competition_id string | problem_id int64 | difficulty int64 | category string | problem_type string | problem string | solutions list | solutions_count int64 | source_file string | competition string |
|---|---|---|---|---|---|---|---|---|---|
1984_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | In $\triangle ABC$, $D$ is on $AC$ and $F$ is on $BC$. Also, $AB\perp AC$, $AF\perp BC$, and $BD=DC=FC=1$. Find $AC$.
$\mathrm{(A) \ }\sqrt{2} \qquad \mathrm{(B) \ }\sqrt{3} \qquad \mathrm{(C) \ } \sqrt[3]{2} \qquad \mathrm{(D) \ }\sqrt[3]{3} \qquad \mathrm{(E) \ } \sqrt[4]{3}$
| [
"[asy] unitsize(4cm); draw((0,0)--(2^(1/3),0)); draw((0,0)--(0,16^(1/3)-4^(1/3))); draw((2^(1/3),0)--(0,16^(1/3)-4^(1/3))); draw((0,16^(1/3)-4^(1/3))--(2^(1/3)-1,0)); draw((0,0)--(0.445861,0.602489)); label(\"$A$\",(0,0),W); label(\"$B$\",(0,16^(1/3)-4^(1/3)),N); label(\"$C$\",(2^(1/3),0),E); label(\"$D$\",(2^(1/3)... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/27.json | AHSME |
1984_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | $\frac{1000^2}{252^2-248^2}$ equals
$\mathrm{(A) \ }62,500 \qquad \mathrm{(B) \ }1,000 \qquad \mathrm{(C) \ } 500\qquad \mathrm{(D) \ }250 \qquad \mathrm{(E) \ } \frac{1}{2}$
| [
"We can use difference of squares to factor the denominator, yielding:\n\n\n$\\frac{1000^2}{252^2-248^2}=\\frac{1000^2}{(252-248)(252+248)}=\\frac{1000^2}{(4)(500)}=\\frac{1000^2}{2000}$.\n\n\nWe see that the $1000$ in the denominator cancels with one of the $1000$s in the numerator, yielding $500, \\boxed{\\text{C... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/1.json | AHSME |
1984_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | A calculator has a key that replaces the displayed entry with its square, and another key which replaces the displayed entry with its reciprocal. Let $y$ be the final result when one starts with a number $x\not=0$ and alternately squares and reciprocates $n$ times each. Assuming the calculator is completely accurate (e.g. no roundoff or overflow), then $y$ equals
$\mathrm{(A) \ }x^{((-2)^n)} \qquad \mathrm{(B) \ }x^{2n} \qquad \mathrm{(C) \ } x^{-2n} \qquad \mathrm{(D) \ }x^{-(2^n)} \qquad \mathrm{(E) \ } x^{((-1)^n2n)}$
| [
"Notice that taking the reciprocal of a number is equivalent to raising the number to the $-1st$ power. Therefore, taking the reciprocal of a number $x$ and then squaring it is $(x^{-1})^2=x^{-2}$. Also, since multiplication is commutative, this is equivalent to squaring the number then taking the reciprocal of it.... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/11.json | AHSME |
1984_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | If $x, y$, and $y-\frac{1}{x}$ are not $0$, then
$\frac{x-\frac{1}{y}}{y-\frac{1}{x}}$ equals
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }\frac{x}{y} \qquad \mathrm{(C) \ } \frac{y}{x}\qquad \mathrm{(D) \ }\frac{x}{y}-\frac{y}{x} \qquad \mathrm{(E) \ } xy-\frac{1}{xy}$
| [
"Multiply the expression by $\\frac{xy}{xy}$ to get rid of the fractional numerator and denominator: $\\frac{x^2y-x}{xy^2-y}$. This can be factored as $\\frac{x(xy-1)}{y(xy-1)}$. The $xy-1$ terms cancel out, leaving $\\frac{x}{y}, \\boxed{\\text{B}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/2.json | AHSME |
1984_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | The number of distinct pairs of integers $(x, y)$ such that $0<x<y$ and $\sqrt{1984}=\sqrt{x}+\sqrt{y}$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ }4 \qquad \mathrm{(E) \ } 7$
| [
"We can simplify $\\sqrt{1984}$ to $8\\sqrt{31}$. Therefore, the only solutions are $a\\sqrt{31}+b\\sqrt{31}$ such that $a+b=8$ and $0<a<b$. The only solutions to this are $a=1, b=7; a=2, b=6; a=3, b=5$. Each of these gives distinct pairs of $(x, y)$, so there are $3$ pairs, $\\boxed{\\text{C}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/28.json | AHSME |
1984_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | If the sequence $\{a_n\}$ is defined by
$a_1=2$
$a_{n+1}=a_n+2n$
where $n\geq1$.
Then $a_{100}$ equals
$\mathrm{(A) \ }9900 \qquad \mathrm{(B) \ }9902 \qquad \mathrm{(C) \ } 9904 \qquad \mathrm{(D) \ }10100 \qquad \mathrm{(E) \ } 10102$
| [
"We begin to evaluate the first couple of terms of the sequence, hoping to find a pattern:\n$2, 4, 8, 14, 22, ....$. We notice that the difference between succesive terms of the sequence are $2, 4, 6, 8, ....$, a clear pattern. We can see that this pattern continues infinitely because of the recursive definition: e... | 2 | ./CreativeMath/AHSME/1984_AHSME_Problems/12.json | AHSME |
1984_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | If $a$ and $b$ are positive real numbers and each of the equations $x^2+ax+2b=0$ and $x^2+2bx+a=0$ has real roots, then the smallest possible value of $a+b$ is
$\mathrm{(A) \ }2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ } 6$
| [
"Since both of the equations have real roots, both of their discriminants are nonnegative. Therefore, we have \n\n\n$a^2-4(1)(2b)=a^2-8b\\geq0\\implies a^2\\geq8b$ from the first equation, and \n\n\n$(2b)^2-4(1)(a)=4b^2-4a\\geq0\\implies 4b^2\\geq4a\\implies b^2\\geq a$ from the second.\n\n\nWe can square the secon... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/24.json | AHSME |
1984_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | The total area of all the faces of a rectangular solid is $22\text{cm}^2$, and the total length of all its edges is $24\text{cm}$. Then the length in cm of any one of its interior diagonals is
$\mathrm{(A) \ }\sqrt{11} \qquad \mathrm{(B) \ }\sqrt{12} \qquad \mathrm{(C) \ } \sqrt{13}\qquad \mathrm{(D) \ }\sqrt{14} \qquad \mathrm{(E) \ } \text{Not uniquely determined}$
| [
"Let the edge lengths be $a, b$, and $c$. Therefore, the total area of all its faces is $2ab+2bc+2ac$. Therefore, $2ab+2bc+2ac=22$ and $ab+bc+ac=11$. Also, the total lengths of all of its edges is $4a+4b+4c$, so $4a+4b+4c=24$, and $a+b+c=6$. Therefore, we have:\n\n\n$ab+bc+ac=11$\n\n\nand\n\n\n$a+b+c=6$.\n\n\nThe l... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/25.json | AHSME |
1984_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | $\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$ equals
$\mathrm{(A) \ }\sqrt{2}+\sqrt{3}-\sqrt{5} \qquad \mathrm{(B) \ }4-\sqrt{2}-\sqrt{3} \qquad \mathrm{(C) \ } \sqrt{2}+\sqrt{3}+\sqrt{6}-5 \qquad$
$\mathrm{(D) \ }\frac{1}{2}(\sqrt{2}+\sqrt{5}-\sqrt{3}) \qquad \mathrm{(E) \ } \frac{1}{3}(\sqrt{3}+\sqrt{5}-\sqrt{2})$
| [
"Multiply the numerator and the denominator by $\\sqrt{2}-(\\sqrt{3}+\\sqrt{5})$ to get $\\frac{2\\sqrt{6}(\\sqrt{2}-\\sqrt{3}-\\sqrt{5})}{(\\sqrt{2}+(\\sqrt{3}+\\sqrt{5}))(\\sqrt{2}-(\\sqrt{3}+\\sqrt{5})}$\n\n\n$=\\frac{2\\sqrt{12}-2\\sqrt{18}-2\\sqrt{30}}{2-(\\sqrt{3}+\\sqrt{5})^2}$\n\n\n$=\\frac{4\\sqrt{3}-6\\sq... | 2 | ./CreativeMath/AHSME/1984_AHSME_Problems/13.json | AHSME |
1984_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | Find the largest value for $\frac{y}{x}$ for pairs of real numbers $(x, y)$ which satisfy $(x-3)^2+(y-3)^2=6$.
$\mathrm{(A) \ }3+2\sqrt{2} \qquad \mathrm{(B) \ }2+\sqrt{3} \qquad \mathrm{(C) \ } 3\sqrt{3} \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 6+2\sqrt{3}$
| [
"Let $\\frac{y}{x}=k$, so that $y=kx$. Substituting this into the given equation $(x-3)^2+(y-3)^2=6$ yields $(x-3)^2+(kx-3)^2=6$. Multiplying this out and forming it into a quadratic yields $(1+k^2)x^2+(-6-6k)x+12=0$. \n\n\nWe want $x$ to be a real number, so we must have the discriminant $\\geq0$. \nThe discrimina... | 3 | ./CreativeMath/AHSME/1984_AHSME_Problems/29.json | AHSME |
1984_AHSME_Problems | 3 | 0 | Number Theory | Multiple Choice | Let $n$ be the smallest nonprime integer greater than $1$ with no prime factor less than $10$. Then
$\mathrm{(A) \ }100<n\leq110 \qquad \mathrm{(B) \ }110<n\leq120 \qquad \mathrm{(C) \ } 120<n\leq130 \qquad \mathrm{(D) \ }130<n\leq140 \qquad \mathrm{(E) \ } 140<n\leq150$
| [
"Since the number isn't prime, it is a product of two primes. If the least integer were a product of more than two primes, then one prime could be removed without making the number prime or introducing any prime factors less than $10$. These prime factors must be greater than $10$, so the least prime factor is $11$... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/3.json | AHSME |
1984_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | Figure $ABCD$ is a trapezoid with $AB||DC$, $AB=5$, $BC=3\sqrt{2}$, $\angle BCD=45^\circ$, and $\angle CDA=60^\circ$. The length of $DC$ is
$\mathrm{(A) \ }7+\frac{2}{3}\sqrt{3} \qquad \mathrm{(B) \ }8 \qquad \mathrm{(C) \ } 9 \frac{1}{2} \qquad \mathrm{(D) \ }8+\sqrt{3} \qquad \mathrm{(E) \ } 8+3\sqrt{3}$
| [
"[asy] unitsize(1.5cm); draw((0,0)--(5,0)--(8,-3)--(-sqrt(3),-3)--cycle); label(\"$A$\",(0,0),NW); label(\"$B$\",(5,0),NE); label(\"$C$\",(8,-3),ENE); label(\"$D$\",(-sqrt(3),-3),WNW); draw((0,0)--(0,-3)); draw((5,0)--(5,-3)); label(\"$E$\",(5,-3),S); label(\"$F$\",(0,-3),S); label(\"$5$\",(2.5,0),N); label(\"$3\\s... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/8.json | AHSME |
1984_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | Let $a$ and $c$ be fixed positive numbers. For each real number $t$ let $(x_t, y_t)$ be the vertex of the parabola $y=ax^2+bx+c$. If the set of the vertices $(x_t, y_t)$ for all real numbers of $t$ is graphed on the plane, the graph is
$\mathrm{(A) \ } \text{a straight line} \qquad \mathrm{(B) \ } \text{a parabola} \qquad \mathrm{(C) \ } \text{part, but not all, of a parabola} \qquad \mathrm{(D) \ } \text{one branch of a hyperbola} \qquad$ $\mathrm{(E) \ } \text{None of these}$
| [
"The x-coordinate of the vertex of a parabola is $-\\frac{b}{2a}$, so $x_t=-\\frac{b}{2a}$. Plugging this into $y=ax^2+bx+c$ yields $y=-\\frac{b^2}{4a^2}+c$, so $y_t=-\\frac{b^2}{4a^2}+c$. Notice that $y_t=-\\frac{b^2}{4a^2}+c=-a(-\\frac{b}{2a})^2+c=-ax_t^2+c$, so all of the vertices are on a parabola. However, we ... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/22.json | AHSME |
1984_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | A point $(x, y)$ is to be chosen in the coordinate plane so that it is equally distant from the x-axis, the y-axis, and the line $x+y=2$. Then $x$ is
$\mathrm{(A) \ }\sqrt{2}-1 \qquad \mathrm{(B) \ }\frac{1}{2} \qquad \mathrm{(C) \ } 2-\sqrt{2} \qquad \mathrm{(D) \ }1 \qquad \mathrm{(E) \ } \text{Not uniquely determined}$
| [
"Consider the triangle bound by the x-axis, the y-axis, and the line $x+y=2$. The point equidistant from the vertices of this triangle is the incenter, the point of intersection of the angle bisectors and the center of the inscribed circle. Now, remove the coordinate system. Let the origin be $O$, the y-intercept o... | 2 | ./CreativeMath/AHSME/1984_AHSME_Problems/18.json | AHSME |
1984_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | A rectangle intersects a circle as shown: $AB=4$, $BC=5$, and $DE=3$. Then $EF$ equals:
[asy]defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, E=(3,0), F=(10,0), G=(12,0), H=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F); draw(D--G--H--A--cycle); draw(Circle(O, abs(O-C))); label("$A$", A, NW); label("$B$", B, NW); label("$C$", C, NE); label("$D$", D, SW); label("$E$", E, SE); label("$F$", F, SW); label("4", (2,0.85), N); label("3", D--E, S); label("5", (6.5,0.85), N); [/asy]
$\mathbf{(A)}\; 6\qquad \mathbf{(B)}\; 7\qquad \mathbf{(C)}\; \frac{20}3\qquad \mathbf{(D)}\; 8\qquad \mathbf{(E)}\; 9$
| [
"[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, E=(3,0), F=(10,0), X=(12,0), Y=(12,1), A=(0,1), B=(4,1), C=(9,1), O=circumcenter(B,C,F), G=foot(E,A,C), H=foot(B,D,F), I=foot(C,D,F); draw(D--X--Y--A--cycle); draw(Circle(O, abs(O-C))); label(\"$A$\", A, NW); label(\"$B$\", B, N); label(\"$C$\", C, NE);... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/4.json | AHSME |
1984_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | The product of all real roots of the equation $x^{\log_{10}{x}}=10$ is
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }-1 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ }10^{-1} \qquad \mathrm{(E) \ } \text{None of these}$
| [
"Take the logarithm base $x$ of both sides to get $\\log_{10}x=\\log_x10$. Use the change of base formula to get $\\log_{10}x=\\frac{\\log_{10}10}{\\log_{10}x}=\\frac{1}{\\log_{10}x}$. Therefore, $(\\log_{10}x)^2=1$ and $\\log_{10}x=\\pm1$. From $\\log_{10}x=1$, we have $x=10^1=10$ and from $\\log_{10}x=-1$ we have... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/14.json | AHSME |
1984_AHSME_Problems | 15 | 0 | Other | Multiple Choice | If $\sin{2x}\sin{3x}=\cos{2x}\cos{3x}$, then one value for $x$ is
$\mathrm{(A) \ }18^\circ \qquad \mathrm{(B) \ }30^\circ \qquad \mathrm{(C) \ } 36^\circ \qquad \mathrm{(D) \ }45^\circ \qquad \mathrm{(E) \ } 60^\circ$
| [
"We divide both sides of the equation by $\\cos{2x}\\times\\cos{3x}$ to get $\\frac{\\sin{2x}\\times\\sin{3x}}{\\cos{2x}\\times\\cos{3x}}=1$, or $\\tan{2x}\\times\\tan{3x}=1$.\n\n\nThis looks a lot like the formula relating the slopes of two perpendicular lines, which is $m_1\\times m_2=-1$, where $m_1$ and $m_2$ a... | 3 | ./CreativeMath/AHSME/1984_AHSME_Problems/15.json | AHSME |
1984_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | The largest integer $n$ for which $n^{200}<5^{300}$ is
$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }9 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ }11 \qquad \mathrm{(E) \ } 12$
| [
"Since both sides are positive, we can take the $100th$ root of both sides to find the largest integer $n$ such that $n^2<5^3$. Fortunately, this is simple to evaluate: $5^3=125$, and the largest square less than $125$ is $11^2=121$, so the largest $n$ is $11, \\boxed{\\text{D}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/5.json | AHSME |
1984_AHSME_Problems | 19 | 0 | Probability | Multiple Choice | A box contains $11$ balls, numbered $1, 2, 3, ... 11$. If $6$ balls are drawn simultaneously at random, what is the probability that the sum of the numbers on the balls drawn is odd?
$\mathrm{(A) \ }\frac{100}{231} \qquad \mathrm{(B) \ }\frac{115}{231} \qquad \mathrm{(C) \ } \frac{1}{2} \qquad \mathrm{(D) \ }\frac{118}{231} \qquad \mathrm{(E) \ } \frac{6}{11}$
| [
"There are exactly $3$ ways the sum can be odd: exactly $5$ are odd, exactly $3$ are odd, or exactly $1$ is odd. There are exactly $6$ odd numbers in the set. For the first case, there are ${6\\choose5}\\times{5\\choose1}=30$ possibilities. For the second case, there are ${6\\choose3}\\times{5\\choose3}=200$ possib... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/19.json | AHSME |
1984_AHSME_Problems | 23 | 0 | Other | Multiple Choice | $\frac{\sin{10^\circ}+\sin{20^\circ}}{\cos{10^\circ}+\cos{20^\circ}}$ equals
$\mathrm{(A) \ }\tan{10^\circ}+\tan{20^\circ} \qquad \mathrm{(B) \ }\tan{30^\circ} \qquad \mathrm{(C) \ } \frac{1}{2}(\tan{10^\circ}+\tan{20^\circ}) \qquad \mathrm{(D) \ }\tan{15^\circ} \qquad \mathrm{(E) \ } \frac{1}{4}\tan{60^\circ}$
| [
"From the sum-to-product formulas, we have $\\sin{a}+\\sin{b}=2\\sin{\\frac{a+b}{2}}\\cos{\\frac{a-b}{2}}$. So, $\\sin{10^\\circ}+\\sin{20^\\circ}=2\\sin{15^\\circ}\\cos{5^\\circ}$. Also from the sum-to-product formulas, we have $\\cos{a}+\\cos{b}=2\\cos{\\frac{a+b}{2}}\\cos{\\frac{a-b}{2}}$, so $\\cos{10^\\circ}+\... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/23.json | AHSME |
1984_AHSME_Problems | 9 | 0 | Number Theory | Multiple Choice | The number of digits in $4^{16}5^{25}$ (when written in the usual base $10$ form) is
$\mathrm{(A) \ }31 \qquad \mathrm{(B) \ }30 \qquad \mathrm{(C) \ } 29 \qquad \mathrm{(D) \ }28 \qquad \mathrm{(E) \ } 27$
| [
"We can rewrite this as $2^{32}5^{25}$. We can also combine some of the factors to introduce factors of $10$, whose digit count is simple to evaluate because it simply adds $0$s. Thus, we have $2^{32}5^{25}=2^72^{25}5^{25}=2^710^{25}$. We can see that this final number is $2^7$ with $25$ $0$s annexed onto it. $2^7=... | 1 | ./CreativeMath/AHSME/1984_AHSME_Problems/9.json | AHSME |
1991_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | The sum of all real $x$ such that $(2^x-4)^3+(4^x-2)^3=(4^x+2^x-6)^3$ is
$\textbf{(A) } \frac32 \qquad \textbf{(B) } 2 \qquad \textbf{(C) } \frac52 \qquad \textbf{(D) } 3 \qquad \textbf{(E) } \frac72$
| [
"Note that $(2^x-4)+(4^x-2)=4^x+2^x-6,$ so we let $a=2^x-4$ and $b=4^x-2.$ The original equation becomes \\[a^3+b^3=(a+b)^3.\\]\nWe expand the right side, then rearrange:\n\\begin{align*} a^3+b^3 &= a^3+3a^2b+3ab^2+b^3 \\\\ 0 &= 3a^2b+3ab^2 \\\\ 0 &= 3ab(a+b). \\end{align*}\n\n\n\\begin{itemize}\n\n\\item If $a=0,$... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/20.json | AHSME |
1991_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | One hundred students at Century High School participated in the AHSME last year, and their mean score was 100. The number of non-seniors taking the AHSME was $50\%$ more than the number of seniors, and the mean score of the seniors was $50\%$ higher than that of the non-seniors. What was the mean score of the seniors?
(A) $100$ (B) $112.5$ (C) $120$ (D) $125$ (E) $150$
| [
"Solution by e_power_pi_times_i\n\n\n\n\nLet $s$ and $\\dfrac{3}{2}s$ denote the numbers of seniors and non-seniors, respectively. Then $\\dfrac{5}{2}s = 100$, so $s = 40$, $\\dfrac{3}{2}s = 60$. Let $m$ and $\\dfrac{2}{3}m$ denote the mean score of seniors and non-seniors, respectively. Then $40m + 60(\\dfrac{2}{3... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/16.json | AHSME |
1991_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | If $x\geq 0$, then $\sqrt{x\sqrt{x\sqrt{x}}}=$
$\textbf{(A) } x\sqrt{x}\qquad \textbf{(B) } x\sqrt[4]{x}\qquad \textbf{(C) } \sqrt[8]{x}\qquad \textbf{(D) } \sqrt[8]{x^3}\qquad \textbf{(E) } \sqrt[8]{x^7}$
| [
"Recall that square roots are one-half powers, namely $\\sqrt y=y^{\\frac12}$ for all $y\\geq0.$\n\n\nWe have\n\\begin{align*} \\sqrt{x\\sqrt{x\\sqrt{x}}} &= \\sqrt{x\\sqrt{x\\cdot x^{\\frac12}}} \\\\ &= \\sqrt{x\\sqrt{x^{\\frac32}}} \\\\ &= \\sqrt{x\\cdot\\left(x^{\\frac32}\\right)^{\\frac12}} \\\\ &= \\sqrt{x^{\\... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/6.json | AHSME |
1991_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | If $x=\frac{a}{b}$, $a\neq b$ and $b\neq 0$, then $\frac{a+b}{a-b}=$
(A) $\frac{x}{x+1}$ (B) $\frac{x+1}{x-1}$ (C) $1$ (D) $x-\frac{1}{x}$ (E) $x+\frac{1}{x}$
| [
"$\\frac{a+b}{a-b}= \\frac{\\frac{a}{b} + 1}{\\frac{a}{b} - 1} = \\frac{x+1}{x-1}$, so the answer is $\\boxed{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/7.json | AHSME |
1991_AHSME_Problems | 17 | 0 | Number Theory | Multiple Choice | A positive integer $N$ is a \textit{palindrome} if the integer obtained by reversing the sequence of digits of $N$ is equal to $N$. The year 1991 is the only year in the current century with the following 2 properties:
(a) It is a palindrome
(b) It factors as a product of a 2-digit prime palindrome and a 3-digit prime palindrome.
How many years in the millenium between 1000 and 2000 have properties (a) and (b)?
$\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$
| [
"Solution by e_power_pi_times_i\n\n\nNotice that all four-digit palindromes are divisible by $11$, so that is our two-digit prime. Because the other factor is a three-digit number, we are looking at palindromes between $1100$ and $2000$, which also means that the last digit of the three-digit number is $1$. Checkin... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/17.json | AHSME |
1991_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | For all real numbers $x$ except $x=0$ and $x=1$ the function $f(x)$ is defined by $f(x/(x-1))=1/x$. Suppose $0\leq t\leq \pi/2$. What is the value of $f(\sec^2t)$?
$\text{(A) } sin^2\theta\quad \text{(B) } cos^2\theta\quad \text{(C) } tan^2\theta\quad \text{(D) } cot^2\theta\quad \text{(E) } csc^2\theta$
| [
"Let $y=\\frac{x}{x-1} \\Rightarrow xy-y=x \\Rightarrow x=\\frac{y}{y-1}$\n\n\n$f(y)=\\frac{1}{x}=\\frac{y-1}{y}=1-\\frac{1}{y}$\n\n\n$f(sec^2t)=sin^2t$\n\n\n$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/21.json | AHSME |
1991_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | Point $P$ is $9$ units from the center of a circle of radius $15$. How many different chords of the circle contain $P$ and have integer lengths?
(A) 11 (B) 12 (C) 13 (D) 14 (E) 29
| [
"Let $O$ be the center of the circle, and let the chord passing through $P$ that is perpendicular to $OP$ intersect the circle at $Q$ and $R$. Then $OP = 9$ and $OQ = 15$, so by the Pythagorean Theorem, $PQ = 12$. By symmetry, $PR = 12$.\n\n\n[asy] import graph; unitsize(0.15 cm); pair O, P, Q, R; O = (0,0); P =... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/10.json | AHSME |
1991_AHSME_Problems | 26 | 0 | Number Theory | Multiple Choice | An $n$-digit positive integer is cute if its $n$ digits are an arrangement of the set $\{1,2,...,n\}$ and its first
$k$ digits form an integer that is divisible by $k$, for $k = 1,2,...,n$. For example, $321$ is a cute $3$-digit integer because $1$ divides $3$, $2$ divides $32$, and $3$ divides $321$. How many cute $6$-digit integers are there?
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 4$
| [
"$\\fbox{C}$ Let the number be $abcdef$. We know $1$ will always divide $a$. $5$ must divide $abcde$, so $e$ must be $5$ or $0$, but we can only use the digits $1$ to $6$, so $e = 5$. $4$ must divide $abcd$, so it must divide $cd$ (the test for divisibility by 4 is that the last two digits form a number divisible b... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/26.json | AHSME |
1991_AHSME_Problems | 30 | 0 | Counting | Multiple Choice | For any set $S$, let $|S|$ denote the number of elements in $S$, and let $n(S)$ be the number of subsets of $S$, including the empty set and the set $S$ itself. If $A$, $B$, and $C$ are sets for which $n(A)+n(B)+n(C)=n(A\cup B\cup C)$ and $|A|=|B|=100$, then what is the minimum possible value of $|A\cap B\cap C|$?
$(A) 96 \ (B) 97 \ (C) 98 \ (D) 99 \ (E) 100$
| [
"$n(A)=n(B)=2^{100}$, so $n(C)$ and $n(A \\cup B \\cup C)$ are integral powers of $2$ $\\Longrightarrow$ $n(C)=2^{101}$ and $n(A \\cup B \\cup C)=2^{102}$. Let $A=\\{s_1,s_2,s_3,...,s_{100}\\}$, $B=\\{s_3,s_4,s_5,...,s_{102}\\}$, and $C=\\{s_1,s_2,s_3,...,s_{k-2},s_{k-1},s_{k+1},s_{k+2},...,s_{100},s_{101},s_{102}... | 2 | ./CreativeMath/AHSME/1991_AHSME_Problems/30.json | AHSME |
1991_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | If \[x+\sqrt{x^2-1}+\frac{1}{x-\sqrt{x^2-1}}=20,\] then \[x^2+\sqrt{x^4-1}+\frac{1}{x^2+\sqrt{x^4-1}}=\]
$\textbf{(A) } 5.05 \qquad \textbf{(B) } 20 \qquad \textbf{(C) } 51.005 \qquad \textbf{(D) } 61.25 \qquad \textbf{(E) } 400$
| [
"We rationalize the denominator in the given equation, then solve for $x:$\n\\begin{align*} x+\\sqrt{x^2-1}+\\frac{x+\\sqrt{x^2-1}}{\\left(x-\\sqrt{x^2-1}\\right)\\left(x+\\sqrt{x^2-1}\\right)} &= 20 \\\\ x+\\sqrt{x^2-1}+x+\\sqrt{x^2-1} &= 20 \\\\ x+\\sqrt{x^2-1} &= 10 \\\\ \\sqrt{x^2-1} &= 10-x \\\\ x^2-1 &= 100-2... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/27.json | AHSME |
1991_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | If for any three distinct numbers $a$, $b$, and $c$ we define $f(a,b,c)=\frac{c+a}{c-b}$, then $f(1,-2,-3)$ is
$\textbf {(A) } -2 \qquad \textbf {(B) } -\frac{2}{5} \qquad \textbf {(C) } -\frac{1}{4} \qquad \textbf {(D) } \frac{2}{5} \qquad \textbf {(E) } 2$
| [
"If we plug in $1$ as $a$, $-2$ as $b$, and $-3$ as $c$ in the expression $\\frac{c+a}{c-b}$, then we get $\\frac{-3+1}{-3-(-2)}=\\frac{-2}{-1}=2$, which is choice $\\boxed{\\textbf{E}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/1.json | AHSME |
1991_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | Jack and Jill run 10 km. They start at the same point, run 5 km up a hill, and return to the starting point by the same route. Jack has a 10 minute head start and runs at the rate of 15 km/hr uphill and 20 km/hr downhill. Jill runs 16 km/hr uphill and 22 km/hr downhill. How far from the top of the hill are they when they pass each other going in opposite directions (in km)?
$\text{(A) } \frac{5}{4}\quad \text{(B) } \frac{35}{27}\quad \text{(C) } \frac{27}{20}\quad \text{(D) } \frac{7}{3}\quad \text{(E) } \frac{28}{49}$
| [
"$\\fbox{B}$ Consider the distance-time graph and use coordinate geometry, with time on the $x$-axis and distance on the $y$-axis. Thus Jack starts at $(0,0)$ and his initial motion, taking time in hours, is $y=15x$. This ends when $y=5$, giving the point $(\\frac{1}{3}, 5)$. He now runs in the opposite direction a... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/11.json | AHSME |
1991_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | $|3-\pi|=$
$\textbf{(A)\ }\frac{1}{7}\qquad\textbf{(B)\ }0.14\qquad\textbf{(C)\ }3-\pi\qquad\textbf{(D)\ }3+\pi\qquad\textbf{(E)\ }\pi-3$
| [
"Since $\\pi>3$, the value of $3-\\pi$ is negative. The absolute value of a negative quantity is the negative quantity multiplied by $-1$, or the negative of that quantity. Therefore $|3-\\pi|=-(3-\\pi)=\\pi-3$, which is choice $\\boxed{\\textbf{E}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/2.json | AHSME |
1991_AHSME_Problems | 28 | 0 | Probability | Multiple Choice | Initially an urn contains 100 white and 100 black marbles. Repeatedly 3 marbles are removed (at random) from the urn and replaced with some marbles from a pile outside the urn as follows: 3 blacks are replaced with 1 black, or 2 blacks and 1 white are replaced with a white and a black, or 1 black and 2 whites are replaced with 2 whites, or 3 whites are replaced with a black and a white. Which of the following could be the contents of the urn after repeated applications of this procedure?
(A) 2 black (B) 2 white (C) 1 black (D) 1 black and 1 white (E) 1 white
| [
"$\\fbox{B}$ The possible operations are $-3B+1B = -2B$, $-2B-W+W+B = -B$, $-B-2W+2W = -B$, or $-3W+B+W = -2W+B$. Notice that the only way the number of whites can change is from $-2W+B$, so it starts at 100 and only ever decreases by 2, so the final number of whites must be even, eliminating $D$ and $E$. Now obser... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/28.json | AHSME |
1991_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | The measures (in degrees) of the interior angles of a convex hexagon form an arithmetic sequence of integers. Let $m$ be the measure of the largest interior angle of the hexagon. The largest possible value of $m$, in degrees, is
(A) 165 (B) 167 (C) 170 (D) 175 (E) 179
| [
"$\\fbox{D}$ The angles must add to $180(6-2) = 720$. Now, let the arithmetic progression have first term $a$ and difference $d$, so by the sum formula, we get $\\frac{6}{2}(2a+5d) = 720 \\implies 2a+5d=240.$ Now, $240$ and $5d$ are both divisible by $5$ (as $d$ is an integer), so $2a$ is divisible by $5$, and thus... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/12.json | AHSME |
1991_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | The graph, $G$ of $y=\log_{10}x$ is rotated $90^{\circ}$ counter-clockwise about the origin to obtain a new graph $G'$. Which of the following is an equation for $G'$?
(A) $y=\log_{10}\left(\frac{x+90}{9}\right)$ (B) $y=\log_{x}10$ (C) $y=\frac{1}{x+1}$ (D) $y=10^{-x}$ (E) $y=10^x$
| [
"$\\fbox{D}$ Rotating a point $(x,y)$ $90^{\\circ}$ anticlockwise about the origin maps it to $(-y,x).$ (You can prove this geometrically, or using matrices if you aren't convinced). Thus $(x, \\log_{10}x)$ maps to $(-\\log_{10}x, x)$, so new $y =$ old $x = 10^{\\log_{10}x} = 10^{-(-\\log_{10}x)} = 10^{-\\text{new}... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/24.json | AHSME |
1991_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | If $T_n=1+2+3+\cdots +n$ and
\[P_n=\frac{T_2}{T_2-1}\cdot\frac{T_3}{T_3-1}\cdot\frac{T_4}{T_4-1}\cdot\cdots\cdot\frac{T_n}{T_n-1}\]
for $n=2,3,4,\cdots,$ then $P_{1991}$ is closest to which of the following numbers?
$\text{(A) } 2.0\quad \text{(B) } 2.3\quad \text{(C) } 2.6\quad \text{(D) } 2.9\quad \text{(E) } 3.2$
| [
"$\\fbox{D}$ It is well known that $T_n = \\frac{n(n+1)}{2}$, so we can put this in and simplify to get that the general term of the product is $\\frac{n(n+1)}{(n+2)(n-1)}.$ Now by writing out the terms, we see that almost everything cancels (telescopes), giving $\\frac{3 \\times 1991}{1993}$ which is almost $3$, s... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/25.json | AHSME |
1991_AHSME_Problems | 13 | 0 | Probability | Multiple Choice | Horses $X,Y$ and $Z$ are entered in a three-horse race in which ties are not possible. The odds against $X$ winning are $3:1$ and the odds against $Y$ winning are $2:3$, what are the odds against $Z$ winning? (By "odds against $H$ winning are $p:q$" we mean the probability of $H$ winning the race is $\frac{q}{p+q}$.)
$\text{(A) } 3:20\quad \text{(B) } 5:6\quad \text{(C) } 8:5\quad \text{(D) } 17:3\quad \text{(E) } 20:3$
| [
"Solution by e_power_pi_times_i\n\n\n\n\nBecause the odds against $X$ are $3:1$, the chance of $X$ losing is $\\dfrac{3}{4}$. Since the chance of $X$ losing is the same as the chance of $Y$ and $Z$ winning, and since the odds against $Y$ are $2:3$, $Y$ wins with a probability of $\\dfrac{3}{5}$. Then the chance of ... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/13.json | AHSME |
1991_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | Equilateral triangle $ABC$ has $P$ on $AB$ and $Q$ on $AC$. The triangle is folded along $PQ$ so that vertex $A$ now rests at $A'$ on side $BC$. If $BA'=1$ and $A'C=2$ then the length of the crease $PQ$ is
$\text{(A) } \frac{8}{5} \text{(B) } \frac{7}{20}\sqrt{21} \text{(C) } \frac{1+\sqrt{5}}{2} \text{(D) } \frac{13}{8} \text{(E) } \sqrt{3}$
| [
"$ABC$ has side length $3$. Let $AP=A'P=x$ and $AQ=A'Q=y$. Thus, $BP=3-x$ and $CQ=3-y$. Applying Law of Cosines on triangles $BPA'$ and $CQA'$ using the $60^{\\circ}$ angles gives $x=\\frac{7}{5}$ and $y=\\frac{7}{4}$. Applying Law of Cosines once again on triangle $APQ$ using the $60^{\\circ}$ angle gives \\[PQ^2=... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/29.json | AHSME |
1991_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | $(4^{-1}-3^{-1})^{-1}=$
(A) $-12$ (B) $-1$ (C) $\frac{1}{12}$ (D) $1$ (E) $12$
| [
"$\\fbox{A}$ This is $\\frac{1}{\\frac{1}{4} - \\frac{1}{3}} = \\frac{1}{\\frac{-1}{12}} = -12.$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/3.json | AHSME |
1991_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | Liquid $X$ does not mix with water. Unless obstructed, it spreads out on the surface of water to form a circular film $0.1$cm thick. A rectangular box measuring $6$cm by $3$cm by $12$cm is filled with liquid $X$. Its contents are poured onto a large body of water. What will be the radius, in centimeters, of the resulting circular film?
$\text{(A)} \frac{\sqrt{216}}{\pi} \quad \text{(B)} \sqrt{\frac{216}{\pi}} \quad \text{(C)} \sqrt{\frac{2160}{\pi}} \quad \text{(D)} \frac{216}{\pi} \quad \text{(E)} \frac{2160}{\pi}$
| [
"$\\fbox{C}$ The volume of liquid $X$ is $6 \\times 3 \\times 12 = 216$, so if the radius is $r$, then using the formula for the volume of a cylinder, we get $\\pi r^2 \\times 0.1 = 216 \\implies \\pi r^2 = 2160 \\implies r = \\sqrt{\\frac{2160}{\\pi}}.$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/8.json | AHSME |
1991_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | [asy] draw(circle((0,6sqrt(2)),2sqrt(2)),black+linewidth(.75)); draw(circle((0,3sqrt(2)),sqrt(2)),black+linewidth(.75)); draw((-8/3,16sqrt(2)/3)--(-4/3,8sqrt(2)/3)--(0,0)--(4/3,8sqrt(2)/3)--(8/3,16sqrt(2)/3),dot); MP("B",(-8/3,16*sqrt(2)/3),W);MP("B'",(8/3,16*sqrt(2)/3),E); MP("A",(-4/3,8*sqrt(2)/3),W);MP("A'",(4/3,8*sqrt(2)/3),E); MP("P",(0,0),S); [/asy]
Two circles are externally tangent. Lines $\overline{PAB}$ and $\overline{PA'B'}$ are common tangents with $A$ and $A'$ on the smaller circle $B$ and $B'$ on the larger circle. If $PA=AB=4$, then the area of the smaller circle is
$\text{(A) } 1.44\pi\quad \text{(B) } 2\pi\quad \text{(C) } 2.56\pi\quad \text{(D) } \sqrt{8}\pi\quad \text{(E) } 4\pi$
| [
"Using the tangent-tangent theorem, $PA=AB=PA'=A'B'=4$. We can then drop perpendiculars from the centers of the circles to the points of tangency and use similar triangles. Let us let the center of the smaller circle be point $S$ and the center of the larger circle be point $L$. If we let the radius of the larger c... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/22.json | AHSME |
1991_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | If $S$ is the set of points $z$ in the complex plane such that $(3+4i)z$ is a real number, then $S$ is a
(A) right triangle
(B) circle
(C) hyperbola
(D) line
(E) parabola
| [
"$\\fbox{D}$\n\n\nSolution 1:\n\n\nWe want $(3+4i)z$ a real number, so we want the $4i$ term to be canceled out. Then, we can make $z$ be in the form $(n-\\frac{4}{3}ni)$ to make sure the imaginary terms cancel out when it's multiplied together. $(n-\\frac{4}{3}ni)$ is a line, so the answer is $\\textbf{(D) } line$... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/18.json | AHSME |
1991_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | Which of the following triangles cannot exist?
(A) An acute isosceles triangle (B) An isosceles right triangle (C) An obtuse right triangle (D) A scalene right triangle (E) A scalene obtuse triangle
| [
"$\\fbox{C}$\nA right triangle has an angle of 90 degrees, so any obtuse angle would make the sum of those two angles over 180. This contradicts the angle sum theorem (all triangles have angles that sum to 180) since negative angles don't exist.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/4.json | AHSME |
1991_AHSME_Problems | 14 | 0 | Number Theory | Multiple Choice | If $x$ is the cube of a positive integer and $d$ is the number of positive integers that are divisors of $x$, then $d$ could be
$\text{(A) } 200\quad\text{(B) } 201\quad\text{(C) } 202\quad\text{(D) } 203\quad\text{(E) } 204$
| [
"Solution by e_power_pi_times_i\n\n\n\n\nNotice that if $x$ is expressed in the form $a^b$, then the number of positive divisors of $x^3$ is $3b+1$. Checking through all the answer choices, the only one that is in the form $3b+1$ is $\\boxed{\\textbf{(C) } 202}$.\n\n\n",
"Solution by e_power_pi_times_i\n\n\n\n\nS... | 2 | ./CreativeMath/AHSME/1991_AHSME_Problems/14.json | AHSME |
1991_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | A circular table has 60 chairs around it. There are $N$ people seated at this table in such a way that the next person seated must sit next to someone. What is the smallest possible value for $N$?
$\text{(A) } 15\quad \text{(B) } 20\quad \text{(C) } 30\quad \text{(D) } 40\quad \text{(E) } 58$
| [
"$\\fbox{B}$ If we fill every third chair with a person, then the condition is satisfied, giving $N=20$. Decreasing $N$ any further means there is at least one gap of $4$, so that the person can sit themselves in the middle (seat $2$ of $4$) and not be next to anyone. Hence the minimum value of $N$ is $20$.\n\n\n"
... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/15.json | AHSME |
1991_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | [asy] draw((0,0)--(2,2)--(2,1)--(5,1)--(5,-1)--(2,-1)--(2,-2)--cycle,dot); MP("A",(0,0),W);MP("B",(2,2),N);MP("C",(2,1),S);MP("D",(5,1),NE);MP("E",(5,-1),SE);MP("F",(2,-1),NW);MP("G",(2,-2),S); MP("5",(2,1.5),E);MP("5",(2,-1.5),E);MP("20",(3.5,1),N);MP("20",(3.5,-1),S);MP("10",(5,0),E); [/asy]
In the arrow-shaped polygon [see figure], the angles at vertices $A,C,D,E$ and $F$ are right angles, $BC=FG=5, CD=FE=20, DE=10$, and $AB=AG$. The area of the polygon is closest to
$\text{(A) } 288\quad \text{(B) } 291\quad \text{(C) } 294\quad \text{(D) } 297\quad \text{(E) } 300$
| [
"$\\fbox{E}$\nSince they tell us that AB=AG and that angle A is a right angle, triangle ABG is a 45-45-90 triangle. Thus we can find the legs of the triangle by dividing $20$ (the hypotenuse) by $\\sqrt2$. Both legs have length $10\\sqrt2$, so the area of the right triangle is $100$. The rectangle is simple, just $... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/5.json | AHSME |
1991_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | [asy] draw((0,0)--(0,3)--(4,0)--cycle,dot); draw((4,0)--(7,0)--(7,10)--cycle,dot); draw((0,3)--(7,10),dot); MP("C",(0,0),SW);MP("A",(0,3),NW);MP("B",(4,0),S);MP("E",(7,0),SE);MP("D",(7,10),NE); [/asy]
Triangle $ABC$ has a right angle at $C, AC=3$ and $BC=4$. Triangle $ABD$ has a right angle at $A$ and $AD=12$. Points $C$ and $D$ are on opposite sides of $\overline{AB}$. The line through $D$ parallel to $\overline{AC}$ meets $\overline{CB}$ extended at $E$. If
\[\frac{DE}{DB}=\frac{m}{n},\]
where $m$ and $n$ are relatively prime positive integers, then $m+n=$
$\text{(A) } 25\quad \text{(B) } 128\quad \text{(C) } 153\quad \text{(D) } 243\quad \text{(E) } 256$
| [
"Solution by e_power_pi_times_i\n\n\n\n\nLet $F$ be the point such that $DF$ and $CF$ are parallel to $CE$ and $DE$, respectively, and let $DE = x$ and $BE^2 = 169-x^2$. Then, $[FDEC] = x(4+\\sqrt{169-x^2}) = [ABC] + [BED] + [ABD] + [AFD] = 6 + \\dfrac{x\\sqrt{169-x^2}}{2} + 30 + \\dfrac{(x-3)(4+\\sqrt{169-x^2})}{2... | 3 | ./CreativeMath/AHSME/1991_AHSME_Problems/19.json | AHSME |
1991_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | [asy] draw((0,0)--(0,2)--(2,2)--(2,0)--cycle,dot); draw((2,2)--(0,0)--(0,1)--cycle,dot); draw((0,2)--(1,0),dot); MP("B",(0,0),SW);MP("A",(0,2),NW);MP("D",(2,2),NE);MP("C",(2,0),SE); MP("E",(0,1),W);MP("F",(1,0),S);MP("H",(2/3,2/3),E);MP("I",(2/5,6/5),N); dot((1,0));dot((0,1));dot((2/3,2/3));dot((2/5,6/5)); [/asy]
If $ABCD$ is a $2\times2$ square, $E$ is the midpoint of $\overline{AB}$,$F$ is the midpoint of $\overline{BC}$,$\overline{AF}$ and $\overline{DE}$ intersect at $I$, and $\overline{BD}$ and $\overline{AF}$ intersect at $H$, then the area of quadrilateral $BEIH$ is
$\text{(A) } \frac{1}{3}\quad \text{(B) } \frac{2}{5}\quad \text{(C) } \frac{7}{15}\quad \text{(D) } \frac{8}{15}\quad \text{(E) } \frac{3}{5}$
| [
"Solution by e_power_pi_times_i\n\n\n\n\nFirst, we find out the coordinates of the vertices of quadrilateral $BEIH$, then use the Shoelace Theorem to solve for the area. Denote $B$ as $(0,0)$. Then $E (0,1)$. Since I is the intersection between lines $DE$ and $AF$, and since the equations of those lines are $y = \\... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/23.json | AHSME |
1991_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | From time $t=0$ to time $t=1$ a population increased by $i\%$, and from time $t=1$ to time $t=2$ the population increased by $j\%$. Therefore, from time $t=0$ to time $t=2$ the population increased by
$\text{(A) (i+j)\%} \quad \text{(B) } ij\%\quad \text{(C) } (i+ij)\%\quad \text{(D) } \left(i+j+\frac{ij}{100}\right)\%\quad \text{(E) } \left(i+j+\frac{i+j}{100}\right)\%$
| [
"The scale factors for the increases are $1+\\frac{i}{100}$ and $1+\\frac{j}{100}$, so the overall scale factor is $(1+\\frac{i}{100})(1+\\frac{j}{100}) = 1 + \\frac{i}{100} + \\frac{j}{100} + \\frac{ij}{100^2}$. To convert this to a percentage, we subtract 1 and then multiply by 100, giving $i + j + \\frac{ij}{100... | 1 | ./CreativeMath/AHSME/1991_AHSME_Problems/9.json | AHSME |
1961_AHSME_Problems | 20 | 0 | Geometry | Multiple Choice | The set of points satisfying the pair of inequalities $y>2x$ and $y>4-x$ is contained entirely in quadrants:
$\textbf{(A)}\ \text{I and II}\qquad \textbf{(B)}\ \text{II and III}\qquad \textbf{(C)}\ \text{I and III}\qquad \textbf{(D)}\ \text{III and IV}\qquad \textbf{(E)}\ \text{I and IV}$
| [
"[asy] fill((-1,5)--(5,5)--(5,-1)--cycle,red); fill((2.5,5)--(-5,5)--(-5,-5)--(-2.5,-5)--cycle,cyan); fill((-1,5)--(4/3,8/3)--(2.5,5)--cycle,magenta); draw((-1,5)--(5,-1),dotted,Arrows); draw((-2.5,-5)--(2.5,5),dotted,Arrows); import graph; size(8.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultp... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/20.json | AHSME |
1961_AHSME_Problems | 36 | 0 | Geometry | Multiple Choice | In $\triangle ABC$ the median from $A$ is given perpendicular to the median from $B$. If $BC=7$ and $AC=6$, find the length of $AB$.
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ \sqrt{17} \qquad \textbf{(C)}\ 4.25\qquad \textbf{(D)}\ 2\sqrt{5} \qquad \textbf{(E)}\ 4.5$
| [
"[asy] draw((-16,0)--(8,0)); draw((-16,0)--(16,-24)); draw((16,-24)--(0,24)--(0,-12)); draw((-16,0)--(0,24)); draw((0,2)--(2,2)--(2,0)); draw((0,-12)--(8,0),dotted); dot((16,-24)); label(\"C\",(16,-24),SE); dot((-16,0)); label(\"A\",(-16,0),W); dot((0,24)); label(\"B\",(0,24),N); label(\"3\",(8,-18),SW); label(\... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/36.json | AHSME |
1961_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | An altitude $h$ of a triangle is increased by a length $m$. How much must be taken from the corresponding base $b$
so that the area of the new triangle is one-half that of the original triangle?
$\textbf{(A)}\ \frac{bm}{h+m}\qquad \textbf{(B)}\ \frac{bh}{2h+2m}\qquad \textbf{(C)}\ \frac{b(2m+h)}{m+h}\qquad \textbf{(D)}\ \frac{b(m+h)}{2m+h}\qquad \textbf{(E)}\ \frac{b(2m+h)}{2(h+m)}$
| [
"Let $x$ be the value reduced from the base. Using the variables from the problem, the equation to solve is\n\\[\\frac{1}{2} (h + m)(b - x) = \\frac{1}{2} \\cdot \\frac{bh}{2}\\]\nDividing both sides by $\\frac{1}{2}$ and $h+m$ results in\n\\[b - x = \\frac{bh}{2(h+m)}\\]\n\\[-x = \\frac{bh}{2(h+m)} - b\\]\n\\[x =... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/16.json | AHSME |
1961_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | When simplified, $\log{8} \div \log{\frac{1}{8}}$ becomes:
$\textbf{(A)}\ 6\log{2} \qquad \textbf{(B)}\ \log{2} \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 0\qquad \textbf{(E)}\ -1$
| [
"First, note that $\\frac{1}{8} = 8^{-1}$. That means the expression can be rewritten as\n\\[\\log{8} \\div \\log{8^{-1}}\\]\n\\[\\log{8} \\cdot \\frac{1}{\\log{8^{-1}}}\\]\n\\[\\log{8} \\cdot \\frac{1}{-1 \\cdot \\log{8}}\\]\nThis simplifies to $-1$, which is answer choice $\\boxed{\\textbf{(E)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/6.json | AHSME |
1961_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | When simplified, the third term in the expansion of $(\frac{a}{\sqrt{x}}-\frac{\sqrt{x}}{a^2})^6$ is:
$\textbf{(A)}\ \frac{15}{x}\qquad \textbf{(B)}\ -\frac{15}{x}\qquad \textbf{(C)}\ -\frac{6x^2}{a^9} \qquad \textbf{(D)}\ \frac{20}{a^3}\qquad \textbf{(E)}\ -\frac{20}{a^3}$
| [
"By the Binomial Theorem, the third term in the expansion is\n\\[\\binom{6}{2}(\\frac{a}{\\sqrt{x}})^{4}(\\frac{-\\sqrt{x}}{a^2})^2\\]\n\\[15 \\cdot \\frac{a^4}{x^2} \\cdot \\frac{x}{a^4}\\]\n\\[\\frac{15}{x}\\]\nThe answer is $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/7.json | AHSME |
1961_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | In the base ten number system the number $526$ means $5 \times 10^2+2 \times 10 + 6$.
In the Land of Mathesis, however, numbers are written in the base $r$.
Jones purchases an automobile there for $440$ monetary units (abbreviated m.u).
He gives the salesman a $1000$ m.u bill, and receives, in change, $340$ m.u. The base $r$ is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 8\qquad \textbf{(E)}\ 12$
| [
"If Jones received $340$ m.u. change after paying $1000$ m.u. for something that costs $440$ m.u., then\n\\[440_r + 340_r = 1000_r\\]\nThis equation can be rewritten as\n\\[4r^2 + 4r + 3r^2 + 4r = r^3\\]\nBring all of the terms to one side to get\n\\[r^3 - 7r^2 - 8r = 0\\]\nFactor to get\n\\[r(r-8)(r+1)=0\\]\nSince... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/17.json | AHSME |
1961_AHSME_Problems | 40 | 0 | Geometry | Multiple Choice | Find the minimum value of $\sqrt{x^2+y^2}$ if $5x+12y=60$.
$\textbf{(A)}\ \frac{60}{13}\qquad \textbf{(B)}\ \frac{13}{5}\qquad \textbf{(C)}\ \frac{13}{12}\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$
| [
"Let $x^2 + y^2 = r^2$, so $r = \\sqrt{x^2 + y^2}$. Thus, this problem is really finding the shortest distance from the origin to the line $5x + 12y = 60$.\n\n\n[asy]import graph; size(8.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-5.2,xmax=13.2,ymin=-5.2,yma... | 3 | ./CreativeMath/AHSME/1961_AHSME_Problems/40.json | AHSME |
1961_AHSME_Problems | 37 | 0 | Algebra | Multiple Choice | In racing over a distance $d$ at uniform speed, $A$ can beat $B$ by $20$ yards, $B$ can beat $C$ by $10$ yards,
and $A$ can beat $C$ by $28$ yards. Then $d$, in yards, equals:
$\textbf{(A)}\ \text{Not determined by the given information} \qquad \textbf{(B)}\ 58\qquad \textbf{(C)}\ 100\qquad \textbf{(D)}\ 116\qquad \textbf{(E)}\ 120$
| [
"Let $a$ be speed of $A$, $b$ be speed of $B$, and $c$ be speed of $C$.\n\n\nPerson $A$ finished the track in $\\frac{d}{a}$ minutes, so $B$ traveled $\\frac{db}{a}$ yards at the same time. Since $B$ is $20$ yards from the finish line, the first equation is\n\\[\\frac{db}{a} + 20 = d\\]\nUsing similar steps, the s... | 2 | ./CreativeMath/AHSME/1961_AHSME_Problems/37.json | AHSME |
1961_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | Medians $AD$ and $CE$ of $\triangle ABC$ intersect in $M$. The midpoint of $AE$ is $N$.
Let the area of $\triangle MNE$ be $k$ times the area of $\triangle ABC$. Then $k$ equals:
$\textbf{(A)}\ \frac{1}{6}\qquad \textbf{(B)}\ \frac{1}{8}\qquad \textbf{(C)}\ \frac{1}{9}\qquad \textbf{(D)}\ \frac{1}{12}\qquad \textbf{(E)}\ \frac{1}{16}$
| [
"[asy] draw((0,0)--(120,0)--(40,60)--(0,0)); draw((40,60)--(60,0)); draw((120,0)--(20,30)); draw((160/3,20)--(30,45)); dot((40,60)); label(\"A\",(40,60),N); dot((0,0)); label(\"B\",(0,0),SW); dot((120,0)); label(\"C\",(120,0),SE); dot((60,0)); label(\"D\",(60,0),S); dot((20,30)); label(\"E\",(20,30),NW); dot((160/... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/21.json | AHSME |
1961_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | Each side of $\triangle ABC$ is $12$ units. $D$ is the foot of the perpendicular dropped from $A$ on $BC$,
and $E$ is the midpoint of $AD$. The length of $BE$, in the same unit, is:
$\textbf{(A)}\ \sqrt{18} \qquad \textbf{(B)}\ \sqrt{28} \qquad \textbf{(C)}\ 6 \qquad \textbf{(D)}\ \sqrt{63} \qquad \textbf{(E)}\ \sqrt{98}$
| [
"[asy] draw((0,0)--(50,0)--(25,43.301)--cycle); draw((25,43.301)--(25,0)); dot((0,0)); label(\"$B$\",(0,0),SW); dot((50,0)); label(\"$C$\",(50,0),SE); dot((25,43.301)); label(\"$A$\",(25,43.301),N); dot((25,0)); label(\"$D$\",(25,0),S); dot((25,21.651)); label(\"$E$\",(25,21.651),E); draw((25,21.651)--(0,0)); labe... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/10.json | AHSME |
1961_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | For a given arithmetic series the sum of the first $50$ terms is $200$, and the sum of the next $50$ terms is $2700$.
The first term in the series is:
$\textbf{(A)}\ -1221 \qquad \textbf{(B)}\ -21.5 \qquad \textbf{(C)}\ -20.5 \qquad \textbf{(D)}\ 3 \qquad \textbf{(E)}\ 3.5$
| [
"Let the first term of the arithmetic sequence be $a$ and the common difference be $d$.\n\n\nThe $50^{\\text{th}}$ term of the sequence is $a+49d$, so the sum of the first $50$ terms is $\\frac{50(a + a + 49d)}{2}$.\n\n\nThe $51^{\\text{th}}$ term of the sequence is $a+50d$ and the $100^{\\text{th}}$ term of the se... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/26.json | AHSME |
1961_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | If $\log_{10}2=a$ and $\log_{10}3=b$, then $\log_{5}12=$?
$\textbf{(A)}\ \frac{a+b}{a+1}\qquad \textbf{(B)}\ \frac{2a+b}{a+1}\qquad \textbf{(C)}\ \frac{a+2b}{1+a}\qquad \textbf{(D)}\ \frac{2a+b}{1-a}\qquad \textbf{(E)}\ \frac{a+2b}{1-a}$
| [
"From the change of base formula, $\\log_{5}12 = \\frac{\\log_{10}12}{\\log_{10}5}$.\n\n\nFor the numerator, $2 \\cdot 2 \\cdot 3 = 12$, so $\\log_{10}12 = \\log_{10}2 + \\log_{10}2 + \\log_{10}3 = 2a+b$.\n\n\nFor the denominator, note that $\\log_{10}10 = 1$. Thus, $\\log_{10}5 = \\log_{10}10 - \\log_{10}2 = 1 - ... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/30.json | AHSME |
1961_AHSME_Problems | 31 | 0 | Geometry | Multiple Choice | In $\triangle ABC$ the ratio $AC:CB$ is $3:4$. The bisector of the exterior angle at $C$ intersects $BA$ extended at $P$
($A$ is between $P$ and $B$). The ratio $PA:AB$ is:
$\textbf{(A)}\ 1:3 \qquad \textbf{(B)}\ 3:4 \qquad \textbf{(C)}\ 4:3 \qquad \textbf{(D)}\ 3:1 \qquad \textbf{(E)}\ 7:1$
| [
"[asy] draw((0,0)--(40,0)--(16,18)--(0,0)); draw((40,0)--(64,72)--(16,18)); draw((40,0)--(160,0)--(64,72),dotted); dot((0,0)); label(\"B\",(0,0),SW); dot((16,18)); label(\"A\",(16,18),NW); dot((40,0)); label(\"C\",(40,0),S); dot((64,72)); label(\"P\",(64,72),N); dot((160,0)); label(\"X\",(160,0),SE); label(\"$4n$\... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/31.json | AHSME |
1961_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | Given two equiangular polygons $P_1$ and $P_2$ with different numbers of sides;
each angle of $P_1$ is $x$ degrees and each angle of $P_2$ is $kx$ degrees,
where $k$ is an integer greater than $1$.
The number of possibilities for the pair $(x, k)$ is:
$\textbf{(A)}\ \infty\qquad \textbf{(B)}\ \text{finite, but greater than 2}\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 0$
| [
"Each angle in each equiangular polygon is $\\frac{180(n-2)}{n} = 180 - \\frac{360}{n}$, where $n$ is the number of sides. As $n$ gets larger, each angle in the equiangular polygon approaches (but does not reach) $180^{\\circ}$. That means $kx < 180$, so $x < \\frac{180}{k}$.\n\n\n\n\nRecall that each angle in an... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/27.json | AHSME |
1961_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | When simplified, $(-\frac{1}{125})^{-2/3}$ becomes:
$\textbf{(A)}\ \frac{1}{25} \qquad \textbf{(B)}\ -\frac{1}{25} \qquad \textbf{(C)}\ 25\qquad \textbf{(D)}\ -25\qquad \textbf{(E)}\ 25\sqrt{-1}$
| [
"To remove the negative exponent, flip the fraction of the base. This results in $(-125)^{2/3}$.\n\n\nThen, cube root $-125$ and and square the result to get the answer of $25$, or answer choice $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/1.json | AHSME |
1961_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | An automobile travels $a/6$ feet in $r$ seconds. If this rate is maintained for $3$ minutes, how many yards does it travel in $3$ minutes?
$\textbf{(A)}\ \frac{a}{1080r}\qquad \textbf{(B)}\ \frac{30r}{a}\qquad \textbf{(C)}\ \frac{30a}{r}\qquad \textbf{(D)}\ \frac{10r}{a}\qquad \textbf{(E)}\ \frac{10a}{r}$
| [
"Use dimensional analysis.\n\\[\\frac{a/6 \\text{ feet}}{r \\text{ seconds}} \\cdot \\frac{1 \\text{ yard}}{3 \\text{ feet}} \\cdot \\frac{60 \\text{ seconds}}{1 \\text{ minute}} \\cdot 3 \\text{ minutes}\\]\n\\[\\frac{10a}{r} \\text{ yards}\\]\nThe answer is $\\boxed{\\textbf{(E)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/2.json | AHSME |
1961_AHSME_Problems | 28 | 0 | Number Theory | Multiple Choice | If $2137^{753}$ is multiplied out, the units' digit in the final product is:
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$
| [
"$7^1$ has a unit digit of $7$.\n$7^2$ has a unit digit of $9$.\n$7^3$ has a unit digit of $3$.\n$7^4$ has a unit digit of $1$.\n$7^5$ has a unit digit of $7$.\n\n\nNotice that the unit digit eventually cycles to itself when the exponent is increased by $4$. It also does not matter what the other digits are in the... | 2 | ./CreativeMath/AHSME/1961_AHSME_Problems/28.json | AHSME |
1961_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | The first three terms of a geometric progression are $\sqrt{2}, \sqrt[3]{2}, \sqrt[6]{2}$. Find the fourth term.
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ \sqrt[7]{2}\qquad \textbf{(C)}\ \sqrt[8]{2}\qquad \textbf{(D)}\ \sqrt[9]{2}\qquad \textbf{(E)}\ \sqrt[10]{2}$
| [
"After rewriting the radicals as fractional exponents, the sequence is $2^{1/2}, 2^{1/3}, 2^{1/6}$.\n\n\nThe common ratio of the geometric sequence is $\\frac{2^{1/3}}{2^{1/2}} = 2^{-1/6}$. Multiplying that by the third term results in $2^0$. It simplifies to $1$, so the answer is $\\boxed{\\textbf{(A)}}$.\n\n\n"... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/12.json | AHSME |
1961_AHSME_Problems | 32 | 0 | Geometry | Multiple Choice | A regular polygon of $n$ sides is inscribed in a circle of radius $R$. The area of the polygon is $3R^2$. Then $n$ equals:
$\textbf{(A)}\ 8\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12\qquad \textbf{(D)}\ 15\qquad \textbf{(E)}\ 18$
| [
"Note that the distance from the center of the circle to each of the vertices of the inscribed regular polygon equals the radius $R$. Since each side of a regular polygon is the same length, all the angles between the two lines from the center to the two vertices of a side is the same.\n\n\nThat means each of thes... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/32.json | AHSME |
1961_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | Thirty-one books are arranged from left to right in order of increasing prices.
The price of each book differs by $\textdollar{2}$ from that of each adjacent book.
For the price of the book at the extreme right a customer can buy the middle book and the adjacent one. Then:
$\textbf{(A)}\ \text{The adjacent book referred to is at the left of the middle book}\qquad \\ \textbf{(B)}\ \text{The middle book sells for \textdollar 36} \qquad \\ \textbf{(C)}\ \text{The cheapest book sells for \textdollar4} \qquad \\ \textbf{(D)}\ \text{The most expensive book sells for \textdollar64 } \qquad \\ \textbf{(E)}\ \text{None of these is correct }$
| [
"Let the leftmost book cost $c$ dollars. The 16th book (the middle book) costs $c+30$ dollars, and the 31st book costs $c+60$ dollars.\n\n\nIf the book to the left of the middle book is used, the equation to solve for the price of the leftmost book is $c+30+c+29=c+60$. Solving yields $c=1$, so the price of the mi... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/24.json | AHSME |
1961_AHSME_Problems | 33 | 0 | Algebra | Multiple Choice | The number of solutions of $2^{2x}-3^{2y}=55$, in which $x$ and $y$ are integers, is:
\[\textbf{(A)} \ 0 \qquad\textbf{(B)} \ 1 \qquad \textbf{(C)} \ 2 \qquad\textbf{(D)} \ 3\qquad \textbf{(E)} \ \text{More than three, but finite}\]
| [
"Let $a = 2^x$ and $b = 3^y$. Substituting these values results in\n\\[a^2 - b^2 = 55\\]\nFactor the difference of squares to get\n\\[(a + b)(a - b) = 55\\]\nIf $y < 0$, then $55 + 3^{2y} < 64$, so $y$ can not be negative. If $x < 0$, then $2^{2x} < 1$. Since $3^{2y}$ is always positive, the result would be way ... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/33.json | AHSME |
1961_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | The symbol $|a|$ means $a$ is a positive number or zero, and $-a$ if $a$ is a negative number.
For all real values of $t$ the expression $\sqrt{t^4+t^2}$ is equal to?
$\textbf{(A)}\ t^3\qquad \textbf{(B)}\ t^2+t\qquad \textbf{(C)}\ |t^2+t|\qquad \textbf{(D)}\ t\sqrt{t^2+1}\qquad \textbf{(E)}\ |t|\sqrt{1+t^2}$
| [
"Factor out the $t^2$ inside the square root.\n\\[\\sqrt{t^2 \\cdot (t^2 + 1)}\\]\n\\[\\sqrt{t^2} \\cdot \\sqrt{t^2 + 1}\\]\nRemember that $\\sqrt{t^2} = |t|$ because square rooting a nonnegative real number will always result in a nonnegative number.\n\\[|t|\\sqrt{t^2 + 1}\\]\nThe answer is $\\boxed{\\textbf{(E)}}... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/13.json | AHSME |
1961_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | Let the roots of $ax^2+bx+c=0$ be $r$ and $s$. The equation with roots $ar+b$ and $as+b$ is:
$\textbf{(A)}\ x^2-bx-ac=0\qquad \textbf{(B)}\ x^2-bx+ac=0 \qquad\\ \textbf{(C)}\ x^2+3bx+ca+2b^2=0 \qquad \textbf{(D)}\ x^2+3bx-ca+2b^2=0 \qquad\\ \textbf{(E)}\ x^2+bx(2-a)+a^2c+b^2(a+1)=0$
| [
"From Vieta's Formulas, $r+s=\\frac{-b}{a}$ and $rs=\\frac{c}{a}$ in the original quadratic.\n\n\nThe sum of the roots in the new quadratic is\n\\[ar + b + as + b\\]\n\\[a(r+s) + 2b\\]\n\\[b\\]\nThe product of the roots in the new quadratic is\n\\[(ar + b)(as + b)\\]\n\\[a^2rs + arb + asb + b^2\\]\n\\[a^2rs + ab(r+... | 2 | ./CreativeMath/AHSME/1961_AHSME_Problems/29.json | AHSME |
1961_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | If the graphs of $2y+x+3=0$ and $3y+ax+2=0$ are to meet at right angles, the value of $a$ is:
$\textbf{(A)}\ \pm \frac{2}{3} \qquad \textbf{(B)}\ -\frac{2}{3}\qquad \textbf{(C)}\ -\frac{3}{2} \qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ -6$
| [
"The slope of the first graph is -1/2. The slope of the second is 2, since it is perpendicular, and it is also -a/3 by rearranging. Thus $a=-6$. Answer is $\\boxed{\\textbf{(E)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/3.json | AHSME |
1961_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | Let S be the set of values assumed by the fraction $\frac{2x+3}{x+2}$.
When $x$ is any member of the interval $x \ge 0$. If there exists a number $M$ such that no number of the set $S$ is greater than $M$,
then $M$ is an upper bound of $S$. If there exists a number $m$ such that such that no number of the set $S$ is less than $m$,
then $m$ is a lower bound of $S$. We may then say:
$\textbf{(A)}\ \text{m is in S, but M is not in S}\qquad\\ \textbf{(B)}\ \text{M is in S, but m is not in S}\qquad\\ \textbf{(C)}\ \text{Both m and M are in S}\qquad\\ \textbf{(D)}\ \text{Neither m nor M are in S}\qquad\\ \textbf{(E)}\ \text{M does not exist either in or outside S}$
| [
"This problem is really finding the range of a function with a restricted domain.\n\n\n\n\n\n\nDividing $x+2$ into $2x+3$ yields $2 - \\frac{1}{x+2}$. Since $x \\ge 0$, as $x$ gets larger, $\\frac{1}{x+2}$ approaches $0$, so $\\frac{2x+3}{x+2}$ approaches $2$ as $x$ gets larger. That means $M = 2$. Since $\\frac... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/34.json | AHSME |
1961_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | Let the two base angles of a triangle be $A$ and $B$, with $B$ larger than $A$.
The altitude to the base divides the vertex angle $C$ into two parts, $C_1$ and $C_2$, with $C_2$ adjacent to side $a$. Then:
$\textbf{(A)}\ C_1+C_2=A+B \qquad \textbf{(B)}\ C_1-C_2=B-A \qquad\\ \textbf{(C)}\ C_1-C_2=A-B \qquad \textbf{(D)}\ C_1+C_2=B-A\qquad \textbf{(E)}\ C_1-C_2=A+B$
| [
"[asy] draw((0,0)--(120,0)--(40,50)--(0,0)); draw((40,50)--(40,0)); draw((40,4)--(44,4)--(44,0)); label(\"$B$\",(10,5)); label(\"$A$\",(100,5)); label(\"$C_1$\",(47,38)); label(\"$C_2$\",(34,33)); [/asy]\n\n\nNoting that side $a$ is opposite of angle $A$, label the diagram as shown.\n\n\nFrom the two right triangle... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/8.json | AHSME |
1961_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | If $3x^3-9x^2+kx-12$ is divisible by $x-3$, then it is also divisible by:
$\textbf{(A)}\ 3x^2-x+4\qquad \textbf{(B)}\ 3x^2-4\qquad \textbf{(C)}\ 3x^2+4\qquad \textbf{(D)}\ 3x-4 \qquad \textbf{(E)}\ 3x+4$
| [
"If $3x^3-9x^2+kx-12$ is divisible by $x-3$, then by the Remainder Theorem, plugging in $3$ in the cubic results in $0$.\n\\[3 \\cdot 3^3 - 9 \\cdot 3^2 + 3k - 12 = 0\\]\nCombine like terms to get\n\\[3k - 12 = 0\\]\nThus, $k=4$. The cubic is $3x^3-9x^2+4x-12$, and it can be factored (by grouping or synthetic divi... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/22.json | AHSME |
1961_AHSME_Problems | 18 | 0 | Arithmetic | Multiple Choice | The yearly changes in the population census of a town for four consecutive years are,
respectively, 25% increase, 25% increase, 25% decrease, 25% decrease.
The net change over the four years, to the nearest percent, is:
$\textbf{(A)}\ -12 \qquad \textbf{(B)}\ -1 \qquad \textbf{(C)}\ 0 \qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 12$
| [
"A 25% increase means the new population is $\\frac{5}{4}$ of the original population. A 25% decrease means the new population is $\\frac{3}{4}$ of the original population.\n\n\nThus, after four years, the population is $1 \\cdot \\frac{5}{4} \\cdot \\frac{5}{4} \\cdot \\frac{3}{4} \\cdot \\frac{3}{4} = \\frac{225... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/18.json | AHSME |
1961_AHSME_Problems | 38 | 0 | Geometry | Multiple Choice | $\triangle ABC$ is inscribed in a semicircle of radius $r$ so that its base $AB$ coincides with diameter $AB$.
Point $C$ does not coincide with either $A$ or $B$. Let $s=AC+BC$. Then, for all permissible positions of $C$:
$\textbf{(A)}\ s^2\le8r^2\qquad \textbf{(B)}\ s^2=8r^2 \qquad \textbf{(C)}\ s^2 \ge 8r^2 \qquad\\ \textbf{(D)}\ s^2\le4r^2 \qquad \textbf{(E)}\ s^2=4r^2$
| [
"[asy] draw((-50,0)--(-30,40)--(50,0)--(-50,0)); draw(Arc((0,0),50,0,180)); draw(rightanglemark((-50,0),(-30,40),(50,0),200)); dot((-50,0)); label(\"A\",(-50,0),SW); dot((-30,40)); label(\"C\",(-30,40),NW); dot((50,0)); label(\"B\",(50,0),SE); [/asy]\nSince $s=AC+BC$, $s^2 = AC^2 + 2 \\cdot AC \\cdot BC + BC^2$. S... | 3 | ./CreativeMath/AHSME/1961_AHSME_Problems/38.json | AHSME |
1961_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | Let the set consisting of the squares of the positive integers be called $u$; thus $u$ is the set $1, 4, 9, 16 \ldots$.
If a certain operation on one or more members of the set always yields a member of the set,
we say that the set is closed under that operation. Then $u$ is closed under:
$\textbf{(A)}\ \text{Addition}\qquad \textbf{(B)}\ \text{Multiplication} \qquad \textbf{(C)}\ \text{Division} \qquad\\ \textbf{(D)}\ \text{Extraction of a positive integral root}\qquad \textbf{(E)}\ \text{None of these}$
| [
"Consider each option, case by case.\n\n\nFor option A, note that $1 + 4 = 5$, so the set is not closed under addition because $5$ is not a perfect square.\n\n\nFor option B, note that $9 \\cdot 4 = 36$ and $4 \\cdot 25 = 100$. Letting one member of the set be $a^2$ and another member be $b^2$ (where $a$ and $b$ a... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/4.json | AHSME |
1961_AHSME_Problems | 14 | 0 | Geometry | Multiple Choice | A rhombus is given with one diagonal twice the length of the other diagonal.
Express the side of the rhombus is terms of $K$, where $K$ is the area of the rhombus in square inches.
$\textbf{(A)}\ \sqrt{K}\qquad \textbf{(B)}\ \frac{1}{2}\sqrt{2K}\qquad \textbf{(C)}\ \frac{1}{3}\sqrt{3K}\qquad \textbf{(D)}\ \frac{1}{4}\sqrt{4K}\qquad \textbf{(E)}\ \text{None of these are correct}$
| [
"[asy] draw((40,0)--(0,20)--(-40,0)--(0,-20)--(40,0)); draw((-40,0)--(40,0)); draw((0,3)--(3,3)--(3,0)); draw((0,-20)--(0,20)); label(\"$\\frac{a}{2}$\",(0,10),W); label(\"$a$\",(-15,0),N); [/asy]\n\n\nLet one of the diagonals of the rhombus be $a$ units long, so the other diagonal is $2a$ units long. That means... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/14.json | AHSME |
1961_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | If $x$ men working $x$ hours a day for $x$ days produce $x$ articles, then the number of articles
(not necessarily an integer) produced by $y$ men working $y$ hours a day for $y$ days is:
$\textbf{(A)}\ \frac{x^3}{y^2}\qquad \textbf{(B)}\ \frac{y^3}{x^2}\qquad \textbf{(C)}\ \frac{x^2}{y^3}\qquad \textbf{(D)}\ \frac{y^2}{x^3}\qquad \textbf{(E)}\ y$
| [
"Let $k$ be the number of articles produced per hour per person. By using dimensional analysis,\n\\[\\frac{x \\text{ hours}}{\\text{day}} \\cdot x \\text{ days} \\cdot \\frac{k \\text{ articles}}{\\text{hours} \\cdot \\text{person}} \\cdot x \\text{ people} = x \\text{ articles}\\]\nSolving this yields $k = \\frac... | 2 | ./CreativeMath/AHSME/1961_AHSME_Problems/15.json | AHSME |
1961_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | Let $S=(x-1)^4+4(x-1)^3+6(x-1)^2+4(x-1)+1$. Then $S$ equals:
$\textbf{(A)}\ (x-2)^4 \qquad \textbf{(B)}\ (x-1)^4 \qquad \textbf{(C)}\ x^4 \qquad \textbf{(D)}\ (x+1)^4 \qquad \textbf{(E)}\ x^4+1$
| [
"Let $y = x-1$. Substitution results in\n\\[S = y^4 + 4y^3 + 6y^2 + 4y + 1\\]\n\\[S = (y+1)^4\\]\nSubstituting back results in\n\\[S = x^4\\]\nThe answer is $\\boxed{\\textbf{(C)}}$. This problem can also be solved by traditionally expanding and combining like terms (though it would take much longer).\n\n\n"
] | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/5.json | AHSME |
1961_AHSME_Problems | 39 | 0 | Geometry | Multiple Choice | Any five points are taken inside or on a square with side length $1$. Let a be the smallest possible number with the
property that it is always possible to select one pair of points from these five such that the distance between them
is equal to or less than $a$. Then $a$ is:
$\textbf{(A)}\ \sqrt{3}/3\qquad \textbf{(B)}\ \sqrt{2}/2\qquad \textbf{(C)}\ 2\sqrt{2}/3\qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ \sqrt{2}$
| [
"Partition the unit square into four smaller squares of sidelength $\\frac{1}{2}$. Each of the five points lies in one of these squares, and so by the Pigeonhole Principle, there exists two points in the same $\\frac{1}{2}\\times \\frac{1}{2}$ square - the maximum possible distance between them being $\\frac{\\sqrt... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/39.json | AHSME |
1961_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | Consider the graphs of $y=2\log{x}$ and $y=\log{2x}$. We may say that:
$\textbf{(A)}\ \text{They do not intersect}\qquad \\ \textbf{(B)}\ \text{They intersect at 1 point only}\qquad \\ \textbf{(C)}\ \text{They intersect at 2 points only} \qquad \\ \textbf{(D)}\ \text{They intersect at a finite number of points but greater than 2} \qquad \\ \textbf{(E)}\ \text{They coincide}$
| [
"Substitute $y$ into the other equation.\n\\[2\\log{x} = \\log{2x}\\]\nUsing a logarithm property,\n\\[2\\log{x} = \\log{2} + \\log{x}\\]\nSubtract both sides by $\\log{x}$ to get\n\\[\\log{x}=\\log{2}\\]\nThus, $x=2$. Since both $2\\log{x}$ and $\\log{2x}$ are functions, there is only one corresponding y-value, s... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/19.json | AHSME |
1961_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | Points $P$ and $Q$ are both in the line segment $AB$ and on the same side of its midpoint. $P$ divides $AB$ in the ratio $2:3$,
and $Q$ divides $AB$ in the ratio $3:4$. If $PQ=2$, then the length of $AB$ is:
$\textbf{(A)}\ 60\qquad \textbf{(B)}\ 70\qquad \textbf{(C)}\ 75\qquad \textbf{(D)}\ 80\qquad \textbf{(E)}\ 85$
| [
"[asy] draw((0,0)--(70,0)); dot((0,0)); dot((28,0)); dot((30,0)); dot((70,0)); label(\"2\",(29,0),N); label(\"x\",(14,0),N); label(\"y\",(50,0),N); label(\"A\",(0,0),S); label(\"P\",(28,0),SW); label(\"Q\",(30,0),SE); label(\"B\",(70,0),S); [/asy]\n\n\nDraw diagram as shown, where $P$ and $Q$ are on the same side. ... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/23.json | AHSME |
1961_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | Let $r$ be the result of doubling both the base and exponent of $a^b$, and $b$ does not equal to $0$.
If $r$ equals the product of $a^b$ by $x^b$, then $x$ equals:
$\textbf{(A)}\ a \qquad \textbf{(B)}\ 2a \qquad \textbf{(C)}\ 4a \qquad \textbf{(D)}\ 2\qquad \textbf{(E)}\ 4$
| [
"From the problem, $r = (2a)^{2b}$, so\n\\[(2a)^{2b} = a^b \\cdot x^b\\]\n\\[(4a^2)^b = (ax)^b\\]\n\\[4a^2 = ax\\]\n\\[x = 4a\\]\nThus, the answer is $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/9.json | AHSME |
1961_AHSME_Problems | 35 | 0 | Number Theory | Multiple Choice | The number $695$ is to be written with a factorial base of numeration, that is, $695=a_1+a_2\times2!+a_3\times3!+ \ldots a_n \times n!$
where $a_1, a_2, a_3 ... a_n$ are integers such that $0 \le a_k \le k,$ and $n!$ means $n(n-1)(n-2)...2 \times 1$. Find $a_4$
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 4$
| [
"This problem can be approached similarly to other base number problems.\n\n\nSince $120 < 695 < 720$, divide $695$ by $120$. The quotient is $5$ and the remainder is $95$, so rewrite the number as\n\\[695 = 5 \\cdot 120 + 95\\]\nSimilarly, dividing $95$ by $24$ results in a quotient of $3$ and a remainder of $23$... | 1 | ./CreativeMath/AHSME/1961_AHSME_Problems/35.json | AHSME |
1974_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | Let
\[T=\frac{1}{3-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2}.\]
Then
$\mathrm{(A)\ } T<1 \qquad \mathrm{(B) \ }T=1 \qquad \mathrm{(C) \ } 1<T<2 \qquad \mathrm{(D) \ } T>2 \qquad$
$\mathrm{(E) \ }T=\frac{1}{(3-\sqrt{8})(\sqrt{8}-\sqrt{7})(\sqrt{7}-\sqrt{6})(\sqrt{6}-\sqrt{5})(\sqrt{5}-2)}$
| [
"Let's try to rationalize $\\frac{1}{\\sqrt{n+1}-\\sqrt{n}}$. Multiplying the numerator and denominator by $\\sqrt{n+1}+\\sqrt{n}$ gives us $\\frac{1}{\\sqrt{n+1}-\\sqrt{n}}=\\sqrt{n+1}+\\sqrt{n}$.\n\n\nTherefore,\n\n\n\\[T=(3+\\sqrt{8})-(\\sqrt{8}+\\sqrt{7})+(\\sqrt{7}+\\sqrt{6})-(\\sqrt{6}+\\sqrt{5})+(\\sqrt{5}+2... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/20.json | AHSME |
1974_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | A circle of radius $r$ is inscribed in a right isosceles triangle, and a circle of radius $R$ is circumscribed about the triangle. Then $R/r$ equals
$\mathrm{(A)\ } 1+\sqrt{2} \qquad \mathrm{(B) \ }\frac{2+\sqrt{2}}{2} \qquad \mathrm{(C) \ } \frac{\sqrt{2}-1}{2} \qquad \mathrm{(D) \ } \frac{1+\sqrt{2}}{2} \qquad \mathrm{(E) \ }2(2-\sqrt{2})$
| [
"[asy] draw((0,0)--(1,0)--(0,1)--cycle); draw(circle((0.5,0.5),sqrt(2)/2)); draw(circle((1-sqrt(2)/2,1-sqrt(2)/2),1-sqrt(2)/2)); label(\"$A$\",(0,0),SW); label(\"$B$\",(0,1),NNW); label(\"$C$\",(1,0),SSE); label(\"$D$\",(0.5,0.5),NE); draw((0,0)--(0.5,0.5)); draw((1-sqrt(2)/2,1-sqrt(2)/2)--(0,1-sqrt(2)/2)); draw((1... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/16.json | AHSME |
1974_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | For positive real numbers $x$ and $y$ define $x*y=\frac{x\cdot y}{x+y}$' then
$\mathrm{(A)\ } \text{``*" is commutative but not associative} \qquad$
$\mathrm{(B) \ }\text{``*" is associative but not commutative} \qquad$
$\mathrm{(C) \ } \text{``*" is neither commutative nor associative} \qquad$
$\mathrm{(D) \ } \text{``*" is commutative and associative} \qquad$
$\mathrm{(E) \ }\text{none of these} \qquad$
| [
"First, let's check for commutivity. We have \\[y*x=\\frac{y\\cdot x}{y+x}=\\frac{x\\cdot y}{x+y}=x*y\\], so $*$ is commutative.\n\n\nNow we check for associativity. We have \\[(x*y)*z=\\left(\\frac{x\\cdot y}{x+y}\\right)*z=\\frac{\\frac{x\\cdot y}{x+y}\\cdot z}{\\frac{x\\cdot y}{x+y}+z}=\\frac{x\\cdot y\\cdot z}{... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/6.json | AHSME |
1974_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | A town's population increased by $1,200$ people, and then this new population decreased by $11\%$. The town now had $32$ less people than it did before the $1,200$ increase. What is the original population?
$\mathrm{(A)\ } 1,200 \qquad \mathrm{(B) \ }11,200 \qquad \mathrm{(C) \ } 9,968 \qquad \mathrm{(D) \ } 10,000 \qquad \mathrm{(E) \ }\text{none of these}$
| [
"Let $n$ be the original population. When $1,200$ people came, the new population was $n+1200$. Then, since the new population decreased by $11\\%$, we must multiply this new population by $.89$ to get the final population. This is equal to $(0.89)(n+1200)=0.89n+1068$. We're given that this is $32$ less than the or... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/7.json | AHSME |
1974_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | 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, \n\n\n\\[(1+i)^{20}-(1-i)^{20}=(2i)^{10}-(-2i)^{10}=(2i)^{10}-(2i)^{10}=0, \\boxed{\\text{C}}.\\]\n\n\n"
] | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/17.json | AHSME |
1974_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | In a geometric series of positive terms the difference between the fifth and fourth terms is $576$, and the difference between the second and first terms is $9$. What is the sum of the first five terms of this series?
$\mathrm{(A)\ } 1061 \qquad \mathrm{(B) \ }1023 \qquad \mathrm{(C) \ } 1024 \qquad \mathrm{(D) \ } 768 \qquad \mathrm{(E) \ }\text{none of these}$
| [
"Let the first term be $a$ and the common ratio be $r$. Therefore, the second term is $ar$, the fourth term is $ar^3$, and the fifth term is $ar^4$. We're given that $ar^4-ar^3=576\\implies ar^3(r-1)=576$ and $ar-a=9\\implies a(r-1)=9$. Dividing this first equation by this second one, we get $r^3=\\frac{576}{9}=64\... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/21.json | AHSME |
1974_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | 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{\\text{B... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/10.json | AHSME |
1974_AHSME_Problems | 26 | 0 | Number Theory | Multiple Choice | The number of distinct positive integral divisors of $(30)^4$ excluding $1$ and $(30)^4$ is
$\mathrm{(A)\ } 100 \qquad \mathrm{(B) \ }125 \qquad \mathrm{(C) \ } 123 \qquad \mathrm{(D) \ } 30 \qquad \mathrm{(E) \ }\text{none of these}$
| [
"$$ (Error compiling LaTeX. Unknown error_msg)The prime factorization of $30$ is $2\\cdot3\\cdot5$, so the prime factorization of $30^4$ is $2^4\\cdot3^4\\cdot5^4$. Therefore, the number of positive divisors of $30^4$ is $(4+1)(4+1)(4+1)=125$. However, we have to subtract $2$ to account for $1$ and $30^4$, so our f... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/26.json | AHSME |
1974_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | 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 $\\left(\\frac{w}{l}\\right)^2+\\left(\\frac{w}{l}\\right)-1=0$. The... | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/30.json | AHSME |
1974_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | If $f(x)=3x+2$ for all real $x$, then the statement:
"$|f(x)+4|<a$ whenever $|x+2|<b$ and $a>0$ and $b>0$"
is true when
$\mathrm{(A)}\ b\le a/3\qquad\mathrm{(B)}\ b > a/3\qquad\mathrm{(C)}\ a\le b/3\qquad\mathrm{(D)}\ a > b/3\\ \qquad\mathrm{(E)}\ \text{The statement is never true.}$
| [
"Plugging in $f(x)=3x+2$ into $|f(x)+4|<a$, we get $|3x+6|=|3(x+2)|=3|x+2|<a$. Therefore, $|x+2|<\\frac{a}{3}$. Therefore, we want $|x+2|<b\\implies |x+2|<\\frac{a}{3}$. This is clearly only true when $b\\le\\frac{a}{3}, \\boxed{\\text{A}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/27.json | AHSME |
1974_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | If $x\not=0$ or $4$ and $y\not=0$ or $6$, then $\frac{2}{x}+\frac{3}{y}=\frac{1}{2}$ is equivalent to
$\mathrm{(A)\ } 4x+3y=xy \qquad \mathrm{(B) \ }y=\frac{4x}{6-y} \qquad \mathrm{(C) \ } \frac{x}{2}+\frac{y}{3}=2 \qquad$
$\mathrm{(D) \ } \frac{4y}{y-6}=x \qquad \mathrm{(E) \ }\text{none of these}$
| [
"To clear out the fractions, we multiply both sides of the equation by $xy$ to get $2y+3x=\\frac{xy}{2}$, or $4y+6x=xy$. From this, we have $4y=xy-6x$. We can factor an $x$ out of the RHS, and so $4y=x(y-6)\\implies \\frac{4y}{y-6}=x, \\boxed{\\text{D}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1974_AHSME_Problems/1.json | AHSME |
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