredangle EBO = \measuredangle OEB = 2\theta$ , so $\measuredangle BOE = 180^\circ-4\theta$ . Notice that $\measuredangle AOB + \measuredangle BOE + 45^\circ = 180^\circ$ . Therefore $\theta+180-4\theta = 135^\circ$ , and $\theta = \boxed{15}$ | B | 15 |
26a60b36e9b66b14919bbb1075de9841 | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_12 | [asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,S);label("$D$",D,E);label("$E$",EE,NE);label("$O$",O,S);label("$45^\circ$",(0.25,0.1),fontsize(10pt)); //Credit to TheMaskedMagician for the diagram[/asy]
In the adjoining figure, $CD$ is the diameter of a semi-circle with center $O$ . Point $A$ lies on the extension of $DC$ past $C$ ;
point $E$ lies on the semi-circle, and $B$ is the point of intersection (distinct from $E$ ) of line segment $AE$ with the semi-circle.
If length $AB$ equals length $OD$ , and the measure of $\measuredangle EOD$ is $45^\circ$ , then the
measure of $\measuredangle BAO$ is
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }25^\circ\qquad \textbf{(E) }30^\circ$ | Draw $BO$ . Let $y = \angle BAO$ . Since $AB = OD = BO$ , triangle $ABO$ is isosceles, so $\angle BOA = \angle BAO = y$ . Angle $\angle EBO$ is exterior to triangle $ABO$ , so $\angle EBO = \angle BAO + \angle BOA = y + y = 2y$
Triangle $BEO$ is isosceles, so $\angle BEO = \angle EBO = 2y$ . Then $\angle EOD$ is external to triangle $AEO$ , so $\angle EOD = \angle EAO + \angle AEO = y + 2y = 3y$ . But $\angle EOD = 45^\circ$ , so $\angle BAO = y = 45^\circ/3 = \boxed{15}$ | B | 15 |
da8285aa172db0a973e083cee36e5180 | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_16 | A circle with area $A_1$ is contained in the interior of a larger circle with area $A_1+A_2$ . If the radius of the larger circle is $3$ ,
and if $A_1 , A_2, A_1 + A_2$ is an arithmetic progression, then the radius of the smaller circle is
$\textbf{(A) }\frac{\sqrt{3}}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }\frac{2}{\sqrt{3}}\qquad \textbf{(D) }\frac{3}{2}\qquad \textbf{(E) }\sqrt{3}$ | Solution by e_power_pi_times_i
The area of the larger circle is $A_1 + A_2 = 9\pi$ . Then $A_1 , 9\pi-A_1 , 9\pi$ are in an arithmetic progression. Thus $9\pi-(9\pi-A_1) = 9\pi-A_1-A_1$ . This simplifies to $3A_1 = 9\pi$ , or $A_1 = 3\pi$ . The radius of the smaller circle is $\boxed{3}$ | E | 3 |
161b16b7ad8b7d3585c11bfadb3c896b | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_19 | Find the sum of the squares of all real numbers satisfying the equation $x^{256}-256^{32}=0$
$\textbf{(A) }8\qquad \textbf{(B) }128\qquad \textbf{(C) }512\qquad \textbf{(D) }65,536\qquad \textbf{(E) }2(256^{32})$ | Solution by e_power_pi_times_i
Notice that the solutions to the equation $x^{256}-1=0$ are the $256$ roots of unity. Then the solutions to the equation $x^{256}-256^{32}=0$ are the $256$ roots of unity dilated by $\sqrt[256]{256^{32}} = \sqrt[256]{2^{256}} = 2$ . However, the only real solutions to the equation are the first root of unity and the root of unity opposite of it, as both are on the real axis in the complex plane. These two roots of unity are $\pm1$ , and dilating by $2$ gives $\pm2$ . The sum of the squares is $(2)^2+(-2)^2 = \boxed{8}$ | A | 8 |
2f7d5669447b6943a937f02ddd82493c | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_20 | If $a=\tfrac{1}{2}$ and $(a+1)(b+1)=2$ then the radian measure of $\arctan a + \arctan b$ equals
$\textbf{(A) }\frac{\pi}{2}\qquad \textbf{(B) }\frac{\pi}{3}\qquad \textbf{(C) }\frac{\pi}{4}\qquad \textbf{(D) }\frac{\pi}{5}\qquad \textbf{(E) }\frac{\pi}{6}$ | Solution by e_power_pi_times_i
Since $a=\frac{1}{2}$ $b=\frac{1}{3}$ . Now we evaluate $\arctan a$ and $\arctan b$ . Denote $x$ and $\theta$ such that $\arctan x = \theta$ . Then $\tan(\arctan(x)) = \tan(\theta)$ , and simplifying gives $x = \tan(\theta)$ . So $a = \tan(\theta_a) = \frac{1}{2}$ and $b = \tan(\theta_b) = \frac{1}{3}$ . The question asks for $\theta_a + \theta_b$ , so we try to find $\tan(\theta_a + \theta_b)$ in terms of $\tan(\theta_a)$ and $\tan(\theta_b)$ . Using the angle addition formula for $\tan(\alpha+\beta)$ , we get that $\tan(\theta_a + \theta_b) = \frac{\tan(\theta_a)+\tan(\theta_b)}{1-\tan(\theta_a)\tan(\theta_b)}$ . Plugging $\tan(\theta_a) = \frac{1}{2}$ and $\tan(\theta_b) = \frac{1}{3}$ in, we have $\tan(\theta_a + \theta_b) = \frac{\frac{1}{2}+\frac{1}{3}}{1-(\frac{1}{2})(\frac{1}{3})}$ . Simplifying, $\tan(\theta_a + \theta_b) = 1$ , so $\theta_a + \theta_b$ in radians is $\boxed{4}$ | C | 4 |
f115fd3e900c9e94aab997c07bea0b4e | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_22 | Find the number of pairs $(m, n)$ of integers which satisfy the equation $m^3 + 6m^2 + 5m = 27n^3 + 27n^2 + 9n + 1$
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }3\qquad \textbf{(D) }9\qquad \textbf{(E) }\infty$ | The equation is equivalent to $m(m+1)(m+5) = 27n^3 + 27n^2 + 9n + 1$ . Taking mod 3, we get $m(m+1)(m+2)=1 (\bmod 3)$ . However, $m(m+1)(m+2)$ is always divisible by $3$ for any integer $m$ . Thus, the answer is $\boxed{0}$ Solution by mickyboy789 | A | 0 |
1e70e2e4ea1a0ad117e04451363d88b8 | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_24 | Sides $AB,~ BC$ , and $CD$ of (simple*) quadrilateral $ABCD$ have lengths $4,~ 5$ , and $20$ , respectively.
If vertex angles $B$ and $C$ are obtuse and $\sin C = - \cos B =\frac{3}{5}$ , then side $AD$ has length
$\textbf{(A) }24\qquad \textbf{(B) }24.5\qquad \textbf{(C) }24.6\qquad \textbf{(D) }24.8\qquad \textbf{(E) }25$ | We know that $\sin(C)=-\cos(B)=\frac{3}{5}$ . Since $B$ and $C$ are obtuse, we have $\sin(180-C)=\cos(180-B)=\frac{3}{5}$ . It is known that $\sin(x)=\cos(90-x)$ , so $180-C=90-(180-C)=180-B$ . We simplify this as follows:
\[-90+C=180-B\]
\[B+C=270^{\circ}\]
Since $B+C=270^{\circ}$ , we know that $A+D=360-(B+C)=90^{\circ}$ . Now extend $AB$ and $CD$ as follows:
[asy] size(10cm); label("A",(-1,0)); dot((0,0)); label("B",(-1,4)); dot((0,4)); label("E",(-1,7)); dot((0,7)); label("C",(4,8)); dot((4,7)); label("D",(24,8)); dot((24,7)); draw((0,0)--(0,4)); draw((0,4)--(4,7)); draw((4,7)--(24,7)); draw((24,7)--(0,0)); draw((0,4)--(0,7), dashed); draw((0,7)--(4,7), dashed); //diagram by WannabeCharmander [/asy]
Let $AB$ and $CD$ intersect at $E$ . We know that $\angle AED=90^{\circ}$ because $\angle E = 180 - (A+D)=180-90 = 90^{\circ}$
Since $\sin BCD = \frac{3}{5}$ , we get $\sin ECB=\sin(180-BCD)=\sin BCD = \frac{3}{5}$ . Thus, $EB=3$ and $EC=4$ from simple sin application.
$AD$ is the hypotenuse of right $\triangle AED$ , with leg lengths $AB+BE=7$ and $EC+CD=24$ . Thus, $AD=\boxed{25}$ | E | 25 |
bc898ff515f0dee024e42670e4f3394f | https://artofproblemsolving.com/wiki/index.php/1979_AHSME_Problems/Problem_29 | For each positive number $x$ , let $f(x)=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+\frac{1}{x^6}\right)-2} {\left(x+\frac{1}{x}\right)^3+\left(x^3+\frac{1}{x^3}\right)}$ .
The minimum value of $f(x)$ is
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad \textbf{(E) }6$ | Let $a = \left( x + \frac{1}{x} \right)^3$ and $b = x^3 + \frac{1}{x^3}$ . Then \begin{align*} f(x) &= \frac{\left( x + \frac{1}{x} \right)^6 - (x^6 + \frac{1}{x^6}) - 2}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} \\ &= \frac{\left( x + \frac{1}{x} \right)^6 - (x^6 + 2 + \frac{1}{x^6})}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} \\ &= \frac{\left( x + \frac{1}{x} \right)^6 - (x^3 + \frac{1}{x^3})^2}{\left( x + \frac{1}{x} \right)^3 + (x^3 + \frac{1}{x^3})} \\ &= \frac{a^2 - b^2}{a + b}. \end{align*}
By difference of squares, \begin{align*} f(x) &= \frac{(a - b)(a + b)}{a + b} \\ &= a - b \\ &= \left( x + \frac{1}{x} \right)^3 - \left( x^3 + \frac{1}{x^3} \right) \\ &= \left( x^3 + 3x + \frac{3}{x} + \frac{1}{x^3} \right) - \left( x^3 + \frac{1}{x^3} \right) \\ &= 3x + \frac{3}{x} \\ &= 3 \left( x + \frac{1}{x} \right). \end{align*}
By the AM-GM inequality, \[x + \frac{1}{x} \ge 2,\] so $f(x) \ge 6$ . Furthermore, when $x = 1$ $f(1) = 6$ , so the minimum value of $f(x)$ is $\boxed{6}$ | null | 6 |
16ae34c94f5266099616f78ba6115a46 | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_1 | If $1-\frac{4}{x}+\frac{4}{x^2}=0$ , then $\frac{2}{x}$ equals
$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad \textbf{(E) }-1\text{ or }-2$ | By guessing and checking, 2 works. $\frac{2}{x} = \boxed{1}$ ~awin | B | 1 |
16ae34c94f5266099616f78ba6115a46 | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_1 | If $1-\frac{4}{x}+\frac{4}{x^2}=0$ , then $\frac{2}{x}$ equals
$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad \textbf{(E) }-1\text{ or }-2$ | Multiplying each side by $x^2$ , we get $x^2-4x+4 = 0$ . Factoring, we get $(x-2)(x-2) = 0$ . Therefore, $x = 2$ $\frac{2}{x} = \boxed{1}$ ~awin | B | 1 |
16ae34c94f5266099616f78ba6115a46 | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_1 | If $1-\frac{4}{x}+\frac{4}{x^2}=0$ , then $\frac{2}{x}$ equals
$\textbf{(A) }-1\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }-1\text{ or }2\qquad \textbf{(E) }-1\text{ or }-2$ | Directly factoring, we get $(1-\frac{2}{x})^2 = 0$ . Thus $\frac{2}{x}$ must equal $\boxed{1}$ | B | 1 |
b23049e9f5f9712b5e62e9412a7eaa19 | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_2 | If four times the reciprocal of the circumference of a circle equals the diameter of the circle, then the area of the circle is
$\textbf{(A) }\frac{1}{\pi^2}\qquad \textbf{(B) }\frac{1}{\pi}\qquad \textbf{(C) }1\qquad \textbf{(D) }\pi\qquad \textbf{(E) }\pi^2$ | Creating equations, we get $4\cdot\frac{1}{2\pi r} = 2r$ . Simplifying, we get $\frac{1}{\pi r} = r$ . Multiplying each side by $r$ , we get $\frac{1}{\pi} = r^2$ . Because the formula of the area of a circle is $\pi r^2$ , we multiply each side by $\pi$ to get $1 = \pi r^2$ .
Therefore, our answer is $\boxed{1}$ | C | 1 |
c541baeed0db736164acab2f795ab844 | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_4 | If $a = 1,~ b = 10, ~c = 100$ , and $d = 1000$ , then $(a+ b+ c-d) + (a + b- c+ d) +(a-b+ c+d)+ (-a+ b+c+d)$ is equal to
$\textbf{(A) }1111\qquad \textbf{(B) }2222\qquad \textbf{(C) }3333\qquad \textbf{(D) }1212\qquad \textbf{(E) }4242$ | Adding all four of the equations up, we can see that it equals \[3(a+b+c+d)\] This is equal to $3(1111) = \boxed{3333}$ ~awin | C | 3333 |
bbe16a0af397224a49d70492c2672792 | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_5 | Four boys bought a boat for $\textdollar 60$ . The first boy paid one half of the sum of the amounts paid by the other boys;
the second boy paid one third of the sum of the amounts paid by the other boys;
and the third boy paid one fourth of the sum of the amounts paid by the other boys. How much did the fourth boy pay?
$\textbf{(A) }\textdollar 10\qquad \textbf{(B) }\textdollar 12\qquad \textbf{(C) }\textdollar 13\qquad \textbf{(D) }\textdollar 14\qquad \textbf{(E) }\textdollar 15$ | If the first boy paid one half of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{3}$ of the total.
If the second boy paid one third of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{4}$ of the total.
If the third boy paid one fourth of the sum of the amounts paid by the other boys, that means that he paid $\frac{1}{5}$ of the total.
Summing it up, we get $\textdollar 20 + \textdollar 15 + \textdollar 12 = \textdollar 47$ .
Therefore, our answer is $\textdollar 60 - \textdollar 47 = \boxed{13}$ ~awin | C | 13 |
f529fe5eba1cf1fdebfa59aa028f6aae | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_6 | The number of distinct pairs $(x,y)$ of real numbers satisfying both of the following equations:
\[x=x^2+y^2 \ \ y=2xy\] is
$\textbf{(A) }0\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }3\qquad \textbf{(E) }4$ | If $x=x^2+y^2$ and $y=2xy$ , then we can break this into two cases.
Case 1: $y = 0$
If $y = 0$ , then $x = x^2$ and $0 = 0$
Therefore, $x = 0$ or $x = 1$
This yields 2 solutions
Case 2: $x = \frac{1}{2}$
If $x = \frac{1}{2}$ , this means that $y = y$ , and $\frac{1}{2} = \frac{1}{4} + y^2$
Because y can be negative or positive, this yields $y = \frac{1}{2}$ or $y = -\frac{1}{2}$
This yields another 2 solutions.
$2+2 = \boxed{4}$ | E | 4 |
cea52b2e09ea4ab8d1925453e5171855 | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_11 | If $r$ is positive and the line whose equation is $x + y = r$ is tangent to the circle whose equation is $x^2 + y ^2 = r$ , then $r$ equals
$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }\sqrt{2}\qquad \textbf{(E) }2\sqrt{2}$ | The circle $x^2 + y^2 = r$ has center $(0,0)$ and radius $\sqrt{r}$ . Therefore, if the line $x + y = r$ is tangent to the circle $x^2 + y^2 = r$ , then the distance between $(0,0)$ and the line $x + y = r$ is $\sqrt{r}$
The distance between $(0,0)$ and the line $x + y = r$ is \[\frac{|0 + 0 - r|}{\sqrt{1^2 + 1^2}} = \frac{r}{\sqrt{2}}.\] Hence, \[\frac{r}{\sqrt{2}} = \sqrt{r}.\] Then $r = \sqrt{r} \cdot \sqrt{2}$ , so $\sqrt{r} = \sqrt{2}$ , which means $r = \boxed{2}$ or (B), $2$ | null | 2 |
0cf4420171b0c3a12bcdadb69c6bf99e | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_13 | If $a,b,c$ , and $d$ are non-zero numbers such that $c$ and $d$ are the solutions of $x^2+ax+b=0$ and $a$ and $b$ are
the solutions of $x^2+cx+d=0$ , then $a+b+c+d$ equals
$\textbf{(A) }0\qquad \textbf{(B) }-2\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }(-1+\sqrt{5})/2$ | By Vieta's formulas, $c + d = -a$ $cd = b$ $a + b = -c$ , and $ab = d$ . From the equation $c + d = -a$ $d = -a - c$ , and from the equation $a + b = -c$ $b = -a - c$ , so $b = d$
Then from the equation $cd = b$ $cb = b$ . Since $b$ is nonzero, we can divide both sides of the equation by $b$ to get $c = 1$ . Similarly, from the equation $ab = d$ $ab = b$ , so $a = 1$ . Then $b = d = -a - c = -2$ . Therefore, $a + b + c + d = 1 + (-2) + 1 + (-2) = \boxed{2}$ . The answer is (B). | null | 2 |
2a7a8c18e76ff4708f1f307f22dfb4da | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_14 | If an integer $n > 8$ is a solution of the equation $x^2 - ax+b=0$ and the representation of $a$ in the base- $n$ number system is $18$ ,
then the base-n representation of $b$ is
$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 80 \qquad \textbf{(D)}\ 81 \qquad \textbf{(E)}\ 280$ | Assuming the solutions to the equation are n and m, by Vieta's formulas, $n_n + m_n = 18_n$
$n_n = 10_n$ , so $10_n + m_n = 18_n$
\[m_n = 8_n\]
Also by Vieta's formulas, $n_n \cdot m_n = b_n$ \[10_n \cdot 8_n = \boxed{80}\] | null | 80 |
81780c8c2bec6dc94a8e439f210f50d4 | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_18 | What is the smallest positive integer $n$ such that $\sqrt{n}-\sqrt{n-1}<.01$
$\textbf{(A) }2499\qquad \textbf{(B) }2500\qquad \textbf{(C) }2501\qquad \textbf{(D) }10,000\qquad \textbf{(E) }\text{There is no such integer}$ | Adding $\sqrt{n - 1}$ to both sides, we get \[\sqrt{n} < \sqrt{n - 1} + 0.01.\] Squaring both sides, we get \[n < n - 1 + 0.02 \sqrt{n - 1} + 0.0001,\] which simplifies to \[0.9999 < 0.02 \sqrt{n - 1},\] or \[\sqrt{n - 1} > 49.995.\] Squaring both sides again, we get \[n - 1 > 2499.500025,\] so $n > 2500.500025$ . The smallest positive integer $n$ that satisfies this inequality is $\boxed{2501}$ | null | 2501 |
5d435f1923903e10d9a09f9ea0cbc96d | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_19 | A positive integer $n$ not exceeding $100$ is chosen in such a way that if $n\le 50$ , then the probability of choosing $n$ is $p$ , and if $n > 50$ , then the probability of choosing $n$ is $3p$ . The probability that a perfect square is chosen is
$\textbf{(A) }.05\qquad \textbf{(B) }.065\qquad \textbf{(C) }.08\qquad \textbf{(D) }.09\qquad \textbf{(E) }.1$ | Let's say that we will have $3$ slips for every number not exceeding $100$ but bigger than $50.$ This is to account for the $3p$ probability part. Let's now say that we will only have one slip for each number below or equal to $50.$ The probability(or $p$ ) will then be $\frac{1}{200}.$ Now let's have all the squares under $50,$ which are $1,4,9,16,25,36,49.$ The probability for these are $\frac{7}{200}.$ The numbers above $50$ that are squares are $64,81,100.$ We then need to multiply the probability by $3$ so the probability of these are $\frac{9}{200}.$ The answer is $\frac{7}{200}+\frac{9}{200}=0.08\implies\boxed{.08}$ | C | .08 |
7ce378cc7c6249107192203e7fe9423f | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_22 | The following four statements, and only these are found on a card: [asy] pair A,B,C,D,E,F,G; A=(0,1); B=(0,5); C=(11,5); D=(11,1); E=(0,4); F=(0,3); G=(0,2); draw(A--B--C--D--cycle); label("On this card exactly one statement is false.", B, SE); label("On this card exactly two statements are false.", E, SE); label("On this card exactly three statements are false.", F, SE); label("On this card exactly four statements are false.", G, SE); [/asy]
(Assume each statement is either true or false.) Among them the number of false statements is exactly
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | There can be at most one true statement on the card, eliminating $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}$ . If there are $0$ true on the card, statement $4$ ("On this card exactly four statements are false") will be correct, causing a contradiction. Therefore, the answer is $\boxed{3}$ , since $3$ are false and only the third statement ("On this card exactly three statements are false") is correct. | D | 3 |
7ce378cc7c6249107192203e7fe9423f | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_22 | The following four statements, and only these are found on a card: [asy] pair A,B,C,D,E,F,G; A=(0,1); B=(0,5); C=(11,5); D=(11,1); E=(0,4); F=(0,3); G=(0,2); draw(A--B--C--D--cycle); label("On this card exactly one statement is false.", B, SE); label("On this card exactly two statements are false.", E, SE); label("On this card exactly three statements are false.", F, SE); label("On this card exactly four statements are false.", G, SE); [/asy]
(Assume each statement is either true or false.) Among them the number of false statements is exactly
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 4$ | If all of them are false, that would mean that the $4$ th one is false too. Therefore, $E$ is not the correct answer. If exactly $3$ of them are false, that would mean that only $1$ statement is true. This is correct since if only $1$ statement is true, the card that is true is the one that has $3$ of these statements are false. If we have $1$ or $2$ false statements, that would mean that there is more than $1$ true statement. Therefore, our answer is $\boxed{3}$ | D | 3 |
61a441e20aad252b4c345451bbba4f8f | https://artofproblemsolving.com/wiki/index.php/1978_AHSME_Problems/Problem_26 | [asy] size(100); real a=4, b=3; // import cse5; pathpen=black; pair A=(a,0), B=(0,b), C=(0,0); D(MP("A",A)--MP("B",B,N)--MP("C",C,SW)--cycle); pair X=IP(B--A,(0,0)--(b,a)); D(CP((X+C)/2,C)); D(MP("R",IP(CP((X+C)/2,C),B--C),NW)--MP("Q",IP(CP((X+C)/2,C),A--C+(0.1,0)))); //Credit to chezbgone2 for the diagram [/asy]
In $\triangle ABC, AB = 10~ AC = 8$ and $BC = 6$ . Circle $P$ is the circle with smallest radius which passes through $C$ and is tangent to $AB$ . Let $Q$ and $R$ be the points of intersection, distinct from $C$ , of circle $P$ with sides $AC$ and $BC$ , respectively. The length of segment $QR$ is
$\textbf{(A) }4.75\qquad \textbf{(B) }4.8\qquad \textbf{(C) }5\qquad \textbf{(D) }4\sqrt{2}\qquad \textbf{(E) }3\sqrt{3}$ | We know that triangle $RCQ$ is similar to triangle $ABC$ . We draw a line to point $D$ on hypotenuse $AB$ such that $\angle QDR$ is $90 ^\circ$ and that $RDQC$ is a rectangle. Since triangle $RCQ$ is similar to triangle $ABC$ , let $RC$ be $4x$ and $RD/CQ$ be $3x$ . Now we have line segment $AQ$ $8-3x$ , and line segment $RB$ $6-4x$ . Since $BD + DA = AB$ , we use simple algebra and Pythagorean Theorem to get $\sqrt {(3x)^2 + (6-4x)^2}$ $\sqrt {(4x)^2 + (8-3x)^2}$ $10$ . Expanding and simplifying gives us $\sqrt {25x^2-48x+36}$ $\sqrt {25x^2-48x+64}$ $10$
Squaring both sides is not an option since it is both messy and time consuming, so we should proceed to subtracting both sides by $\sqrt {25x^2-48x+36}$ . Now, we can square both sides and simplify to get $0 = 72 - 20 \sqrt{25x^2-48x+36}$ . Dividing both sides by $4$ , we get $18 - 5 \sqrt {25x^2-48x+36}$ $0$ . We then add $5 \sqrt {25x^2-48x+36}$ to both sides to get $18 = 5 \sqrt {25x^2-48x+36}$ . Since this is very messy, let $25x^2 - 48x = y$ . Squaring both sides, we get $324 = 25y + 900, 25y = -576$ . Solving for $y$ , we have $y = -23.04$ . Plugging in $y$ as $25x^2-48x$ , we have $25x^2-48x+23.04 = 0$ . Using the quadratic equation, we get $\frac {48+0}{50}$ . Therefore, $x = \frac {48}{50}$
Remember that our desired answer is the hypotenuse of the triangle $3x - 4x - 5x$ . Since $5x$ is the hypotenuse, our answer is $\boxed{4.8}$ | B | 4.8 |
21d068c205ca634e015b01c134bc2b1c | https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_1 | If $y = 2x$ and $z = 2y$ , then $x + y + z$ equals
$\text{(A)}\ x \qquad \text{(B)}\ 3x \qquad \text{(C)}\ 5x \qquad \text{(D)}\ 7x \qquad \text{(E)}\ 9x$ | Solution by e_power_pi_times_i
$x+y+z = x+(2x)+(4x) = \boxed{7}$ | D | 7 |
03f165002cfd6641bd904762548ec1e8 | https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_8 | For every triple $(a,b,c)$ of non-zero real numbers, form the number $\frac{a}{|a|}+\frac{b}{|b|}+\frac{c}{|c|}+\frac{abc}{|abc|}$ .
The set of all numbers formed is
$\textbf{(A)}\ {0} \qquad \textbf{(B)}\ \{-4,0,4\} \qquad \textbf{(C)}\ \{-4,-2,0,2,4\} \qquad \textbf{(D)}\ \{-4,-2,2,4\}\qquad \textbf{(E)}\ \text{none of these}$ | Solution by e_power_pi_times_i
$\dfrac{x}{|x|} = 1$ or $-1$ depending whether $x$ is positive or negative. If $a$ $b$ , and $c$ are positive, then the entire thing amounts to $4$ . If one of the three is negative and the other two positive, the answer is $0$ . If two of the three is negative and one is positive, the answer is $0$ . If all three are negative, the answer is $-4$ . Therefore the set is $\boxed{4,0,4}$ | B | 4,0,4 |
d1f5c492d185abd55b21eab9e8638f11 | https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_9 | [asy] size(120); path c = Circle((0, 0), 1); pair A = dir(20), B = dir(130), C = dir(240), D = dir(330); draw(c); pair F = 3(A-B) + B; pair G = 3(D-C) + C; pair E = intersectionpoints(B--F, C--G)[0]; draw(B--E--C--A); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, SW); label("$D$", D, SE); label("$E$", E, E); //Credit to MSTang for the diagram [/asy]
In the adjoining figure $\measuredangle E=40^\circ$ and arc $AB$ , arc $BC$ , and arc $CD$ all have equal length. Find the measure of $\measuredangle ACD$
$\textbf{(A) }10^\circ\qquad \textbf{(B) }15^\circ\qquad \textbf{(C) }20^\circ\qquad \textbf{(D) }\left(\frac{45}{2}\right)^\circ\qquad \textbf{(E) }30^\circ$ | Solution by e_power_pi_times_i
If arcs $AB$ $BC$ , and $CD$ are congruent, then $\measuredangle ACB = \measuredangle BDC = \measuredangle CBD = \theta$ . Because $ABCD$ is cyclic, $\measuredangle CAD = \measuredangle CBD = \theta$ , and $\measuredangle ADB = \measuredangle ACB = \theta$ . Then, $\measuredangle EAD = \measuredangle EDA = \dfrac{180^\circ - 40^\circ}{2} = 70^\circ$ $\theta = 55^\circ$ $\measuredangle ACD = 180^\circ - 55^\circ - 110^\circ = \boxed{15}$ | B | 15 |
45c327453f9a266941db17ee50287061 | https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_10 | If $(3x-1)^7 = a_7x^7 + a_6x^6 + \cdots + a_0$ , then $a_7 + a_6 + \cdots + a_0$ equals
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 64 \qquad \text{(D)}\ -64 \qquad \text{(E)}\ 128$ | Solution by e_power_pi_times_i
Notice that if $x=1$ , then $a_7x^7 + a_6x^6 + \cdots + a_0 = a_7 + a_6 + \cdots + a_0$ . Therefore the answer is $(3(1)-1)^7) = \boxed{128}$ | D | 128 |
7bc1674e0cd88cb2d10270725302c11e | https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_12 | Al's age is $16$ more than the sum of Bob's age and Carl's age, and the square of Al's age is $1632$ more than the square of the sum of
Bob's age and Carl's age. What is the sum of the ages of Al, Bob, and Carl?
$\text{(A)}\ 64 \qquad \text{(B)}\ 94 \qquad \text{(C)}\ 96 \qquad \text{(D)}\ 102 \qquad \text{(E)}\ 140$ | Solution by e_power_pi_times_i
Denote Al's age, Bob's age, and Carl's age by $a$ $b$ , and $c$ , respectively. Then, $a = 16 + b + c$ and $a^2 = 1632 + b^2 + c^2$ . Substituting the first equation into the second, $(16 + b + c)^2 = b^2 + c^2 + 2bc + 32b + 32c + 256 = b^2 + c^2 + 1632$ . Thus, $bc + 16b + 16c = 688$ , and $(b+16)(c+16) = 944$ . Since $944 = 2^4\cdot59$ $(b,c) = (0,43)$ or $(43,0)$ . Then $a + b + c = 2b + 2c + 16 = \boxed{102}$ | D | 102 |
903c429db4de1108fb147c1724fdc0dc | https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_14 | How many pairs $(m,n)$ of integers satisfy the equation $m+n=mn$
$\textbf{(A) }1\qquad \textbf{(B) }2\qquad \textbf{(C) }3\qquad \textbf{(D) }4\qquad \textbf{(E) }\text{more than }4$ | Solution by e_power_pi_times_i
If $m+n=mn$ $mn-m-n = (m-1)(n-1)-1 = 0$ . Then $(m-1)(n-1) = 1$ , and $(m,n) = (2,2) or (0,0)$ . The answer is $\boxed{2}$ | B | 2 |
48f6395388fb31f36ef5c5f9d004276d | https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_18 | If $y=(\log_23)(\log_34)\cdots(\log_n[n+1])\cdots(\log_{31}32)$ then
$\textbf{(A) }4<y<5\qquad \textbf{(B) }y=5\qquad \textbf{(C) }5<y<6\qquad \textbf{(D) }y=6\qquad \\ \textbf{(E) }6<y<7$ | Solution by e_power_pi_times_i
Note that $\log_{a}b = \dfrac{\log{b}}{\log{a}}$ . Then $y=(\dfrac{\log3}{\log2})(\dfrac{\log4}{\log3})\cdots(\dfrac{\log32}{\log31}) = \dfrac{\log32}{\log2} = \log_232 = \boxed{5}$ | B | 5 |
6ff4171d312850f588c07f00287e7b6c | https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_25 | Determine the largest positive integer $n$ such that $1005!$ is divisible by $10^n$
$\textbf{(A) }102\qquad \textbf{(B) }112\qquad \textbf{(C) }249\qquad \textbf{(D) }502\qquad \textbf{(E) }\text{none of the above}\qquad$ | We first observe that since there will be more 2s than 5s in $1005!$ , we are looking for the largest $n$ such that $5^n$ divides $1005!$ . We will use the fact that:
\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor \cdots\]
(This is an application of Legendre's formula).
From $k=5$ and onwards, $\left \lfloor {\frac{1005}{5^k}}\right \rfloor = 0$ . Thus, our calculation becomes
\[n = \left \lfloor {\frac{1005}{5^1}}\right \rfloor + \left \lfloor {\frac{1005}{5^2}}\right \rfloor + \left \lfloor {\frac{1005}{5^3}}\right \rfloor + \left \lfloor {\frac{1005}{5^4}}\right \rfloor\]
\[n = \left \lfloor {\frac{1005}{5}}\right \rfloor + \left \lfloor {\frac{1005}{25}}\right \rfloor + \left \lfloor {\frac{1005}{125}}\right \rfloor + \left \lfloor {\frac{1005}{625}}\right \rfloor\]
\[n = 201 + 40 + 8 + 1 = \boxed{250}\] | null | 250 |
fe6274db0c234102e2f21240d3f5e5cf | https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_28 | Let $g(x)=x^5+x^4+x^3+x^2+x+1$ . What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$ | Let $r(x)$ be the remainder when $g(x^{12})$ is divided by $g(x)$ . Then $r(x)$ is the unique polynomial such that \[g(x^{12}) - r(x) = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} + 1 - r(x)\] is divisible by $g(x) = x^5 + x^4 + x^3 + x^2 + x + 1$ , and $\deg r(x) < 5$
Note that $(x - 1)(x^5 + x^4 + x^3 + x^2 + 1) = x^6 - 1$ is a multiple of $g(x)$ . Also, \[g(x^{12}) - 6 = x^{60} + x^{48} + x^{36} + x^{24} + x^{12} - 5 \\ = (x^{60} - 1) + (x^{48} - 1) + (x^{36} - 1) + (x^{24} - 1) + (x^{12} - 1).\] Each term is a multiple of $x^6 - 1$ . For example, \[x^{60} - 1 = (x^6 - 1)(x^{54} + x^{48} + \dots + x^6 + 1).\] Hence, $g(x^{12}) - 6$ is a multiple of $x^6 - 1$ , which means that $g(x^{12}) - 6$ is a multiple of $g(x)$ . Therefore, the remainder is $\boxed{6}$ . The answer is (A). | null | 6 |
fe6274db0c234102e2f21240d3f5e5cf | https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_28 | Let $g(x)=x^5+x^4+x^3+x^2+x+1$ . What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$ | We express the quotient and remainder as follows. \[g(x^{12}) = Q(x) g(x) + R(x)\] Note that the solutions to $g(x)$ correspond to the 6th roots of unity, excluding $1$ . Hence, we have $x^6 = 1$ , allowing us to set: \[g(x^{12}) = 6\] \[g(x) = 0\] We have $5$ values of $x$ that return $R(x) = 6$ . However, $g(x)$ is quintic, implying the remainder is of degree at most $4$ . Since there are $5$ solutions, the only possibility is that the remainder is a constant $\boxed{6}$ | null | 6 |
fe6274db0c234102e2f21240d3f5e5cf | https://artofproblemsolving.com/wiki/index.php/1977_AHSME_Problems/Problem_28 | Let $g(x)=x^5+x^4+x^3+x^2+x+1$ . What is the remainder when the polynomial $g(x^{12})$ is divided by the polynomial $g(x)$
$\textbf{(A) }6\qquad \textbf{(B) }5-x\qquad \textbf{(C) }4-x+x^2\qquad \textbf{(D) }3-x+x^2-x^3\qquad \\ \textbf{(E) }2-x+x^2-x^3+x^4$ | We can use the Chinese remainder theorem over $\mathbb{Q}[x].$ Since $g(x)=(x+1)(x^2-x+1)(x^2+x+1)$ $\mathbb{Q}[x]/g\cong \mathbb{Q}[x]/(x+1)\times \mathbb{Q}[x]/(x^2-x+1)\times \mathbb{Q}[x]/(x^2+x+1).$ This means that if we can find the remainder of $g(x^{12})$ modulo $x+1,x^2-x+1,x^2+x+1$ , we can reconstruct the remainder modulo $g.$ We can further use that each factor is irreducible and that if $p(x)$ is an irreducible polynomial over $\mathbb{Q}$ with root $\alpha$ $\mathbb{Q}[x]/p\cong \mathbb{Q}(\alpha)$ so to evaluate the remainders of $g(x^{12})$ , we just need to evaluate it on one of the roots of the irreducible factors. The first factor has root $-1$ , the second has roots the primitive sixth roots of unity, and the third as roots the primitive cube roots of unity (this is easily seen as $g(x)(x-1)=x^6-1$ ). Evaluating $g(x^{12})$ on each of these values yields $g(1)=6$ so the remainder is $6$ on each factor on the right of the isomorphism. Hence, by the Chinese remainder theorem, the remainder modulo $g$ must be $\boxed{6}$ as well. | null | 6 |
36ed05b49f80da2a7720d17865c48741 | https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_12 | A supermarket has $128$ crates of apples. Each crate contains at least $120$ apples and at most $144$ apples.
What is the largest integer $n$ such that there must be at least $n$ crates containing the same number of apples?
$\textbf{(A) }4\qquad \textbf{(B) }5\qquad \textbf{(C) }6\qquad \textbf{(D) }24\qquad \textbf{(E) }25$ | To find the largest number of "repeated" crates necessary, we must account for all the possibilities of the number of apples in each crate. Since each crate contains a minimum of $120$ apples and a maximum of $144$ apples, there are $144 - 120 + 1 = 25$ different amounts possible for the number of apples per crate.
Now, we have to count for the worst case scenario: the $25$ amounts are repeated as many times as possible.
$25$ can go into $128$ exactly $5$ times because $5 \cdot 25 = 125$ , which is less than $128$ . This leaves a remainder of $3$ crates.
The worst case scenario would be that these $3$ crates have a different number of apples each. It doesn't actually matter how many apples are in these $3$ crates because any of the $25$ values would be repeated again anyway. So, the answer is $5 + 1 = \boxed{6}$ jiang147369 | C | 6 |
5404e16bbe558b2262a444301dbdd0f3 | https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_14 | The measures of the interior angles of a convex polygon are in arithmetic progression.
If the smallest angle is $100^\circ$ , and the largest is $140^\circ$ , then the number of sides the polygon has is
$\textbf{(A) }6\qquad \textbf{(B) }8\qquad \textbf{(C) }10\qquad \textbf{(D) }11\qquad \textbf{(E) }12$ | Let $n$ equal the number of sides the polygon has. The sum of all the interior angles of a polygon is: $180(n-2)$
The formula for an arithmetic series is $\frac{n(a_1 + a_n)}{2}$ . Set this equal to $180(n-2)$ and solve. In this case, $a_1=100$ and $a_n=140$
Our equation becomes $\frac{n(100+140)}{2} = 180(n-2) \Rightarrow 240n = 360(n-2) \Rightarrow 120n = 720$
Simplifying, we get $n = \boxed{6}$ jiang147369 | A | 6 |
f8f0b690315fdb1f416540e52bd8ddb2 | https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_15 | If $r$ is the remainder when each of the numbers $1059,~1417$ , and $2312$ is divided by $d$ , where $d$ is an integer greater than $1$ , then $d-r$ equals
$\textbf{(A) }1\qquad \textbf{(B) }15\qquad \textbf{(C) }179\qquad \textbf{(D) }d-15\qquad \textbf{(E) }d-1$ | We are given these congruences:
Let's make a new congruence by subtracting (i) from (ii), which results in \[358 \equiv 0 \pmod{d}.\] Subtract (ii) from (iii) to get \[895 \equiv 0 \pmod{d}.\]
Now we know that $358$ and $895$ are both multiples of $d$ . Their prime factorizations are $358=2 \cdot 179$ and $895=5 \cdot 179$ , so their common factor is $179$ , which means $d=179$
Plug $d=179$ back into any of the original congruences to get $r=164$ . Then, $d-r=179-164= \boxed{15}$ . ~ jiang147369 | B | 15 |
835d6e66b4890215ec04346a27714338 | https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_17 | If $\theta$ is an acute angle, and $\sin 2\theta=a$ , then $\sin\theta+\cos\theta$ equals
$\textbf{(A) }\sqrt{a+1}\qquad \textbf{(B) }(\sqrt{2}-1)a+1\qquad \textbf{(C) }\sqrt{a+1}-\sqrt{a^2-a}\qquad\\ \textbf{(D) }\sqrt{a+1}+\sqrt{a^2-a}\qquad \textbf{(E) }\sqrt{a+1}+a^2-a$ | Let $x = \sin\theta+\cos\theta$ , so we want to find $x$ . First, square the expression to get $x^2 = \sin^2 \theta + 2\sin\theta\cos\theta + \cos^2 \theta$ . Recall that $\sin^2 \theta + \cos^2 \theta = 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$ . Plugging these in, the equation simplifies to $x^2 = 1 + \sin 2\theta$ . Given that $\sin 2\theta=a$ , the equation becomes $x^2=1+a$ . Take the square root of both sides to get $x = \sqrt{1+a}$
Hence, the answer is $\boxed{1}$ jiang147369 | A | 1 |
9354a105fb62bcacfdc4c55c22f0d355 | https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_21 | What is the smallest positive odd integer $n$ such that the product $2^{1/7}2^{3/7}\cdots2^{(2n+1)/7}$ is greater than $1000$ ?
(In the product the denominators of the exponents are all sevens, and the numerators are the successive odd integers from $1$ to $2n+1$ .)
$\textbf{(A) }7\qquad \textbf{(B) }9\qquad \textbf{(C) }11\qquad \textbf{(D) }17\qquad \textbf{(E) }19$ | Combine the terms in the product to get $2^{\frac{1+3+5+ \dots +(2n-1)+(2n+1)}{7}}$
The exponent can be simplified to \[\frac{1+3+5+ \dots +(2n-1)+(2n+1)}{7} \Rightarrow \frac{\frac{n(1+(2n+1))}{2}}{7} \Rightarrow \frac{n^2}{7}.\]
We want this inequality to be true with the smallest positive odd integer value of $n$ \[2^{\frac{n^2}{7}} > 1000.\]
Now, let's test the answer choices. For $n=7$ , we have $2^{49/7}=2^{7}<1000$ . For $n=9$ , we have $2^{81/7}>2^{10}>1000$
So our answer is $\boxed{9}$ . ~ jiang147369 | B | 9 |
0e51765d1e2bfded0a97567480841a52 | https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_24 | In the adjoining figure, circle $K$ has diameter $AB$ ; circle $L$ is tangent to circle $K$ and to $AB$ at the center of circle $K$ ; and circle $M$ tangent to circle $K$ , to circle $L$ and $AB$ . The ratio of the area of circle $K$ to the area of circle $M$ is [asy] /* Made by Klaus-Anton, Edited by MRENTHUSIASM */ size(150); pair K=(0,0),B=(1,0),A=(-1,0),L=(0,0.5),M=(sqrt(2)/2,.25); draw(circle(K,1)^^A--B); draw(circle(L,0.5)^^circle(M,.25)); label("$A$", A, W); label("$K$", K, S); label("$B$", B, E); label("$L$", L); label("$M$", M); [/asy] $\textbf{(A) }12\qquad \textbf{(B) }14\qquad \textbf{(C) }16\qquad \textbf{(D) }18\qquad \textbf{(E) }\text{not an integer}$ | Let $R$ and $r$ be the radius of $\odot K$ and the radius of $\odot M,$ respectively. It follows that the radius of $\odot L$ is $\frac{R}{2}.$
Suppose $P$ is the foot of the perpendicular from $M$ to $\overline{KL}.$ We construct the auxiliary lines, as shown below: [asy] /* Made by Klaus-Anton, Edited by MRENTHUSIASM */ size(200); pair K=(0,0),B=(1,0),A=(-1,0),L=(0,0.5),M=(sqrt(2)/2,.25),I=(2*sqrt(2)/3,1/3),E=(sqrt(2)/3,1/3),P=(0,0.25); draw(circle(K,1)^^A--B); draw(circle(L,0.5)^^circle(M,.25)); draw(L--K,red); draw(L--M,red); draw(K--I,red); draw(P--M,red); label("$A$", A, (-5/4,0)); label("$K$", K, (0,-5/4)); label("$B$", B, (5/4,0)); label("$L$", L, (0,5/4)); label("$M$", M, (0,5/4)); label("$P$", P, (-5/4,0)); dot(K,linewidth(4)); dot(L,linewidth(4)); dot(M,linewidth(4)); dot(I,linewidth(4)); dot(E,linewidth(4)); dot(P,linewidth(4)); [/asy] In right $\triangle KPM,$ we have $KP=r$ and $KM=R-r.$ By the Pythagorean Theorem, we get $PM^2=(R-r)^2-r^2.$
In right $\triangle LPM,$ we have $LP=\frac{R}{2}-r$ and $LM=\frac{R}{2}+r.$ By the Pythagorean Theorem, we get $PM^2=\left(\frac{R}{2}+r\right)^2-\left(\frac{R}{2}-r\right)^2.$
We equate the expressions for $PM^2,$ then simplify: \begin{align*} (R-r)^2-r^2&=\left(\frac{R}{2}+r\right)^2-\left(\frac{R}{2}-r\right)^2 \\ \left(R^2-2Rr+r^2\right)-r^2&=\left(\frac{R^2}{4}+Rr+r^2\right)-\left(\frac{R^2}{4}-Rr+r^2\right) \\ R^2-2Rr&=2Rr \\ R^2&=4Rr \\ R&=4r. \end{align*} Therefore, the ratio of the area of $\odot K$ to the area of $\odot M$ is $\frac{\pi R^2}{\pi r^2}=\left(\frac{R}{r}\right)^2=\boxed{16}.$ | C | 16 |
071c0d04f2d33561505c2b41d6bd2902 | https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_27 | If \[N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}},\] then $N$ equals
$\textbf{(A) }1\qquad \textbf{(B) }2\sqrt{2}-1\qquad \textbf{(C) }\frac{\sqrt{5}}{2}\qquad \textbf{(D) }\sqrt{\frac{5}{2}}\qquad \textbf{(E) }\text{none of these}$ | Let $x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}$ and $y=\sqrt{3-2\sqrt{2}}.$
Note that \begin{align*} x^2&=\frac{\left(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}\right)^2}{\left(\sqrt{\sqrt{5}+1}\right)^2} \\ &=\frac{\left(\sqrt{5}+2\right)+2\cdot\left(\sqrt{\sqrt{5}+2}\cdot\sqrt{\sqrt{5}-2}\right)+\left(\sqrt{5}-2\right)}{\sqrt{5}+1} \\ &=\frac{\left(\sqrt{5}+2\right)+2+\left(\sqrt{5}-2\right)}{\sqrt{5}+1} \\ &=\frac{2\sqrt{5}+2}{\sqrt{5}+1} \\ &=2. \end{align*} Since $x>0,$ we have $x=\sqrt{2}.$
On the other hand, note that \begin{align*} y^2&=3-2\sqrt{2} \\ &=2-2\sqrt{2}+1 \\ &=\left(\sqrt{2}-1\right)^2. \end{align*} Since $y>0,$ we have $y=\sqrt{2}-1.$
Finally, the answer is \[N=x-y=\boxed{1}.\] | A | 1 |
071c0d04f2d33561505c2b41d6bd2902 | https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_27 | If \[N=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2\sqrt{2}},\] then $N$ equals
$\textbf{(A) }1\qquad \textbf{(B) }2\sqrt{2}-1\qquad \textbf{(C) }\frac{\sqrt{5}}{2}\qquad \textbf{(D) }\sqrt{\frac{5}{2}}\qquad \textbf{(E) }\text{none of these}$ | Let $x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}$ and $y=\sqrt{3-2\sqrt{2}}.$
Note that \[x=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}\cdot\frac{\sqrt{\sqrt{5}-1}}{\sqrt{\sqrt{5}-1}}=\frac{\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}}{2}. \hspace{15mm} (\bigstar)\] We rewrite each term in the numerator separately:
Substituting these results into $(\bigstar),$ we have \[x=\frac{\left(\frac{\sqrt2}{2}+\frac{\sqrt{10}}{2}\right)+\left(\frac{3\sqrt2}{2}-\frac{\sqrt{10}}{2}\right)}{2}=\sqrt2.\] On the other hand, we have \[y=\sqrt2-1\] by the argument of either Solution 1 or Solution 2.
Finally, the answer is \[N=x-y=\boxed{1}.\] | A | 1 |
5bbb0d3bcbc8c53b35cd19a0459c3c2f | https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_28 | Lines $L_1,L_2,\dots,L_{100}$ are distinct. All lines $L_{4n}, n$ a positive integer, are parallel to each other.
All lines $L_{4n-3}, n$ a positive integer, pass through a given point $A.$ The maximum number of points of intersection of pairs of lines from the complete set $\{L_1,L_2,\dots,L_{100}\}$ is
$\textbf{(A) }4350\qquad \textbf{(B) }4351\qquad \textbf{(C) }4900\qquad \textbf{(D) }4901\qquad \textbf{(E) }9851$ | We partition $\{L_1,L_2,\dots,L_{100}\}$ into three sets. Let \begin{align*} X &= \{L_n\mid n\equiv0\pmod{4}\}, \\ Y &= \{L_n\mid n\equiv1\pmod{4}\}, \\ Z &= \{L_n\mid n\equiv2,3\pmod{4}\}, \\ \end{align*} from which $|X|=|Y|=25$ and $|Z|=50.$
Any two distinct lines can intersect at most once. To maximize the number of points of intersection, note that each point must be passed by exactly two lines. If three or more lines pass through the same point, then we can create more points of intersection by translating the lines.
We construct the sets one by one:
Together, the answer is $626+3725=\boxed{4351}.$ | B | 4351 |
72e3ff4c076c85933503f4432f5f7c3a | https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_29 | Ann and Barbara were comparing their ages and found that Barbara is as old as Ann was when Barbara was as old as
Ann had been when Barbara was half as old as Ann is. If the sum of their present ages is $44$ years, then Ann's age is
$\textbf{(A) }22\qquad \textbf{(B) }24\qquad \textbf{(C) }25\qquad \textbf{(D) }26\qquad \textbf{(E) }28$ | This problem is very wordy. Nonetheless, let $a$ and $b$ be Ann and Barbara's current ages, respectively. We are given that $a+b=44$ . Let $y$ equal the difference between their ages, so $y=a-b$ . Know that $y$ is constant because the difference between their ages will always be the same.
Now, let's tackle the equation: $b=$ Ann's age when Barbara was Ann's age when Barbara was $\frac{a}{2}$ . When Barbara was $\frac{a}{2}$ years old, Ann was $\frac{a}{2}+y$ years old. So the equation becomes $b=$ Ann's age when Barbara was $\frac{a}{2}+y$ . Adding on their age difference again, we get $b = \frac{a}{2} + y + y \Rightarrow b = \frac{a}{2} + 2y$ . Substitute $a-b$ back in for $y$ to get $b = \frac{a}{2} + 2(a-b)$ . Simplify: $2b = a + 4(a-b) \Rightarrow 6b = 5a$ . Solving $b$ in terms of $a$ , we have $b = \frac{5a}{6}$ . Substitute that back into the first equation of $a+b=44$ to get $\frac{11a}{6}=44$ . Solve for $a$ , and the answer is $\boxed{24}$ . ~ jiang147369 | B | 24 |
0eff129931eb91288682c23fc51e2c29 | https://artofproblemsolving.com/wiki/index.php/1976_AHSME_Problems/Problem_30 | How many distinct ordered triples $(x,y,z)$ satisfy the following equations? \begin{align*} x + 2y + 4z &= 12 \\ xy + 4yz + 2xz &= 22 \\ xyz &= 6 \end{align*} $\textbf{(A) }\text{none}\qquad \textbf{(B) }1\qquad \textbf{(C) }2\qquad \textbf{(D) }4\qquad \textbf{(E) }6$ | The first equation suggests the substitution $(a,b,c)=(x,2y,4z),$ from which $(x,y,z)=\left(a,\frac b2,\frac c4\right).$
We rewrite the given equations in terms of $a,b,$ and $c:$ \begin{align*} a + b + c &= 12, \\ \frac{ab}{2} + \frac{bc}{2} + \frac{ac}{2} &= 22, \\ \frac{abc}{8} &= 6. \end{align*} We clear fractions in these equations: \begin{align*} a + b + c &= 12, \\ ab + ac + bc &= 44, \\ abc &= 48. \end{align*} By Vieta's Formulas, note that $a,b,$ and $c$ are the roots of the equation \[r^3 - 12r^2 + 44r - 48 = 0,\] which factors as \[(r - 2)(r - 4)(r - 6) = 0.\] It follows that $\{a,b,c\}=\{2,4,6\}.$ Since the substitution $(x,y,z)=\left(a,\frac b2,\frac c4\right)$ is not symmetric with respect to $x,y,$ and $z,$ we conclude that different ordered triples $(a,b,c)$ generate different ordered triples $(x,y,z),$ as shown below: \[\begin{array}{c|c|c||c|c|c} & & & & & \\ [-2.5ex] \boldsymbol{a} & \boldsymbol{b} & \boldsymbol{c} & \boldsymbol{x} & \boldsymbol{y} & \boldsymbol{z} \\ [0.5ex] \hline & & & & & \\ [-2ex] 2 & 4 & 6 & 2 & 2 & 3/2 \\ 2 & 6 & 4 & 2 & 3 & 1 \\ 4 & 2 & 6 & 4 & 1 & 3/2 \\ 4 & 6 & 2 & 4 & 3 & 1/2 \\ 6 & 2 & 4 & 6 & 1 & 1 \\ 6 & 4 & 2 & 6 & 2 & 1/2 \end{array}\] So, there are $\boxed{6}$ such ordered triples $(x,y,z).$ | E | 6 |
0058ff53d45e2fce596f3fcdfc076dcf | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_2 | For which real values of m are the simultaneous equations
\begin{align*}y &= mx + 3 \\ y& = (2m - 1)x + 4\end{align*}
satisfied by at least one pair of real numbers $(x,y)$
$\textbf{(A)}\ \text{all }m\qquad \textbf{(B)}\ \text{all }m\neq 0\qquad \textbf{(C)}\ \text{all }m\neq 1/2\qquad \textbf{(D)}\ \text{all }m\neq 1\qquad \textbf{(E)}\ \text{no values of }m$ | Solution by e_power_pi_times_i
Solving the systems of equations, we find that $mx+3 = (2m-1)x+4$ , which simplifies to $(m-1)x+1 = 0$ . Therefore $x = \dfrac{1}{1-m}$ $x$ is only a real number if $\boxed{1}$ | D | 1 |
1b598e0be0cdbff244b608599b9a6011 | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_4 | If the side of one square is the diagonal of a second square, what is the ratio of the area of the first square to the area of the second?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ \sqrt2 \qquad \textbf{(C)}\ 1/2 \qquad \textbf{(D)}\ 2\sqrt2 \qquad \textbf{(E)}\ 4$ | Solution by e_power_pi_times_i
Denote the side of one square as $s$ . Then the diagonal of the second square is $s$ , so the side of the second square is $\dfrac{s\sqrt{2}}{2}$ . The area of the second square is $\dfrac{1}{2}s^2$ , so the ratio of the areas is $\dfrac{s^2}{\dfrac{1}{2}s^2} = \boxed{2}$ | A | 2 |
9c486b35ba9caa05d4cfaa7a97bbb0d5 | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_6 | The sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 40 \qquad \textbf{(D)}\ 60 \qquad \textbf{(E)}\ 80$ | Solution by e_power_pi_times_i
When the $n$ th odd positive integer is subtracted from the $n$ th even positive integer, the result is $1$ . Therefore the sum of the first eighty positive odd integers subtracted from the sum of the first eighty positive even integers is $80\cdot1 = \boxed{80}$ | E | 80 |
c734954e1e49cd906d32737c7ba5cf39 | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_9 | Let $a_1, a_2, \ldots$ and $b_1, b_2, \ldots$ be arithmetic progressions such that $a_1 = 25, b_1 = 75$ , and $a_{100} + b_{100} = 100$ .
Find the sum of the first hundred terms of the progression $a_1 + b_1, a_2 + b_2, \ldots$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 100 \qquad \textbf{(C)}\ 10,000 \qquad \textbf{(D)}\ 505,000 \qquad \\ \textbf{(E)}\ \text{not enough information given to solve the problem}$ | Notice that $a_{100}$ and $b_{100}$ are $25+99k_1$ and $75+99k_2$ , respectively. Therefore $k_2 = -k_1$ . Now notice that $a_n + b_n = 25+k_1(n-1)+75+k_2(n-1) = 100+k_1(n-1)-k_1(n-1) = 100$ . The sum of the first $100$ terms is $100\cdot100 = \boxed{10,000}$ | C | 10,000 |
1cd2ba918e20a7189856a264024e46b8 | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_10 | The sum of the digits in base ten of $(10^{4n^2+8}+1)^2$ , where $n$ is a positive integer, is
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 4n \qquad \textbf{(C)}\ 2+2n \qquad \textbf{(D)}\ 4n^2 \qquad \textbf{(E)}\ n^2+n+2$ | We see that the result of this expression will always be in the form $(100\text{ some number of zeros }001)^2.$ Multiplying these together yields: \[110\text{ some number of zeros }011.\] This works because of the way they are multiplied. Therefore, the answer is $\boxed{4}$ | A | 4 |
ae7477f44ac6cab86b8a325139e57985 | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_15 | In the sequence of numbers $1, 3, 2, \ldots$ each term after the first two is equal to the term preceding it minus the term preceding that. The sum of the first one hundred terms of the sequence is
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ -1$ | First, write a few terms of the sequence: $1, 3, 2, -1, -3, -2, 1, 3, 2, \ldots$ Notice how the pattern repeats every six terms and every six terms have a sum of 0. Then, find that the $16*6=96$ th term is $-2$ and the sum of the all those previous terms is $0$ . Then, write the 97th to the 100th terms down: $1, 3, 2,-1$ and add them up to get the sum of $\boxed{5}$ | A | 5 |
3c005d99e0f3928b63016fe89a490d8c | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_17 | A man can commute either by train or by bus. If he goes to work on the train in the morning, he comes home on the bus in the afternoon; and if he comes home in the afternoon on the train, he took the bus in the morning. During a total of $x$ working days, the man took the bus to work in the morning $8$ times, came home by bus in the afternoon $15$ times, and commuted by train (either morning or afternoon) $9$ times. Find $x$
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 16 \qquad \\ \textbf{(E)}\ \text{ not enough information given to solve the problem}$ | The man has three possible combinations of transportation: \[\text{Morning train, Afternoon bus (m.t., a.b.)}\] \[\text{Morning bus, Afternoon train (m.b., a.t.)}\] \[\text{Morning bus, Afternoon bus (m.b, a.b.)}\]
Let $y$ be the number of times the man takes the $\text{a.t.}$ . Then, $9-y$ is the number of times he takes the $\text{m.t.}$ . Keep in mind that $\text{m.b.}=y$ and $\text{a.b.}=9-y$
Let $z$ be the number of times the man takes the $\text{m.b.}$ and $\text{a.b.}$ . Now, we get the two equations \[y+z=8\] and \[9-y+z=15.\]
Solving the system of equations, we get $y=1$ and $z=7$
So during the $x$ working days, the man took the $\text{(m.t., a.b.)}$ on $9-1=8$ days, the $\text{(m.b., a.t.)}$ on $1$ day, and the $\text{(m.b., a.b.)}$ on $7$ days.
Therefore, $x=8+1+7= \boxed{16}$ . ~ jiang147369 | D | 16 |
4030f94fe85e08c0b10c295acfd47af1 | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_19 | Which positive numbers $x$ satisfy the equation $(\log_3x)(\log_x5)=\log_35$
$\textbf{(A)}\ 3 \text{ and } 5 \text{ only} \qquad \textbf{(B)}\ 3, 5, \text{ and } 15 \text{ only} \qquad \\ \textbf{(C)}\ \text{only numbers of the form } 5^n \cdot 3^m, \text{ where } n \text{ and } m \text{ are positive integers} \qquad \\ \textbf{(D)}\ \text{all positive } x \neq 1 \qquad \textbf{(E)}\ \text{none of these}$ | By the change-of-base formula, we can simplify the left side of the equation: $(\log_3x)(\log_x5) = (\frac{\log_x}{\log_3})(\frac{\log_5}{\log_x}) = \frac{\log_5}{\log_3}$
We see that this in fact simplifies to $\log_35$ , which will always equal the right side of the equation, since they are the same exact expressions.
But we have to be careful because $x \neq 1$ . Plugging in $x=1$ , the left side would equal $(\log_31)(\log_x5) = 0 \cdot \log_x5 = 0$ , and $\log_35$ definitely does not equal $0$
Besides $1$ $x$ can take on any positive value, and the equation would work. Therefore, the answer is $\boxed{1}$ jiang147369 | D | 1 |
1408028a070d4feac1efbc0fc92bbd3f | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_27 | If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$ , then $p^3+q^3+r^3$ equals
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$ | If $p$ is a root of $x^3 - x^2 + x - 2 = 0$ , then $p^3 - p^2 + p - 2 = 0$ , or \[p^3 = p^2 - p + 2.\] Similarly, $q^3 = q^2 - q + 2$ , and $r^3 = r^2 - r + 2$ , so \[p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.\]
By Vieta's formulas $p + q + r = 1$ $pq + pr + qr = 1$ , and $pqr = 2$ . Squaring the equation $p + q + r = 1$ , we get \[p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.\] Subtracting $2pq + 2pr + 2qr = 2$ , we get \[p^2 + q^2 + r^2 = -1.\]
Therefore, $p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6 = (-1) - 1 + 6 = \boxed{4}$ | E | 4 |
1408028a070d4feac1efbc0fc92bbd3f | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_27 | If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$ , then $p^3+q^3+r^3$ equals
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$ | We know that $p^3+q^3+r^3=(p+q+r)(p^2+q^2+r^2-pq-qr-pr)+3pqr$ . By Vieta's formulas, $p+q+r=1$ $pqr=2$ , and $pq+qr+pr=1$ .
So if we can find $p^2+q^2+r^2$ , we are done. Notice that $(p+q+r)^2=p^2+q^2+r^2+2pq+2qr+2pr$ , so $p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+pr)=1^2-2\cdot1=-1$ , which means that $p^3+q^3+r^3=1\cdot-2+3\cdot2=\boxed{4}$ | E | 4 |
1408028a070d4feac1efbc0fc92bbd3f | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_27 | If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$ , then $p^3+q^3+r^3$ equals
$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$ | Use Vieta's formulas to get $p+q+r=1$ $pq+qr+pr=1$ , and $pqr=2$
Square $p+q+r=1$ , and get $p^2+q^2+r^2+2pq+2pr+2qr=1$
Substitute $pq+qr+pr=1$ and simplify to get $-1=p^2+q^2+r^2$
After that, multiply both sides by $1=p+q+r$ , to get $-1=p^3+q^3+r^3+p^2q+q^2r+p^2r+q^2r+r^2p+r^2q$
Then, factor out $pq$ $qr$ , and $pr$ $-1=p^3+q^3+r^3+pq(p+q)+qr(q+r)+pr(p+r)$
Then, substitute the first equation into $p+q$ $q+r$ , and $p+r$ $-1=p^3+q^3+r^3+pq(1-r)+qr(1-p)+pr(1-q)$
Then, multiply it out: $-1=p^3+q^3+r^3+pq+qr+pr-3pqr$
After that, substitute the equations $pq+qr+pr=1$ and $pqr=2$ $-1=p^3+q^3+r^3+1-6$
Solving that, you get $p^3+q^3+r^3=\boxed{4}$ ~EZ PZ Ms.Lemon SQUEEZY | E | 4 |
3c3470f1f2eb24dab001f36769e743c8 | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_29 | What is the smallest integer larger than $(\sqrt{3}+\sqrt{2})^6$
$\textbf{(A)}\ 972 \qquad \textbf{(B)}\ 971 \qquad \textbf{(C)}\ 970 \qquad \textbf{(D)}\ 969 \qquad \textbf{(E)}\ 968$ | $(\sqrt{3}+\sqrt{2})^6=(5+2\sqrt{6})^3=(5+2\sqrt{6})(49+20\sqrt{6})=(485+198\sqrt{6})$ Then, find that $\sqrt{6}$ is about $2.449$ . Finally, multiply and add to find that the smallest integer higher is $\boxed{970}$ | C | 970 |
3c3470f1f2eb24dab001f36769e743c8 | https://artofproblemsolving.com/wiki/index.php/1975_AHSME_Problems/Problem_29 | What is the smallest integer larger than $(\sqrt{3}+\sqrt{2})^6$
$\textbf{(A)}\ 972 \qquad \textbf{(B)}\ 971 \qquad \textbf{(C)}\ 970 \qquad \textbf{(D)}\ 969 \qquad \textbf{(E)}\ 968$ | Let's evaluate $(\sqrt{3}+\sqrt{2})^6 + (\sqrt{3}-\sqrt{2})^6$ . We see that all the irrational terms cancel. Then, using binomial theorem, we evaluate all the rational terms in the first expression to get 485. Then, the sum of the rational parts of the 2nd term will be 485 as well. Then, we get a total of 970 and since $(\sqrt{3}-\sqrt{2})^6) < 1$ , the greatest integer greater than our original expression is $\boxed{970}$ | C | 970 |
762022367f036015ed03382e40c29ef3 | https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_2 | Let $x_1$ and $x_2$ be such that $x_1\not=x_2$ and $3x_i^2-hx_i=b$ $i=1, 2$ . Then $x_1+x_2$ equals
$\mathrm{(A)\ } -\frac{h}{3} \qquad \mathrm{(B) \ }\frac{h}{3} \qquad \mathrm{(C) \ } \frac{b}{3} \qquad \mathrm{(D) \ } 2b \qquad \mathrm{(E) \ }-\frac{b}{3}$ | Notice that $x_1$ and $x_2$ are the distinct solutions to the quadratic $3x^2-hx-b=0$ . By Vieta, the sum of the roots of this quadratic is the negation of the coefficient of the linear term divided by the coefficient of the quadratic term, so in this case $-\frac{-h}{3}=\frac{h}{3}, \boxed{3}$ | B | 3 |
1b518fe8829494f0861f4cb7fb70a427 | https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_4 | What is the remainder when $x^{51}+51$ is divided by $x+1$
$\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 49 \qquad \mathrm{(D) \ } 50 \qquad \mathrm{(E) \ }51$ | From the Remainder Theorem , the remainder when $x^{51}+51$ is divided by $x+1$ is $(-1)^{51}+51=-1+51=50, \boxed{50}$ | D | 50 |
92e3933b8909ac8628b76d732d259fe5 | https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_5 | Given a quadrilateral $ABCD$ inscribed in a circle with side $AB$ extended beyond $B$ to point $E$ , if $\measuredangle BAD=92^\circ$ and $\measuredangle ADC=68^\circ$ , find $\measuredangle EBC$
$\mathrm{(A)\ } 66^\circ \qquad \mathrm{(B) \ }68^\circ \qquad \mathrm{(C) \ } 70^\circ \qquad \mathrm{(D) \ } 88^\circ \qquad \mathrm{(E) \ }92^\circ$ | Since $ABCD$ is cyclic, opposite angles must sum to $180^\circ$ . Therefore, $\angle ADC+\angle ABC=180^\circ$ , and $\angle ABC=180^\circ-\angle ADC=180^\circ-68^\circ=112^\circ$ . Notice also that $\angle ABC$ and $\angle CBE$ form a linear pair, and so they sum to $180^\circ$ . Therefore, $\angle EBC=180^\circ-\angle ABC=180^\circ-112^\circ=68^\circ, \boxed{68}$ . Notice that the answer didn't even depend on $\angle BAD$ | B | 68 |
0d1b64d6ac8e3ad5ae4df5ad666c1589 | https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_10 | What is the smallest integral value of $k$ such that \[2x(kx-4)-x^2+6=0\] has no real roots?
$\mathrm{(A)\ } -1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ }5$ | Expanding, we have $2kx^2-8x-x^2+6=0$ , or $(2k-1)x^2-8x+6=0$ . For this quadratic not to have real roots, it must have a negative discriminant. Therefore, $(-8)^2-4(2k-1)(6)<0\implies 64-48k+24<0\implies k>\frac{11}{6}$ . From here, we can easily see that the smallest integral value of $k$ is $2, \boxed{2}$ | B | 2 |
bc4a4e156be9c2fb0569e51deebd4301 | https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_17 | If $i^2=-1$ , then $(1+i)^{20}-(1-i)^{20}$ equals
$\mathrm{(A)\ } -1024 \qquad \mathrm{(B) \ }-1024i \qquad \mathrm{(C) \ } 0 \qquad \mathrm{(D) \ } 1024 \qquad \mathrm{(E) \ }1024i$ | Notice that $(1+i)^2=2i$ and $(1-i)^2=-2i$ . Therefore,
\[(1+i)^{20}-(1-i)^{20}=(2i)^{10}-(-2i)^{10}=(2i)^{10}-(2i)^{10}=0, \boxed{0}.\] | C | 0 |
66ba260d0ae5eeaa87fe9262cc9185f0 | https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_29 | For $p=1, 2, \cdots, 10$ let $S_p$ be the sum of the first $40$ terms of the arithmetic progression whose first term is $p$ and whose common difference is $2p-1$ ; then $S_1+S_2+\cdots+S_{10}$ is
$\mathrm{(A)\ } 80000 \qquad \mathrm{(B) \ }80200 \qquad \mathrm{(C) \ } 80400 \qquad \mathrm{(D) \ } 80600 \qquad \mathrm{(E) \ }80800$ | The $40\text{th}$ term of an arithmetic progression with a first term $p$ and a common difference $2p-1$ is $p+39(2p-1)=79p-39$ . Therefore, the sum of the first $40$ terms of such a progression is $\frac{40}{2}(79p-39+p)=1600p-780$
We now want to evaluate $\sum_{p=1}^{10}(1600p-780)$ \[\sum_{p=1}^{10}(1600p-780)=1600\sum_{p=1}^{10}(p)-\sum_{p=1}^{10}(780)\] \[=(1600)\left(\frac{10\cdot11}{2}\right)-(780)(10)=88000-7800=80200, \boxed{80200}.\] | B | 80200 |
e81703ae149e2f0cfa8260fa3e9666b2 | https://artofproblemsolving.com/wiki/index.php/1974_AHSME_Problems/Problem_30 | A line segment is divided so that the lesser part is to the greater part as the greater part is to the whole. If $R$ is the ratio of the lesser part to the greater part, then the value of
\[R^{[R^{(R^2+R^{-1})}+R^{-1}]}+R^{-1}\]
is
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }2R \qquad \mathrm{(C) \ } R^{-1} \qquad \mathrm{(D) \ } 2+R^{-1} \qquad \mathrm{(E) \ }2+R$ | Let $w$ be the length of the shorter segment and $l$ be the length of the longer segment. We're given that $\frac{w}{l}=\frac{l}{w+l}$ . Cross-multiplying, we find that $w^2+wl=l^2\implies w^2+wl-l^2=0$ . Now we divide both sides by $l^2$ to get $\left(\frac{w}{l}\right)^2+\left(\frac{w}{l}\right)-1=0$ . Therefore, $R^2+R-1=0$
From this, we have $R^2=-R+1$ . Dividing both sides by $R$ , we get $R=-1+\frac{1}{R}\implies R^{-1}=R+1$ . Therefore, $R^2+R^{-1}=-R+1+R+1=2$ . Finally, we have \[R^{[R^{(R^2+R^{-1})}+R^{-1}]}+R^{-1}=R^{[R^2+R^{-1}]}+R^{-1}=R^2+R^{-1}= \boxed{2}.\] | A | 2 |
d217a7d5eb009e4284ce1734ea8e94b7 | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_2 | One thousand unit cubes are fastened together to form a large cube with edge length 10 units; this is painted and then separated into the original cubes. The number of these unit cubes which have at least one face painted is
$\textbf{(A)}\ 600\qquad\textbf{(B)}\ 520\qquad\textbf{(C)}\ 488\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 400$ | The total number of cubes is $10^3$ or $1000$ . Because each surface of the large cube is one cube deep, the number of the unpainted cubes is $8^3 = 512$ , since we subtract two from the side lengths of the cube itself, and cube it to find the volume of that cube. So there are $1000-512=\boxed{488}$ cubes that have at least one face painted. | C | 488 |
d217a7d5eb009e4284ce1734ea8e94b7 | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_2 | One thousand unit cubes are fastened together to form a large cube with edge length 10 units; this is painted and then separated into the original cubes. The number of these unit cubes which have at least one face painted is
$\textbf{(A)}\ 600\qquad\textbf{(B)}\ 520\qquad\textbf{(C)}\ 488\qquad\textbf{(D)}\ 480\qquad\textbf{(E)}\ 400$ | Each face has $100$ cubes, so multiply by six to get $600$ . However, we overcounted each small cube on the edge but not on corner of the big cube once and each small cube on the corner of the big cube twice. Thus, there are $600 - (12 \cdot 8 + 2 \cdot 8) = \boxed{488}$ cubes that have at least one face painted. | C | 488 |
1371b00661a328f65b30db75c0c948fb | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_3 | The stronger Goldbach conjecture states that any even integer greater than 7 can be written as the sum of two different prime numbers. For such representations of the even number 126, the largest possible difference between the two primes is
$\textbf{(A)}\ 112\qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 92\qquad\textbf{(D)}\ 88\qquad\textbf{(E)}\ 80$ | We can guess and check small primes, subtract it from $126$ , and see if the result is a prime because the further away the two numbers are, the greater the difference will be. Since $126 = 2 \cdot 3^2 \cdot 7$ , we can eliminate $2$ $3$ , and $7$ as an option because subtracting these would result in a composite number.
If we subtract $5$ , then the resulting number is $121$ , which is not prime. If we subtract $11$ , then the resulting number is $115$ , which is also not prime. But when we subtract $13$ , the resulting number is $113$ , a prime number. The largest possible difference is $113-13=\boxed{100}$ | B | 100 |
ad5924f74b82785c37c3907741b74b0f | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_6 | If 554 is the base $b$ representation of the square of the number whose base $b$ representation is 24, then $b$ , when written in base 10, equals
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 8\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 14\qquad\textbf{(E)}\ 16$ | Write out the numbers using the definition of base numbers \[554_b = 5b^2 + 5b + 4\] \[24_b = 2b+4\] Since $554_b = (24_b)^2$ , we can write an equation. \[5b^2 + 5b + 4 = (2b+4)^2\] \[5b^2 + 5b + 4 = 4b^2 + 16b + 16\] \[b^2 - 11b - 12 = 0\] \[(b-12)(b+1) = 0\] Since base numbers must be positive, $b$ in base 10 equals $\boxed{12}$ | C | 12 |
44e3fd8a71e69cf4dfbf51943b2c279e | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_7 | The sum of all integers between 50 and 350 which end in 1 is
$\textbf{(A)}\ 5880\qquad\textbf{(B)}\ 5539\qquad\textbf{(C)}\ 5208\qquad\textbf{(D)}\ 4877\qquad\textbf{(E)}\ 4566$ | The numbers that we are adding are $51,61,71 \cdots 341$ . The numbers are part of an arithmetic series with first term $51$ , last term $341$ , common difference $10$ , and $30$ terms. Using the arithmetic series formula, the sum of the terms is $\tfrac{30 \cdot 392}{2} = \boxed{5880}$ | A | 5880 |
eaf01afe7cc0cc9c39bd4159797ea01f | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_8 | If 1 pint of paint is needed to paint a statue 6 ft. high, then the number of pints it will take to paint (to the same thickness) 540 statues similar to the original but only 1 ft. high is
$\textbf{(A)}\ 90\qquad\textbf{(B)}\ 72\qquad\textbf{(C)}\ 45\qquad\textbf{(D)}\ 30\qquad\textbf{(E)}\ 15$ | The statues are similar, and since the height if the smaller statue is $\tfrac{1}{6}$ of the original statue, the surface area is $\tfrac{1}{36}$ of the original statue. Thus, $\tfrac{1}{36}$ pints of paint is needed for one 1 ft. statue, so painting 540 of these statues requires $\tfrac{540}{36} = \boxed{15}$ pints of paint. | E | 15 |
9bb1f9a33bf6e002c420859474ab83af | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_9 | In $\triangle ABC$ with right angle at $C$ , altitude $CH$ and median $CM$ trisect the right angle. If the area of $\triangle CHM$ is $K$ , then the area of $\triangle ABC$ is
$\textbf{(A)}\ 6K\qquad\textbf{(B)}\ 4\sqrt3\ K\qquad\textbf{(C)}\ 3\sqrt3\ K\qquad\textbf{(D)}\ 3K\qquad\textbf{(E)}\ 4K$ | [asy] pair A=(-6,0),B=(6,0),C=(-3,5.196),M=(0,0),H=(-3,0); draw((-6,0)--(6,0)--(-3,5.196)--(-6,0)); draw((-3,5.196)--(0,0)); draw(C--H); dot(A); label("$A$",A,SW); dot(B); label("$B$",B,SE); dot(C); label("$C$",C,N); dot(M); label("$M$",M,S); dot(H); label("$H$",H,S); markscalefactor=0.1; draw(anglemark((-6,0),C,(6,0))); draw((-3,0.5)--(-2.5,0.5)--(-2.5,0)); [/asy]
Draw diagram as shown (note that $A$ and $B$ can be interchanged, but it doesn’t change the solution).
Note that because $CM$ is a median, $AM = BM$ . Also, by ASA Congruency, $\triangle CHA = \triangle CHM$ , so $AH = HM$ . That means $HM = \tfrac{1}{4} \cdot AB$ , and since $\triangle CHM$ and $\triangle ABC$ share an altitude, $[ABC] = \boxed{4}$ | E | 4 |
12a52557cdfc2af5c3b3df1a0642386a | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_14 | Each valve $A$ $B$ , and $C$ , when open, releases water into a tank at its own constant rate. With all three valves open, the tank fills in 1 hour, with only valves $A$ and $C$ open it takes 1.5 hours, and with only valves $B$ and $C$ open it takes 2 hours. The number of hours required with only valves $A$ and $B$ open is
$\textbf{(A)}\ 1.1\qquad\textbf{(B)}\ 1.15\qquad\textbf{(C)}\ 1.2\qquad\textbf{(D)}\ 1.25\qquad\textbf{(E)}\ 1.75$ | Let the rate of water flowing through valve $A$ be $a$ , the rate of water flowing through valve $B$ be $b$ , and the rate of water flowing through valve $C$ be $c$ WLOG , let the volume of the tank be 1 liter, and let the units for the rates be liters per hour. With this information, we can write three equations. \[\frac{1}{a+b+c} = 1\] \[\frac{1}{a+c} = \frac{3}{2}\] \[\frac{1}{b+c} = 2\] Manipulate each equation to get \[1 = a+b+c\] \[\frac{2}{3} = a+c\] \[\frac{1}{2} = b+c\] Solving for $a$ yields $a = \tfrac{1}{2}$ , and solving for $b$ yields $b = \tfrac{1}{3}$ . The number of hours to fill the tub with only valves $A$ and $B$ on is $\frac{1}{\frac{1}{2} + \frac{1}{3}} = \frac{6}{5} = \boxed{1.2}$ | C | 1.2 |
f0529fa6de081e85a6001f0b8716fa76 | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_16 | If the sum of all the angles except one of a convex polygon is $2190^{\circ}$ , then the number of sides of the polygon must be
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 15 \qquad \textbf{(C)}\ 17 \qquad \textbf{(D)}\ 19 \qquad \textbf{(E)}\ 21$ | Let $n$ be the number of sides in the polygon. The number of interior angles in the polygon is $180(n-2)$ . We know that the sum of all but one of them is $2190^{\circ}$ , so the sum of all the angles is more than that. \[180(n-2) > 2190\] \[n-2 > 12 \tfrac{1}{6}\] \[n > 14 \tfrac{1}{6}\]
The sum of the angles in a 15-sided polygon is $2340^{\circ}$ , making the remaining angle $150^{\circ}$ . The angles of a convex polygon are all less than $180^{\circ}$ , and since adding one more side means adding $180^{\circ}$ to the measure of the remaining angle, we can confirm that there are $\boxed{15}$ sides in the polygon. | B | 15 |
d4843f348332f7cd1c73a1e587b51cb4 | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_21 | The number of sets of two or more consecutive positive integers whose sum is 100 is
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | If the first number of a group of $n$ consecutive numbers is $a$ , the $n^\text{th}$ number is $a+n-1$ . We know that the sum of the group of numbers is $100$ , so \[\frac{n(2a+n-1)}{2} = 100\] \[2a+n-1=\frac{200}{n}\] \[2a = 1-n + \frac{200}{n}\] We know that $n$ and $a$ are positive integers, so we check values of $n$ that are a factor of $200$ . Of these values, the only ones that result in a positive integer $a$ is when $n = 5$ or when $n = 8$ , so there are $\boxed{2}$ sets of two or more consecutive positive integers whose sum is $100$ | B | 2 |
b4df296a4e87fd6278c06a1887360e47 | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_23 | There are two cards; one is red on both sides and the other is red on one side and blue on the other. The cards have the same probability (1/2) of being chosen, and one is chosen and placed on the table. If the upper side of the card on the table is red, then the probability that the under-side is also red is
$\textbf{(A)}\ \frac14 \qquad \textbf{(B)}\ \frac13 \qquad \textbf{(C)}\ \frac12 \qquad \textbf{(D)}\ \frac23 \qquad \textbf{(E)}\ \frac34$ | There are three red faces, and two are on the card that is completely red, so our answer is $\frac{2}{3}$ , which is $\boxed{23}$ | D | 23 |
d81b6a18d85b475171f9c82ffd37410a | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_24 | The check for a luncheon of 3 sandwiches, 7 cups of coffee and one piece of pie came to $$3.15$ . The check for a luncheon consisting of 4 sandwiches, 10 cups of coffee and one piece of pie came to $$4.20$ at the same place. The cost of a luncheon consisting of one sandwich, one cup of coffee, and one piece of pie at the same place will come to
$\textbf{(A)}\ $1.70 \qquad \textbf{(B)}\ $1.65 \qquad \textbf{(C)}\ $1.20 \qquad \textbf{(D)}\ $1.05 \qquad \textbf{(E)}\ $0.95$ | Let $s$ be the cost of one sandwich, $c$ be the cost of one cup of coffee, and $p$ be the price of one piece of pie. With the information,
\[3s+7c+p=3.15\] \[4s+10c+p=4.20\]
Subtract the first equation from the second to get
\[s+3c=1.05\]
That means $s=1.05-3c$ . Substituting it back in the second equation results in.
\[4.20-12c+10c+p=4.20\]
Solving for $p$ yields $p=2c$ . With the substitutions, the cost of one sandwich, one cup of coffee, and one slice of pie is $(1.05-3c)+c+(2c) = \boxed{1.05}$ | D | 1.05 |
cd65b01f0b2199df1dd31ddd54796d7a | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_26 | The number of terms in an A.P. (Arithmetic Progression) is even. The sum of the odd and even-numbered terms are 24 and 30, respectively. If the last term exceeds the first by 10.5, the number of terms in the A.P. is
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 18 \qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 8$ | Let $a$ be the first term, $n$ be the number of terms, and $d$ be the common difference. That means the last term is $a+r(n-1)$
We can write an equation on the difference between the last and first term based on the conditions. \[a+r(n-1)-a =10.5\] \[rn-r=10.5\] Also, half of the terms add up to $24$ while the other half of the terms add up to $30$ , so \[24 + r\frac{n}{2} = 30\] \[nr = 12\] Substituting the value back to a previous equation, \[12-r=10.5\] \[r=1.5\] Substituting to a previous equation again, \[1.5n-1.5=10.5\] \[n=8\] Thus, there are $\boxed{8}$ terms in the arithmetic sequence. | E | 8 |
378f4fbbbcdc4dcecef4b9e84908ef74 | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_29 | Two boys start moving from the same point A on a circular track but in opposite directions. Their speeds are 5 ft. per second and 9 ft. per second. If they start at the same time and finish when they first meet at the point A again, then the number of times they meet, excluding the start and finish, is
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 25 \qquad \textbf{(C)}\ 44 \qquad \textbf{(D)}\ \text{infinity} \qquad \textbf{(E)}\ \text{none of these}$ | Let $d$ be the length of the track in feet and $x$ be the number of laps that one of the boys did, so time one of the boys traveled before the two finish is $\tfrac{dx}{5}$ . Since the time elapsed for both boys is equal, one boy ran $5(\tfrac{dx}{5})$ feet while the other boy ran $9(\tfrac{dx}{5})$ feet. Because both finished at the starting point, both ran an integral number of laps, so $5(\tfrac{dx}{5})$ and $9(\tfrac{dx}{5})$ are multiples of $d$ . Because both stopped when both met at the start for the first time, $x = 5$
Note that between the time a runner finishes a lap and a runner (can be same) finishes a lap, both runners must meet each other. When $0 < x \le 5$ and either $5(\tfrac{dx}{5})$ or $9(\tfrac{dx}{5})$ is a multiple of $d$ , one of the runners completed a lap. This is achieved when $x = \tfrac59, 1, \tfrac{10}{9}, \tfrac{15}{9}, 2, \tfrac{20}{9}, \tfrac{25}{9}, 3, \tfrac{30}{9}, \tfrac{35}{9}, 4, \tfrac{40}{9}, 5$ , so the two meet each other (excluding start and finish) a total of $\boxed{13}$ times. | A | 13 |
892fb36261ecd96de1561f0c25f39684 | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_31 | In the following equation, each of the letters represents uniquely a different digit in base ten:
\[(YE) \cdot (ME) = TTT\]
The sum $E+M+T+Y$ equals
$\textbf{(A)}\ 19 \qquad \textbf{(B)}\ 20 \qquad \textbf{(C)}\ 21 \qquad \textbf{(D)}\ 22 \qquad \textbf{(E)}\ 24$ | The right side of the equation can be rewritten as $111T = 37 \cdot 3T$ . With trial and error and prime factorization as a guide, we can test different digits of $T$ to see if we can find two two-digit numbers that have the same units digit and multiply to $111T$
The only possibility that works is $37 \cdot 27 = 999$ . That means $E+M+T+Y = \boxed{21}$ | C | 21 |
f344e9a6b5d3a8393842361bf0382142 | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_32 | The volume of a pyramid whose base is an equilateral triangle of side length 6 and whose other edges are each of length $\sqrt{15}$ is
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 9/2 \qquad \textbf{(C)}\ 27/2 \qquad \textbf{(D)}\ \frac{9\sqrt3}{2} \qquad \textbf{(E)}\ \text{none of these}$ | [asy] import three; unitsize(1cm); size(200); draw((0,0,0)--(6,0,0)--(3,5.196,0)--(0,0,0)); draw((3,1.732,1.732)--(0,0,0)); draw((3,1.732,1.732)--(6,0,0)); draw((3,1.732,1.732)--(3,5.196,0)); draw((3,1.732,1.732)--(3,1.732,0)--(0,0,0),dotted); label("6",(4.5,2.598,0),SW); label("$\sqrt{15}$",(4.5,0.866,0.866),N); currentprojection=orthographic(1/6,1/2,1/3); [/asy]
Draw an altitude towards the equilateral triangle base. By symmetry (this can also be proved by HL), the base of the altitude is equidistant from the three points of the equilateral triangle. This means that the distance from the base of the altitude to one of the points of the equilateral triangle is $2\sqrt{3}$
[asy] draw((0,1.732)--(0,0)--(3.464,0),dotted); draw((0,1.732)--(3.464,0)); label("$2\sqrt{3}$",(1.732,0),S); label("$\sqrt{15}$",(1.732,0.866),NE); [/asy]
Using the Pythagorean Theorem , the length of the altitude is $\sqrt{3}$ , so the volume of the triangular pyramid is $\tfrac13 \cdot \tfrac{6^2 \cdot \sqrt{3}}{4} \cdot \sqrt{3} = \boxed{9}$ | A | 9 |
33246f8bc8a41b809e44e05a132b2b04 | https://artofproblemsolving.com/wiki/index.php/1973_AHSME_Problems/Problem_33 | When one ounce of water is added to a mixture of acid and water, the new mixture is $20\%$ acid. When one ounce of acid is added to the new mixture, the result is $33\frac13\%$ acid. The percentage of acid in the original mixture is
$\textbf{(A)}\ 22\% \qquad \textbf{(B)}\ 24\% \qquad \textbf{(C)}\ 25\% \qquad \textbf{(D)}\ 30\% \qquad \textbf{(E)}\ 33\frac13 \%$ | Let $a$ be the original number of ounces of acid and $w$ be the original number of ounces of water. We can write two equations since we know the percentage of acid after some water and acid. \[\frac{a}{a+w+1} = \frac{1}{5}\] \[\frac{a+1}{a+w+2} = \frac{1}{3}\] Cross-multiply to get rid of the fractions. \[5a = a+w+1\] \[3a+3=a+w+1\] Solve the system to get $a=1$ and $w=3$ . The percentage of acid in the original mixture is $\tfrac{1}{1+3} = \boxed{25}$ | C | 25 |
c670021124b1295a0fe4626fd69c1450 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_3 | If $x=\dfrac{1-i\sqrt{3}}{2}$ where $i=\sqrt{-1}$ , then $\dfrac{1}{x^2-x}$ is equal to
$\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1+i\sqrt{3}\qquad \textbf{(D) }1\qquad \textbf{(E) }2$ | Using DeMoivre's theorem, we can calculate $x^2=\frac{1+i\sqrt{3}}{2}$ The denominator is therefore $-1$ which makes the answer \[\boxed{1}.\] ~lopkiloinm | C | 1 |
e49fde5d9f1a73ba06c2c8205524a6c8 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_4 | 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 | D | 8 |
4ca2bf599f1693b695997dd52d9e4774 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_24 | 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,
\[\frac{4t}{5}(x+\frac{1}{2})=xt\]
we note that we can solve for $x$ . Solving for $x$ , we get $x=2.$
Now we look at the last two parts of the equation:
\[xt=(t+\frac{5}{2})(x-\frac{1}{2})\]
we note that we can solve for $t$ and we get $t=\frac{15}{2}.$ We want the find the distance, which is $xt= \boxed{15}.$ | null | 15 |
967c7919df1b7c29bd74d08ae0eb2354 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_25 | 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 $\boxed{65}$ | C | 65 |
e32d39ef4b569c1bf20ce7e6703993d9 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_28 | 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 total. | null | 32 |
86b53c8a0b379c7622195ac1895534e0 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_31 | 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 $\boxed{3}$ | C | 3 |
f5d412643d4e3c24c025e3ef5d09459f | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_33 | The minimum value of the quotient of a (base ten) number of three different non-zero digits divided by the sum of its digits is
$\textbf{(A) }9.7\qquad \textbf{(B) }10.1\qquad \textbf{(C) }10.5\qquad \textbf{(D) }10.9\qquad \textbf{(E) }20.5$ | The answer we are looking for can be expressed as $\dfrac{100a+10b+c}{a+b+c}$ . This is equivalent to $1 + \dfrac{99a+9b}{a+b+c}$ . Because we are trying to minimize our solution, we set $c$ $9$ , so we have $1 + \dfrac{99a+9b}{a+b+9}$ . This is equal to $1 + \dfrac{9a+9b+81}{a+b+9} + \dfrac{90a-81}{a+b+9}$ , which simplifies to $10+ \dfrac{90a-81}{a+b+9}$ . Since each digit is unique, we set $b$ to $8$ , leaving us with $10 + \dfrac{90a-81}{a+17}$ . Clearly, $a$ should be minimized, so $a = 1$ and our answer is \[\boxed{10.5}.\] | C | 10.5 |
8787784c6e22d5b4a4790f145bb068d7 | https://artofproblemsolving.com/wiki/index.php/1972_AHSME_Problems/Problem_34 | 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
$\textbf{(A) }42\qquad \textbf{(B) }46\qquad \textbf{(C) }122\qquad \textbf{(D) }290\qquad \textbf{(E) }326$ | \[t=2h-3d\] \[3d^3+t^3=2h^3\]
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 answer is thus $1+16+25=\boxed{42}.$ ~lopkiloinm | A | 42 |