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 |
|---|---|---|---|---|---|---|---|---|---|
1963_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | The expression $x^2-y^2-z^2+2yz+x+y-z$ has:
$\textbf{(A)}\ \text{no linear factor with integer coefficients and integer exponents} \qquad \\ \textbf{(B)}\ \text{the factor }-x+y+z \qquad \\ \textbf{(C)}\ \text{the factor }x-y-z+1 \qquad \\ \textbf{(D)}\ \text{the factor }x+y-z+1 \qquad \\ \textbf{(E)}\ \text{the fact... | [
"Factor the perfect square trinomial.\n\\[x^2 - (y-z)^2 + x + y - z\\]\nFactor the difference of squares.\n\\[(x+y-z)(x-y+z) + x + y - z\\]\nFactor by grouping.\n\\[(x+y-z)(x-y+z+1)\\]\nThe answer is $\\boxed{\\textbf{(E)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/21.json | AHSME |
1963_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | Point $P$ is taken interior to a square with side-length $a$ and such that is it equally distant from two
consecutive vertices and from the side opposite these vertices. If $d$ represents the common distance, then $d$ equals:
$\textbf{(A)}\ \frac{3a}{5}\qquad \textbf{(B)}\ \frac{5a}{8}\qquad \textbf{(C)}\ \frac{3a}{... | [
"[asy] draw((0,0)--(16,0)--(16,16)--(0,16)--(0,0)); draw((0,0)--(8,6)); draw((16,0)--(8,6)); draw((8,16)--(8,6)); draw((8,0)--(8,6),dotted); label(\"$d$\",(4,3),NW); label(\"$d$\",(12,3),NE); label(\"$d$\",(8,11),W); label(\"$\\frac{a}{2}$\",(4,0),S); label(\"$\\frac{a}{2}$\",(12,0),S); label(\"$a-d$\",(8,2),W); [/... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/10.json | AHSME |
1963_AHSME_Problems | 26 | 0 | Other | Multiple Choice | Consider the statements:
$\textbf{(1)}\ p\text{ }\wedge\sim q\wedge r\qquad\textbf{(2)}\ \sim p\text{ }\wedge\sim q\wedge r\qquad\textbf{(3)}\ p\text{ }\wedge\sim q\text{ }\wedge\sim r\qquad\textbf{(4)}\ \sim p\text{ }\wedge q\text{ }\wedge r$
where $p,q$, and $r$ are propositions. How many of these imply the truth... | [
"Statement $1$ states that $p$ is true and $q$ is false. Therefore, $p \\rightarrow q$ is false, because a premise being true and a conclusion being false is, itself, false. This means that $(p \\rightarrow q) \\rightarrow X$, where $X$ is any logical statement (or series of logical statements) must be true - if ... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/26.json | AHSME |
1963_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | Let $F=\log\dfrac{1+x}{1-x}$. Find a new function $G$ by replacing each $x$ in $F$ by $\dfrac{3x+x^3}{1+3x^2}$, and simplify.
The simplified expression $G$ is equal to:
$\textbf{(A)}\ -F \qquad \textbf{(B)}\ F\qquad \textbf{(C)}\ 3F \qquad \textbf{(D)}\ F^3 \qquad \textbf{(E)}\ F^3-F$
| [
"Replace the $x$ in $F$ with $\\frac{3x+x^3}{1+3x^2}$.\n\\[\\log\\dfrac{1 + \\frac{3x+x^3}{1+3x^2} }{1 - \\frac{3x+x^3}{1+3x^2} }\\]\n\\[\\log\\dfrac{\\frac{1+3x^2}{1+3x^2} + \\frac{3x+x^3}{1+3x^2} }{\\frac{1+3x^2}{1+3x^2} - \\frac{3x+x^3}{1+3x^2} }\\]\n\\[\\log (\\frac{1+3x+3x^2+x^3}{1+3x^2} \\div \\frac{1-3x+3x^2... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/30.json | AHSME |
1963_AHSME_Problems | 31 | 0 | Number Theory | Multiple Choice | The number of solutions in positive integers of $2x+3y=763$ is:
$\textbf{(A)}\ 255 \qquad \textbf{(B)}\ 254\qquad \textbf{(C)}\ 128 \qquad \textbf{(D)}\ 127 \qquad \textbf{(E)}\ 0$
| [
"Solving for $x$ in the equation yields $x =rfthe meaning of theta 0$. Solving the inequality results in $y \\le 254 \\frac{1}{3}$. From the two conditions, $y$ can be an odd number from $1$ to $253$, so there are $127$ solutions where $x$ and $y$ are integers. The answer is $\\boxed{\\textbf{(D)}}$.\n\n\n",
"... | 3 | ./CreativeMath/AHSME/1963_AHSME_Problems/31.json | AHSME |
1963_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | Which one of the following points is not on the graph of $y=\dfrac{x}{x+1}$?
$\textbf{(A)}\ (0,0)\qquad \textbf{(B)}\ \left(-\frac{1}{2},-1\right)\qquad \textbf{(C)}\ \left(\frac{1}{2},\frac{1}{3}\right)\qquad \textbf{(D)}\ (-1,1)\qquad \textbf{(E)}\ (-2,2)$
| [
"Looking at the domain of the graph, $x$ cannot be $-1$. Therefore, the answer is $\\boxed{\\textbf{(D)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/1.json | AHSME |
1963_AHSME_Problems | 11 | 0 | Arithmetic | Multiple Choice | The arithmetic mean of a set of $50$ numbers is $38$. If two numbers of the set, namely $45$ and $55$, are discarded,
the arithmetic mean of the remaining set of numbers is:
$\textbf{(A)}\ 38.5 \qquad \textbf{(B)}\ 37.5 \qquad \textbf{(C)}\ 37 \qquad \textbf{(D)}\ 36.5 \qquad \textbf{(E)}\ 36$
| [
"If the arithmetic mean of a set of $50$ numbers is $38$, then the sum of the $50$ numbers equals $1900$. Since $45$ and $55$ are being removed, subtract $100$ to get the sum of the remaining $48$ numbers, which is $1800$. Therefore, the new mean is $37.5$, which is answer choice $\\boxed{\\textbf{(B)}}$.\n\n\n"
... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/11.json | AHSME |
1963_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | let $n=x-y^{x-y}$. Find $n$ when $x=2$ and $y=-2$.
$\textbf{(A)}\ -14 \qquad \textbf{(B)}\ 0 \qquad \textbf{(C)}\ 1 \qquad \textbf{(D)}\ 18 \qquad \textbf{(E)}\ 256$
| [
"Substitute the variables to determine the value of $n$.\n\\[n = 2 - (-2)^{2-(-2)}\\]\n\\[n = 2 - (-2)^4\\]\n\\[n = 2 - 16\\]\n\\[n = -14\\]\nThe answer is $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/2.json | AHSME |
1963_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | Given the equation $3x^2 - 4x + k = 0$ with real roots. The value of $k$ for which the product of the roots of the equation is a maximum is:
$\textbf{(A)}\ \frac{16}{9}\qquad \textbf{(B)}\ \frac{16}{3}\qquad \textbf{(C)}\ \frac{4}{9}\qquad \textbf{(D)}\ \frac{4}{3}\qquad \textbf{(E)}\ -\frac{4}{3}$
| [
"By Vieta's Formulas, the product of the roots is $\\frac{k}{3}$. This value increases as $k$ increases.\n\n\nAlso, the quadratic’s roots are real, then the discriminant is greater than or equal to zero, so\n\\[16-12k \\ge 0\\]\n\\[-12k \\ge -16\\]\n\\[k \\le \\frac{4}{3}\\]\n\n\nThus, the $k$ value that maximizes... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/28.json | AHSME |
1963_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | Three vertices of parallelogram $PQRS$ are $P(-3,-2), Q(1,-5), R(9,1)$ with $P$ and $R$ diagonally opposite.
The sum of the coordinates of vertex $S$ is:
$\textbf{(A)}\ 13 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 11 \qquad \textbf{(D)}\ 10 \qquad \textbf{(E)}\ 9$
| [
"[asy] import graph; size(7.22 cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.2,xmax=10.2,ymin=-6.2,ymax=5.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=1,gy=1... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/12.json | AHSME |
1963_AHSME_Problems | 32 | 0 | Algebra | Multiple Choice | The dimensions of a rectangle $R$ are $a$ and $b$, $a < b$. It is required to obtain a rectangle with dimensions $x$ and $y$, $x < a, y < a$,
so that its perimeter is one-third that of $R$, and its area is one-third that of $R$. The number of such (different) rectangles is:
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1\qq... | [
"Using the perimeter and area formulas,\n\\[2(x+y) = \\frac{2}{3}(a+b)\\]\n\\[x+y = \\frac{a+b}{3}\\]\n\\[xy = \\frac{ab}{3}\\]\nDividing the second equation by the last equation results in\n\\[\\frac1y + \\frac1x = \\frac1b + \\frac1a\\]\nSince $x,y < a$, $\\tfrac1a < \\tfrac1x, \\tfrac1y$. Since $a < b$, $\\tfra... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/32.json | AHSME |
1963_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | Consider equations of the form $x^2 + bx + c = 0$. How many such equations have real roots and have coefficients $b$ and $c$ selected
from the set of integers $\{1,2,3, 4, 5,6\}$?
$\textbf{(A)}\ 20 \qquad \textbf{(B)}\ 19 \qquad \textbf{(C)}\ 18 \qquad \textbf{(D)}\ 17 \qquad \textbf{(E)}\ 16$
| [
"The discriminant of the quadratic is $b^2 - 4c$. Since the quadratic has real roots,\n\\[b^2 - 4c \\ge 0\\]\n\\[b^2 \\ge 4c\\]\nIf $b = 6$, then $c$ can be from $1$ to $6$. If $b = 5$, then $c$ can also be from $1$ to $6$. If $b=4$, then $c$ can be from $1$ to $4$. If $b=3$, then $c$ can be $1$ or $2$. If $b=... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/24.json | AHSME |
1963_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | Point $F$ is taken in side $AD$ of square $ABCD$. At $C$ a perpendicular is drawn to $CF$, meeting $AB$ extended at $E$.
The area of $ABCD$ is $256$ square inches and the area of $\triangle CEF$ is $200$ square inches. Then the number of inches in $BE$ is:
[asy] size(6cm); pair A = (0, 0), B = (1, 0), C = (1, 1), D ... | [
"Because $ABCD$ is a square, $DC = CB = 16$, $DC \\perp DA$, and $CB \\perp BA$. Also, because $\\angle DCF + \\angle FCB = \\angle FCB + \\angle BCE = 90^\\circ$, $\\angle DCF = \\angle BCE$. Thus, by ASA Congruency, $\\triangle DCF \\cong \\triangle BCF$.\n\n\nFrom the congruency, $CF = CE$. Using the area for... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/25.json | AHSME |
1963_AHSME_Problems | 33 | 0 | Geometry | Multiple Choice | Given the line $y = \dfrac{3}{4}x + 6$ and a line $L$ parallel to the given line and $4$ units from it. A possible equation for $L$ is:
$\textbf{(A)}\ y =\frac{3}{4}x+1\qquad \textbf{(B)}\ y =\frac{3}{4}x\qquad \textbf{(C)}\ y =\frac{3}{4}x-\frac{2}{3}\qquad \\ \textbf{(D)}\ y = \dfrac{3}{4}x -1 \qquad \textbf{(E)}\... | [
"[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=-10.2,xmax=10.2,ymin=-10.2,ymax=10.2; pen cqcqcq=rgb(0.75,0.75,0.75), evevff=rgb(0.9,0.9,1), zzttqq=rgb(0.6,0.2,0); /*grid*/ pen gs=linewidth(0.7)+cqcqcq+linetype(\"2 2\"); real gx=1,g... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/33.json | AHSME |
1963_AHSME_Problems | 13 | 0 | Number Theory | Multiple Choice | If $2^a+2^b=3^c+3^d$, the number of integers $a,b,c,d$ which can possibly be negative, is, at most:
$\textbf{(A)}\ 4 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ 1 \qquad \textbf{(E)}\ 0$
| [
"Assume $c,d \\ge 0$, and WLOG, assume $a<0$ and $a \\le b$. This also takes into account when $b$ is negative. That means\n\\[\\frac{1}{2^{-a}} + 2^b = 3^c + 3^d\\]\nMultiply both sides by $2^{-a}$ to get\n\\[1 + 2^{-a+b} = 2^{-a} (3^c + 3^d)\\]\nNote that both sides are integers. If $a \\ne b$, then the right ... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/13.json | AHSME |
1963_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | A particle projected vertically upward reaches, at the end of $t$ seconds, an elevation of $s$ feet where $s = 160 t - 16t^2$. The highest elevation is:
$\textbf{(A)}\ 800 \qquad \textbf{(B)}\ 640\qquad \textbf{(C)}\ 400 \qquad \textbf{(D)}\ 320 \qquad \textbf{(E)}\ 160$
| [
"The highest elevation a particle can reach is the vertex of the quadratic. The x-value that can get the maximum is $\\frac{-160}{-2 \\cdot 16} = 5$, so the highest elevation is $160(5) - 16(5^2) = 400$ feet, which is answer choice $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/29.json | AHSME |
1963_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | If the reciprocal of $x+1$ is $x-1$, then $x$ equals:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ \pm 1\qquad \textbf{(E)}\ \text{none of these}$
| [
"We form the equation $x+1=\\frac{1}{x-1}$. \n\n\nGetting rid of the fraction yields: $x^2-1=1$ $\\implies$ $x^2=2$ $\\implies$ $x=\\pm{\\sqrt{2}}=\\boxed{\\text{E}}$\n\n\n~mathsolver101\n\n\n"
] | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/3.json | AHSME |
1963_AHSME_Problems | 34 | 0 | Geometry | Multiple Choice | In $\triangle ABC$, side $a = \sqrt{3}$, side $b = \sqrt{3}$, and side $c > 3$. Let $x$ be the largest number such that the magnitude,
in degrees, of the angle opposite side $c$ exceeds $x$. Then $x$ equals:
$\textbf{(A)}\ 150^{\circ} \qquad \textbf{(B)}\ 120^{\circ}\qquad \textbf{(C)}\ 105^{\circ} \qquad \textbf{(D... | [
"Using the Law of Cosines,\n\\[\\sqrt{3 + 3 - 2\\cdot 3 \\cdot \\cos{x^\\circ}}>3\\]\n\n\nBoth sides are positive, so squaring both sides will not affect the inequality.\n\n\n\\[6 - 6 \\cos{x^\\circ}>9\\]\n\\[\\cos{x^\\circ} < -\\frac{1}{2}\\]\n\n\nNote that $\\cos{120^\\circ} = -\\frac{1}{2}$. As $x$ gets closer ... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/34.json | AHSME |
1963_AHSME_Problems | 8 | 0 | Number Theory | Multiple Choice | The smallest positive integer $x$ for which $1260x=N^3$, where $N$ is an integer, is:
$\textbf{(A)}\ 1050 \qquad \textbf{(B)}\ 1260 \qquad \textbf{(C)}\ 1260^2 \qquad \textbf{(D)}\ 7350 \qquad \textbf{(6)}\ 44100$
| [
"Factoring $1260$ results in $2^2 \\cdot 3^2 \\cdot 5 \\cdot 7$. If an integer $N$ is a perfect cube, then the exponents of all the primes in its prime factorization are multiples of 3. Thus, the smallest positive integer that can be multiplied by $1260$ to result in a perfect cube is $2 \\cdot 3 \\cdot 5^2 \\cdo... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/8.json | AHSME |
1963_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | Acute-angled $\triangle ABC$ is inscribed in a circle with center at $O$; $\stackrel \frown {AB} = 120^\circ$ and $\stackrel \frown {BC} = 72^\circ$.
A point $E$ is taken in minor arc $AC$ such that $OE$ is perpendicular to $AC$. Then the ratio of the magnitudes of $\angle OBE$ and $\angle BAC$ is:
$\textbf{(A)}\ \... | [
"[asy] draw(circle((0,0),1)); dot((-1,0)); pair A=(-1,0),B=(0.5,0.866),C=(0.978,-0.208),O=(0,0),E=(-0.105,-0.995); label(\"A\",(-1,0),W); dot((0.5,0.866)); label(\"B\",(0.5,0.866),NE); dot((0.978,-0.208)); label(\"C\",(0.978,-0.208),SE); dot((0,0)); label(\"O\",(0,0),NE); dot(E); label(\"E\",E,S); draw(A--B--C--A);... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/22.json | AHSME |
1963_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | Chord $EF$ is the perpendicular bisector of chord $BC$, intersecting it in $M$.
Between $B$ and $M$ point $U$ is taken,
and $EU$ extended meets the circle in $A$. Then, for any selection of $U$, as described, $\triangle EUM$ is similar to:
[asy] pair B = (-0.866, -0.5); pair C = (0.866, -0.5); pair E = (0, -1); pai... | [
"Note that $\\triangle EUM$ is a right triangle with one of the angles being $\\angle FEA$. This leads to prediction that $\\triangle FEA$ is the similar triangle as it shares an angle, and to prove this, we need to show that $FE$ is a diameter.\n\n\n[asy] pair B = (-0.866, -0.5); pair C = (0.866, -0.5); pair E = ... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/18.json | AHSME |
1963_AHSME_Problems | 38 | 0 | Geometry | Multiple Choice | Point $F$ is taken on the extension of side $AD$ of parallelogram $ABCD$. $BF$ intersects diagonal $AC$ at $E$ and side $DC$ at $G$.
If $EF = 32$ and $GF = 24$, then $BE$ equals:
[asy] size(7cm); pair A = (0, 0), B = (7, 0), C = (10, 5), D = (3, 5), F = (5.7, 9.5); pair G = intersectionpoints(B--F, D--C)[0]; pair E ... | [
"Let $BE = x$ and $BC = y$. Since $AF \\parallel BC$, by AA Similarity, $\\triangle AFE \\sim \\triangle CBE$. That means $\\frac{AF}{CB} = \\frac{FE}{BE}$. Substituting in values results in\n\\[\\frac{AF}{y} = \\frac{32}{x}\\]\nThus, $AF = \\frac{32y}{x}$, so $FD = \\frac{32y - xy}{x}$.\n\n\n\n\nIn addition, $D... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/38.json | AHSME |
1963_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | For what value(s) of $k$ does the pair of equations $y=x^2$ and $y=3x+k$ have two identical solutions?
$\textbf{(A)}\ \frac{4}{9}\qquad \textbf{(B)}\ -\frac{4}{9}\qquad \textbf{(C)}\ \frac{9}{4}\qquad \textbf{(D)}\ -\frac{9}{4}\qquad \textbf{(E)}\ \pm\frac{9}{4}$
| [
"If the system of equations has two identical solutions, then only one $(x,y)$ pair will satisfy both equations.\n\n\nSubstitute $y$ in one equation into another equation.\n\\[x^2 = 3x + k\\]\n\\[x^2 - 3x = k\\]\nComplete the square to get\n\\[x^2 - 3x + \\frac{9}{4} = k + \\frac{9}{4}\\]\n\\[(x - \\frac{3}{2})^2 =... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/4.json | AHSME |
1963_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | Given the equations $x^2+kx+6=0$ and $x^2-kx+6=0$. If, when the roots of the equation are suitably listed,
each root of the second equation is $5$ more than the corresponding root of the first equation, then $k$ equals:
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ -5 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ -7 \qquad \t... | [
"Let the two roots of $x^2 + kx + 6$ be $r$ and $s$. By Vieta's Formulas,\n\\[r+s=-k\\]\nEach root of $x^2 - kx + 6$ is five more than each root of the original, so using Vieta's Formula again yields\n\\[r+5+s+5 = k\\]\n\\[r+s+10=k\\]\nSubstitute $r+s$ to get\n\\[-k + 10 = k\\]\n\\[10 = 2k\\]\n\\[5 = k\\]\nThe ans... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/14.json | AHSME |
1963_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | A circle is inscribed in an equilateral triangle, and a square is inscribed in the circle.
The ratio of the area of the triangle to the area of the square is:
$\textbf{(A)}\ \sqrt{3}:1\qquad \textbf{(B)}\ \sqrt{3}:\sqrt{2}\qquad \textbf{(C)}\ 3\sqrt{3}:2\qquad \textbf{(D)}\ 3:\sqrt{2}\qquad \textbf{(E)}\ 3:2\sqrt{2}... | [
"[asy] draw(circle((0,0),100)); draw((173.205,-100)--(-173.205,-100)--(0,200)--(173.205,-100)); draw((-100,0)--(0,100)--(100,0)--(0,-100)--(-100,0)); draw((0,0)--(0,-100),dotted); draw((0,0)--(-173.205,-100),dotted); [/asy]\nLet the radius of the circle be $r$. That means the diameter of the circle is $2r$, so the... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/15.json | AHSME |
1963_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | If $x$ and $\log_{10} x$ are real numbers and $\log_{10} x<0$, then:
$\textbf{(A)}\ x<0 \qquad \textbf{(B)}\ -1<x<1 \qquad \textbf{(C)}\ 0<x\le 1 \\ \textbf{(D)}\ -1<x<0 \qquad \textbf{(E)}\ 0<x<1$
| [
"Let $\\log_{10} x = a$, so $10^a = x$. Note that $10^0 = 1$ and that $10^a$ increases as $a$ increases. Since $a < 0$, $10^a < 1$. However, for all $a$, $10^a > 0$, so $0 < x < 1$. The answer is $\\boxed{\\textbf{(E)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/5.json | AHSME |
1963_AHSME_Problems | 39 | 0 | Geometry | Multiple Choice | In $\triangle ABC$ lines $CE$ and $AD$ are drawn so that
$\dfrac{CD}{DB}=\dfrac{3}{1}$ and $\dfrac{AE}{EB}=\dfrac{3}{2}$. Let $r=\dfrac{CP}{PE}$
where $P$ is the intersection point of $CE$ and $AD$. Then $r$ equals:
[asy] size(8cm); pair A = (0, 0), B = (9, 0), C = (3, 6); pair D = (7.5, 1.5), E = (6.5, 0); pair P ... | [
"[asy] size(8cm); pair A = (0, 0), B = (9, 0), C = (3, 6); pair D = (7.5, 1.5), E = (6.5, 0); pair P = intersectionpoints(A--D, C--E)[0]; draw(A--B--C--cycle); draw(A--D); draw(C--E); label(\"$A$\", A, SW); label(\"$B$\", B, SE); label(\"$C$\", C, N); label(\"$D$\", D, NE); label(\"$E$\", E, S); label(\"$P$\", P, S... | 2 | ./CreativeMath/AHSME/1963_AHSME_Problems/39.json | AHSME |
1963_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | In counting $n$ colored balls, some red and some black, it was found that $49$ of the first $50$ counted were red.
Thereafter, $7$ out of every $8$ counted were red. If, in all, $90$ % or more of the balls counted were red, the maximum value of $n$ is:
$\textbf{(A)}\ 225 \qquad \textbf{(B)}\ 210 \qquad \textbf{(C)}\... | [
"The desired percentage of red balls is more than $90$ percent, so write an inequality.\n\n\n\\[\\frac{49+7x}{50+8x} \\ge 0.9\\]\n\n\nSince $x >0$, the sign does not need to be swapped after multiplying both sides by $50+8x$.\n\n\n\\[49+7x \\ge 45+7.2x\\]\n\\[4 \\ge 0.2x\\]\n\\[20 \\ge x\\]\n\n\nThus, up to $20$ ba... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/19.json | AHSME |
1963_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | A gives $B$ as many cents as $B$ has and $C$ as many cents as $C$ has. Similarly, $B$ then gives $A$ and $C$ as many cents as each then has.
$C$, similarly, then gives $A$ and $B$ as many cents as each then has. If each finally has $16$ cents, with how many cents does $A$ start?
$\textbf{(A)}\ 24 \qquad \textbf{(B)}... | [
"Let $a$ be number of cents $A$ originally had, $b$ be number of cents $B$ originally had, and $c$ be number of cents $C$ originally had.\n\n\nAfter $A$ gave his money away, $A$ has $a-b-c$ cents, $B$ has $2b$ cents, and $C$ has $2c$ cents.\n\n\nAfter $B$ gave his money away, $A$ has $2a-2b-2c$ cents, $B$ has $-a+3... | 2 | ./CreativeMath/AHSME/1963_AHSME_Problems/23.json | AHSME |
1963_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | In the expansion of $\left(a-\dfrac{1}{\sqrt{a}}\right)^7$ the coefficient of $a^{-\dfrac{1}{2}}$ is:
$\textbf{(A)}\ -7 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ -21 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 35$
| [
"By the Binomial Theorem, each term of the expansion is $\\binom{7}{n}(a)^{7-n}(\\frac{-1}{\\sqrt{a}})^n$.\n\n\nWe want the exponent of $a$ to be $-\\frac{1}{2}$, so \n\\[(7-n)-\\frac{1}{2}n=-\\frac{1}{2}\\]\n\\[-\\frac{3}{2}n = -\\frac{15}{2}\\]\n\\[n = 5\\]\n\n\nIf $n=5$, then the corresponding term is\n\\[\\bino... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/9.json | AHSME |
1963_AHSME_Problems | 35 | 0 | Geometry | Multiple Choice | The lengths of the sides of a triangle are integers, and its area is also an integer.
One side is $21$ and the perimeter is $48$. The shortest side is:
$\textbf{(A)}\ 8 \qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 12 \qquad \textbf{(D)}\ 14 \qquad \textbf{(E)}\ 16$
| [
"Let $b$ and $c$ be the other two sides of the triangle. The perimeter of the triangle is $48$ units, so $c = 27-b$ and the semiperimeter equals $24$ units.\n\n\nBy Heron's Formula, the area of the triangle is $\\sqrt{24 \\cdot 3(24-b)(b-3)}$. Plug in the answer choices for $b$ and write the prime factorization o... | 1 | ./CreativeMath/AHSME/1963_AHSME_Problems/35.json | AHSME |
1976_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | Let $a,~b$, and $x$ be positive real numbers distinct from one. Then $4(\log_ax)^2+3(\log_bx)^2=8(\log_ax)(\log_bx)$
$\textbf{(A) }\text{for all values of }a,~b,\text{ and }x\qquad\\ \textbf{(B) }\text{if and only if }a=b^2\qquad\\ \textbf{(C) }\text{if and only if }b=a^2\qquad\\ \textbf{(D) }\text{if and only if }x=... | [
"Because $\\log_{m} n = \\dfrac{\\log n}{\\log m}$, $4(\\log_{a} x)^2+3(\\log_{b} x)^2 =$ $\\dfrac{4(\\log x)^2}{(\\log a)^2}+\\dfrac{3(\\log x)^2}{(\\log b)^2} = \\dfrac{(\\log x)^2(4(\\log a)^2+3(\\log b)^2)}{(\\log a \\log b)^2}$.\n\n\nTherefore,\n\n\n\\[\\dfrac{(\\log x)^2(4(\\log a)^2+3(\\log b)^2)}{(\\log a \... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/20.json | AHSME |
1976_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | If $c$ is a real number and the negative of one of the solutions of $x^2-3x+c=0$ is a solution of $x^2+3x-c=0$, then the solutions of $x^2-3x+c=0$ are
$\textbf{(A) }1,~2\qquad \textbf{(B) }-1,~-2\qquad \textbf{(C) }0,~3\qquad \textbf{(D) }0,~-3\qquad \textbf{(E) }\frac{3}{2},~\frac{3}{2}$
\section{Solution}
We let... | [
"We let the roots of the first equation be $r,s$ and the roots of the second equation be $s, -t$. By Vieta's Formulas, $r+s=3$ and $s-t=-3$, $rs=c$ and $-st=c$. So, $r=t$. Thus, $t+s=3$, $s-t=3$, so $t=0$, and $s=3\\Rightarrow \\textbf{(C)}$.~MathJams\n\n"
] | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/6.json | AHSME |
1976_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | If $x$ is a real number, then the quantity $(1-|x|)(1+x)$ is positive if and only if
$\textbf{(A) }|x|<1\qquad \textbf{(B) }|x|>1\qquad \textbf{(C) }x<-1\text{ or }-1<x<1\qquad\\ \textbf{(D) }x<1\qquad \textbf{(E) }x<-1$
| [
"We divide our solution into three cases: that of $x < -1$, that of $-1 < x < 1$, and that of $x > 1$. (When $x = -1$ or $x = 1$, the expression is zero, therefore not positive.)\n\n\nIf $x < -1$, then the first factor is negative, and the second factor is also negative.\n\n\nIf $-1 < x < 1$, then the first factor ... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/7.json | AHSME |
1976_AHSME_Problems | 17 | 0 | Other | Multiple Choice | 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... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/17.json | AHSME |
1976_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | 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{(... | [
"Combine the terms in the product to get $2^{\\frac{1+3+5+ \\dots +(2n-1)+(2n+1)}{7}}$.\n\n\nThe 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}.\\]\n\n\nWe want this inequality to be true with the smallest positive ... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/21.json | AHSME |
1976_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | If $m,~n,~p$, and $q$ are real numbers and $f(x)=mx+n$ and $g(x)=px+q$, then the equation $f(g(x))=g(f(x))$ has a solution
$\textbf{(A) }\text{for all choices of }m,~n,~p, \text{ and } q\qquad\\ \textbf{(B) }\text{if and only if }m=p\text{ and }n=q\qquad\\ \textbf{(C) }\text{if and only if }mq-np=0\qquad\\ \textbf{(D... | [
"\\[f(g(x) = g(f(x)) \\qquad \\implies \\qquad m(px + q) + n = p(mx + n) + q\\]\n\n\nThis simplifies to $mq + n = pn + q$ or $n(1 - p) - q(1 - m) = 0.$\n\n\n\n\nThe answer is $\\boxed{\\textbf{(D)}}.$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/10.json | AHSME |
1976_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | 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 $(x,y,z)=\\left(a,\\frac b2,\\frac c4\\right).$\n\n\nWe rewrite the given equations in terms of $a,b,$ and $c:$\n\\begin{align*} a + b + c &= 12, \\\\ \\frac{ab}{2} + \\frac{bc}{2} + \\frac{ac}{2} &= 22, \\\\ \\frac{abc}{8} &= 6. \\end{al... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/30.json | AHSME |
1976_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | 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 $y=\\sqrt{3-2\\sqrt{2}}.$\n\n\nNote that\n\\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... | 2 | ./CreativeMath/AHSME/1976_AHSME_Problems/27.json | AHSME |
1976_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | If one minus the reciprocal of $(1-x)$ equals the reciprocal of $(1-x)$, then $x$ equals
$\textbf{(A) }-2\qquad \textbf{(B) }-1\qquad \textbf{(C) }1/2\qquad \textbf{(D) }2\qquad \textbf{(E) }3$
| [
"The reciprocal of $(1-x)$ is $\\frac{1}{1-x}$, so our equation is \\[1-\\frac{1}{1-x}=\\frac{1}{1-x},\\] which is equivalent to $\\frac{1}{1-x}=\\frac{1}{2}$. So, $1-x=2$ and $x=-1\\Rightarrow \\textbf{(B)}$.~MathJams\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/1.json | AHSME |
1976_AHSME_Problems | 11 | 0 | Other | Multiple Choice | Which of the following statements is (are) equivalent to the statement
"If the pink elephant on planet alpha has purple eyes, then the wild pig on planet beta does not have a long nose"?
$\begin{array}{l}\textbf{I. }\text{"If the wild pig on planet beta has a long nose, then the pink elephant on planet alpha has pur... | [
"The following statements are all logically equivalent.\n\n\nI. P implies Q (pink elephant on planet alpha has purple eyes ---> wild pig on planet beta does not have a long nose)\n\n\n\n\nIII. Not Q implies not P (contrapositive)\n\n\nIV. Not P nor Q (inverse)\n\n\n\n\nI is the given statement, so III and IV work. ... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/11.json | AHSME |
1976_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | For how many real numbers $x$ is $\sqrt{-(x+1)^2}$ a real number?
$\textbf{(A) }\text{none}\qquad \textbf{(B) }\text{one}\qquad \textbf{(C) }\text{two}\qquad\\ \textbf{(D) }\text{a finite number greater than two}\qquad \textbf{(E) }\infty$
| [
"$\\sqrt{-(x+1)^2}$ is a real number, if and only if $-(x+1)^2$ is nonnegative. Since $(x+1)^2$ is always nonnegative, $-(x+1)^2$ is nonnegative only when $-(x+1)^2=0$, or when $x=-1 \\Rightarrow \\textbf{(B)}$.~MathJams\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/2.json | AHSME |
1976_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | 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... | [
"We partition $\\{L_1,L_2,\\dots,L_{100}\\}$ into three sets. Let\n\\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*}\nfrom which $|X|=|Y|=25$ and $|Z|=50.$\n\n\nAny two distinct lines can intersect a... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/28.json | AHSME |
1976_AHSME_Problems | 12 | 0 | Number Theory | Multiple Choice | 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 \textb... | [
"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 cra... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/12.json | AHSME |
1976_AHSME_Problems | 24 | 0 | Geometry | Multiple Choice | 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(... | [
"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 $\\frac{R}{2}.$\n\n\nSuppose $P$ is the foot of the perpendicular from $M$ to $\\overline{KL}.$ We construct the auxiliary lines, as shown below:\n[asy] /* Made by Klaus-Anton, Edited... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/24.json | AHSME |
1976_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | For a sequence $u_1,u_2\dots$, define $\Delta^1(u_n)=u_{n+1}-u_n$ and, for all integer
$k>1, \Delta^k(u_n)=\Delta^1(\Delta^{k-1}(u_n))$.
If $u_n=n^3+n$, then $\Delta^k(u_n)=0$ for all $n$
$\textbf{(A) }\text{if }k=1\qquad \\ \textbf{(B) }\text{if }k=2,\text{ but not if }k=1\qquad \\ \textbf{(C) }\text{if }k=3,\text... | [
"Note that\n\\begin{align*} \\Delta^1(u_n) &= \\left[(n+1)^3+(n+1)\\right]-\\left[n^3+n\\right] \\\\ &= \\left[n^3+3n^2+4n+2\\right]-\\left[n^3+n\\right] \\\\ &= 3n^2+3n+2, \\\\ \\Delta^2(u_n) &= \\left[3(n+1)^2+3(n+1)+2\\right]-\\left[3n^2+3n+2\\right] \\\\ &= \\left[3n^2+9n+8\\right]-\\left[3n^2+3n+2\\right] \\\\... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/25.json | AHSME |
1976_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | If $x$ cows give $x+1$ cans of milk in $x+2$ days, how many days will it take $x+3$ cows to give $x+5$ cans of milk?
$\textbf{(A) }\frac{x(x+2)(x+5)}{(x+1)(x+3)}\qquad \textbf{(B) }\frac{x(x+1)(x+5)}{(x+2)(x+3)}\qquad\\ \textbf{(C) }\frac{(x+1)(x+3)(x+5)}{x(x+2)}\qquad \textbf{(D) }\frac{(x+1)(x+3)}{x(x+2)(x+5)}\qqua... | [
"First, the problem states:\n\n\n\\begin{center}\n $x$ cows $\\qquad x+1$ cans $\\qquad x+2$ days \\end{center}\nMultiply the current number of cows by $\\frac{x+3}{x}$ to get the number of cows you want, and divide the number of days by the same amount because an increase in cows will cause a decrease in time.\n\n... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/13.json | AHSME |
1976_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | 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\qq... | [
"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.\n\n\nNow, let's tackle the ... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/29.json | AHSME |
1976_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | The sum of the distances from one vertex of a square with sides of length $2$ to the midpoints of each of the sides of the square is
$\textbf{(A) }2\sqrt{5}\qquad \textbf{(B) }2+\sqrt{3}\qquad \textbf{(C) }2+2\sqrt{3}\qquad \textbf{(D) }2+\sqrt{5}\qquad \textbf{(E) }2+2\sqrt{5}$
| [
"The lengths to the side are $1, \\sqrt{2^2+1^2}, \\sqrt{2^2+1^2}, 1$, respectively. Therefore, the sum is $\\boxed{\\textbf{(E) } 2+2\\sqrt{5}}$.\n~MathJams\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/3.json | AHSME |
1976_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | Given an equilateral triangle with side of length $s$, consider the locus of all points $\mathit{P}$ in the plane of the
triangle such that the sum of the squares of the distances from $\mathit{P}$ to the vertices of the triangle is a fixed number $a$. This locus
$\textbf{(A) }\text{is a circle if }a>s^2\qquad\\ \te... | [
"We can consider the locus of points $\\mathit{P}$ as the set of points satisfying the equation:\n$[(x_1-x_2)^2+(y_1-y_2)^2+(x_1-x_3)^2+(y_1-y_3)^2+(x_2-x_3)^2+(y_2-y_3)^2=a]$\nwhere $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ are the coordinates of the three vertices of the equilateral triangle.\n\n\nIf we simplify ... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/22.json | AHSME |
1976_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | [asy] //size(100);//local size(200); real r1=2; pair O=(0,0), D=(.5,.5*sqrt(3)), C=(D.x+.5*3,D.y), B, B_prime=endpoint(arc(D, 3, 0,-2)); B=B_prime; path c1=circle(O, r1); pair C=midpoint(D--B_prime); path arc2=arc(B_prime, 6/2, 158.25,250); draw(c1); draw(O--D); draw(D--C); draw(C--B_prime); pair A=beginpoint(arc2); dr... | [
"Extend $\\overline{BD}$ until it touches the opposite side of the circle, say at point $E.$ By power of a point, we have $AB^2=(BC)(BE),$ so $BE=\\frac{6^2}{3}=12.$ Therefore, $DE=12-3-3=6.$\n\n\nNow extend $\\overline{OD}$ in both directions so that it intersects the circle in two points (in other words, draw the... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/18.json | AHSME |
1976_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | Let a geometric progression with n terms have first term one, common ratio $r$ and sum $s$, where $r$ and $s$ are not zero. The sum of the geometric progression formed by replacing each term of the original progression by its reciprocal is
$\textbf{(A) }\frac{1}{s}\qquad \textbf{(B) }\frac{1}{r^ns}\qquad \textbf{(C) ... | [
"The sum of a geometric series with $n$ terms, initial term $a$, and ratio $r$ is $\\frac{a(1-r^n)}{1-r}$. So, $s=\\frac{(1-r^n)}{1-r}$. Our initial sequence is $1, r, r^2, \\dots, r^n$, and replacing each terms with its reciprocal gives us the sequence $1, \\frac{1}{r}, \\frac{1}{r^2}, \\dots, \\frac{1}{r^n}$. The... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/4.json | AHSME |
1976_AHSME_Problems | 14 | 0 | Geometry | Multiple Choice | 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: $180(n-2)$.\n\n\nThe 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 $a_n=140$.\n\n\nOur equation becomes $\\frac{n(100+140)}{2} ... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/14.json | AHSME |
1976_AHSME_Problems | 15 | 0 | Number Theory | Multiple Choice | 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:\n\n\n\\begin{center}\n$1059 \\equiv r \\pmod{d} \\qquad$ (i)\\end{center}\n\\begin{center}\n$1417 \\equiv r \\pmod{d} \\qquad$ (ii)\\end{center}\n\\begin{center}\n$2312 \\equiv r \\pmod{d} \\qquad$ (iii)\\end{center}\n\n\nLet's make a new congruence by subtracting (i) from (ii), whi... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/15.json | AHSME |
1976_AHSME_Problems | 5 | 0 | Number Theory | Multiple Choice | How many integers greater than $10$ and less than $100$, written in base-$10$ notation, are increased by $9$ when their digits are reversed?
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9 \qquad \textbf{(E)}\ 10$
\section{Solution}
Let our two digit number be $\overline{ab}$, ... | [
"Let our two digit number be $\\overline{ab}$, where $a$ is the tens digit, and $b$ is the ones digit. So, $\\overline{ab}=10a+b$. When we reverse our digits, it becomes $10b+a$. So, $10a+b+9=10b+a\\implies a-b=1$. So, our numbers are $12, 23, 34, 45, 56, 67, 78, 89\\Rightarrow \\textbf{(C)}$.~MathJams\n\n"
] | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/5.json | AHSME |
1976_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | A polynomial $p(x)$ has remainder three when divided by $x-1$ and remainder five when divided by $x-3$.
The remainder when $p(x)$ is divided by $(x-1)(x-3)$ is
$\textbf{(A) }x-2\qquad \textbf{(B) }x+2\qquad \textbf{(C) }2\qquad \textbf{(D) }8\qquad \textbf{(E) }15$
| [
"We know that $p(1)=3$ and $p(3)=5$. We write $p(x)$ as $p(x)=q(x)(x-1)(x-3)+r(x)$, where $r(x)=ax+b$. Plugging in $x=1$, we get $a+b=3$. Plugging in $x=3$, we know $3a+b=5$. We have a systems of equations, where we can solve that $a=1$ and $b=2$. So, our answer is $1(x)+(2)=x+2\\Rightarrow \\textbf{(B)}$. ~MathJam... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/19.json | AHSME |
1976_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | For integers $k$ and $n$ such that $1\le k<n$, let $C^n_k=\frac{n!}{k!(n-k)!}$. Then $\left(\frac{n-2k-1}{k+1}\right)C^n_k$ is an integer
$\textbf{(A) }\text{for all }k\text{ and }n\qquad \\ \textbf{(B) }\text{for all even values of }k\text{ and }n,\text{ but not for all }k\text{ and }n\qquad \\ \textbf{(C) }\text{fo... | [
"We know $C^n_k = \\binom{n}{k}$, so let's rewrite the expression as $\\left(\\frac{n-2k-1}{k+1}\\right) \\binom{n}{k}$. Notice that \\[n-2k-1 = n-2(k+1)+1 = (n+1)-2(k+1).\\]\n\n\nThis allows us to rewrite the expression as \\[\\left(\\frac{(n+1)-2(k+1)}{k+1}\\right) \\binom{n}{k}.\\]\n\n\nFrom here, we just have t... | 1 | ./CreativeMath/AHSME/1976_AHSME_Problems/23.json | AHSME |
1989_AHSME_Problems | 20 | 0 | Probability | Multiple Choice | Let $x$ be a real number selected uniformly at random between 100 and 200. If $\lfloor {\sqrt{x}} \rfloor = 12$, find the probability that $\lfloor {\sqrt{100x}} \rfloor = 120$. ($\lfloor {v} \rfloor$ means the greatest integer less than or equal to $v$.)
$\text{(A)} \ \frac{2}{25} \qquad \text{(B)} \ \frac{241}{2500... | [
"Since $\\lfloor\\sqrt{x}\\rfloor=12$, $12\\leq\\sqrt{x}<13$ and thus $144\\leq x<169$.\n\n\nThe successful region is when $120\\leq10\\sqrt{x}<121$ in which case $12\\leq\\sqrt{x}<12.1$ Thus, the successful region is when\n\\[144\\leq x<146.41\\]\n\n\nThe successful region consists of a 2.41 long segment, while t... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/20.json | AHSME |
1989_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | A lattice point is a point in the plane with integer coordinates. How many lattice points are on the line segment whose endpoints are $(3,17)$ and $(48,281)$? (Include both endpoints of the segment in your count.)
$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 16\qquad\textbf{(E)}\ 46$
... | [
"The difference in the $y$-coordinates is $281 - 17 = 264$, and the difference in the $x$-coordinates is $48 - 3 = 45$.\nThe gcd of 264 and 45 is 3, so the line segment joining $(3,17)$ and $(48,281)$ has slope\n\\[\\frac{88}{15}.\\]\nThe points on the line have coordinates\n\\[\\left(3+t,17+\\frac{88}{15}t\\right)... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/16.json | AHSME |
1989_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | If $a,b>0$ and the triangle in the first quadrant bounded by the coordinate axes and the graph of $ax+by=6$ has area 6, then $ab=$
$\mathrm{(A) \ 3 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 108 } \qquad \mathrm{(E) \ 432 }$
| [
"Setting $y=0$ we have that the $x-$intercept of the line is $x= \\frac{6}{a}$. Similarly setting $x=0$ we find the $y-$intercept to be $y= \\frac{6}{b}$. Then $\\frac{18}{ab}=\\frac{1}{2}\\cdot\\frac{6}{a}\\cdot\\frac{6}{b}$ so that $\\frac{18}{ab} = 6$, simplifying we would get $ab=3$. Hence the answer is $\\fbox... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/6.json | AHSME |
1989_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | In $\triangle ABC$, $\angle A = 100^\circ$, $\angle B = 50^\circ$, $\angle C = 30^\circ$, $\overline{AH}$ is an altitude, and $\overline{BM}$ is a median. Then $\angle MHC=$
[asy] draw((0,0)--(16,0)--(6,6)--cycle); draw((6,6)--(6,0)--(11,3)--(0,0)); dot((6,6)); dot((0,0)); dot((11,3)); dot((6,0)); dot((16,0)); label(... | [
"We are told that $\\overline{BM}$ is a median, so $\\overline{AM}=\\overline{MC}$. Drop an altitude from $M$ to $\\overline{HC}$, adding point $N$, and you can see that $\\triangle ACH$ and $\\triangle MCN$ are similar, implying $\\overline{HN}=\\overline{NC}$, implying that $\\triangle MNH$ and $\\triangle MNC$ a... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/7.json | AHSME |
1989_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | The perimeter of an equilateral triangle exceeds the perimeter of a square by $1989 \ \text{cm}$. The length of each side of the triangle exceeds the length of each side of the square by $d \ \text{cm}$. The square has perimeter greater than 0. How many positive integers are NOT possible value for $d$?
$\text{(A)} \ ... | [
"If t is the side of the triangle, and s is the side of the square, then\n\n\n\\[3t-4s=1989\\]\n\\[t-s=d\\]\n\n\nSolving the first equation for t gives\n\n\n\\[t = \\frac{4s+1989}{3}\\]\n\n\nSubstituting into the second equation,\n\n\n\\[\\frac{4s+1989}{3} - s = d\\]\n\\[\\frac{s+1989}{3} = d\\]\n\\[s+1989 = 3d\\]\... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/17.json | AHSME |
1989_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | A square flag has a red cross of uniform width with a blue square in the center on a white background as shown. (The cross is symmetric with respect to each of the diagonals of the square.) If the entire cross (both the red arms and the blue center) takes up 36% of the area of the flag, what percent of the area of the ... | [
"The diagram can be quartered as shown:\n[asy] draw((0,0)--(0,5)--(5,5)--(5,0)--(0,0)); draw((0,1)--(4,5)); draw((1,0)--(5,4)); draw((0,4)--(4,0)); draw((1,5)--(5,1)); draw((0,0)--(5,5),dotted); draw((0,5)--(5,0),dotted); [/asy]\nand reassembled into two smaller squares of side $k$, each of which looks like this:\n... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/21.json | AHSME |
1989_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | Consider the sequence defined recursively by $u_1=a$ (any positive number), and $u_{n+1}=-1/(u_n+1)$, $n=1,2,3,...$ For which of the following values of $n$ must $u_n=a$?
$\mathrm{(A) \ 14 } \qquad \mathrm{(B) \ 15 } \qquad \mathrm{(C) \ 16 } \qquad \mathrm{(D) \ 17 } \qquad \mathrm{(E) \ 18 }$
| [
"Repeatedly applying the function, and simplifying, we get \\[a,\\quad-\\frac1{a+1},\\quad-\\frac{a+1}a,\\]and then $a$ again. So $a$ must appear at every third term after $u_1$. The only option given of the form $1+3k$ is $\\boxed{\\mathrm{(C)}\\,16}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/10.json | AHSME |
1989_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | A regular octahedron is formed by joining the centers of adjoining faces of a cube. The ratio of the volume of the octahedron to the volume of the cube is
$\mathrm{(A) \frac{\sqrt{3}}{12} } \qquad \mathrm{(B) \frac{\sqrt{6}}{16} } \qquad \mathrm{(C) \frac{1}{6} } \qquad \mathrm{(D) \frac{\sqrt{2}}{8} } \qquad \mathrm... | [
"Call the length of a side of the cube x. Thus, the volume of the cube is $x^3$. We can then find that a side of this regular octahedron is the square root of $(\\frac{x}{2})^2$+$(\\frac{x}{2})^2$ which is equivalent to $\\frac{x\\sqrt{2}}{2}$. Using our general formula for the volume of a regular octahedron of sid... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/26.json | AHSME |
1989_AHSME_Problems | 30 | 0 | Probability | Multiple Choice | Suppose that 7 boys and 13 girls line up in a row. Let $S$ be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row $\text{GBBGGGBGBGGGBGBGGBGG}$ we have that $S=12$. The average value of $S$ (if all possible orders of these 20 people are considered) is closest... | [
"We approach this problem using Linearity of Expectation. Consider a pair of two people standing next to each other. Ignoring all other people, the probability that a boy is standing on the left position and a girl is standing on the right position is $\\frac7{20}\\cdot\\frac{13}{19}$. Similarly, if a girl is stand... | 2 | ./CreativeMath/AHSME/1989_AHSME_Problems/30.json | AHSME |
1989_AHSME_Problems | 27 | 0 | Counting | Multiple Choice | Let $n$ be a positive integer. If the equation $2x+2y+z=n$ has 28 solutions in positive integers $x$, $y$, and $z$, then $n$ must be either
$\mathrm{(A)}\ 14 \text{ or } 15\ \mathrm{(B)}\ 15 \text{ or } 16\ \mathrm{(C)}\ 16 \text{ or } 17\ \mathrm{(D)}\ 17 \text{ or } 18\ \mathrm{(E)}\ 18 \text{ or } 19$
| [
"This is equivalent to finding the powers of $k$ with coefficient $28$ in the expansion of $(k^2+k^4+k^6+k^8+...)^2(k+k^2+k^3+k^4+...)$.\n\n\nBut this is $\\left(\\frac{k^2}{1-k^2}\\right)^2\\cdot\\frac{k}{1-k}\\ =\\ k^5\\cdot\\left(\\frac{1}{1-k}\\right)^3\\cdot\\left(\\frac{1}{1+k}\\right)^2\\ =\\ k^5\\cdot(1+k)\... | 2 | ./CreativeMath/AHSME/1989_AHSME_Problems/27.json | AHSME |
1989_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | $(-1)^{5^{2}}+1^{2^{5}}=$
$\textrm{(A)}\ -7\qquad\textrm{(B)}\ -2\qquad\textrm{(C)}\ 0\qquad\textrm{(D)}\ 1\qquad\textrm{(E)}\ 57$
| [
"$(-1)^{5^2} + 1^{2^5} = (-1)^{25}+1^{32}=-1+1=0$ thus the answer is C.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/1.json | AHSME |
1989_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | Let $a$, $b$, $c$, and $d$ be positive integers with $a < 2b$, $b < 3c$, and $c<4d$. If $d<100$, the largest possible value for $a$ is
$\textrm{(A)}\ 2367\qquad\textrm{(B)}\ 2375\qquad\textrm{(C)}\ 2391\qquad\textrm{(D)}\ 2399\qquad\textrm{(E)}\ 2400$
| [
"Each of these integers is bounded above by the next one.\n\n\n\\begin{itemize}\n\\item $d<100$, so the maximum $d$ is $99$.\n\n\\item $c<4d\\le396$, so the maximum $c$ is $395$.\n\n\\item $b<3c\\le1185$, so the maximum $b$ is $1184$.\n\n\\item $a<2b\\le2368$, so the maximum $a$ is $\\boxed{2367}$.\n\\end{itemize}\... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/11.json | AHSME |
1989_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | $\sqrt{\frac{1}{9}+\frac{1}{16}}=$
$\textrm{(A)}\ \frac{1}5\qquad\textrm{(B)}\ \frac{1}4\qquad\textrm{(C)}\ \frac{2}7\qquad\textrm{(D)}\ \frac{5}{12}\qquad\textrm{(E)}\ \frac{7}{12}$
| [
"$\\sqrt{\\frac{1}{9}+\\frac{1}{16}}=\\sqrt{\\frac{25}{9\\cdot 16}}=\\frac{5}{12}$ and hence the answer is $\\fbox{D}$.\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/2.json | AHSME |
1989_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | Find the sum of the roots of $\tan^2x-9\tan x+1=0$ that are between $x=0$ and $x=2\pi$ radians.
$\mathrm{(A) \frac{\pi}{2} } \qquad \mathrm{(B) \pi } \qquad \mathrm{(C) \frac{3\pi}{2} } \qquad \mathrm{(D) 3\pi } \qquad \mathrm{(E) 4\pi }$
| [
"The roots of $t^2-9t+1=0$ are positive and distinct, so by considering the graph of $y=\\tan x$, the smallest two roots of the original equation $x_1,\\ x_2$ are between $0$ and $\\tfrac\\pi{2}$, and the two other roots are $\\pi+x_1, \\pi+x_2$.\n\n\nThen, from the quadratic equation, we discover that the product ... | 2 | ./CreativeMath/AHSME/1989_AHSME_Problems/28.json | AHSME |
1989_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | The traffic on a certain east-west highway moves at a constant speed of 60 miles per hour in both directions. An eastbound driver passes 20 west-bound vehicles in a five-minute interval. Assume vehicles in the westbound lane are equally spaced. Which of the following is closest to the number of westbound vehicles prese... | [
"At the beginning of the five-minute interval, say the eastbound driver is at the point $x=0$, and at the end of the interval at $x=5$, having travelled five miles. Because both lanes are travelling at the same speed, the last westbound car to be passed by the eastbound driver was just west of the position $x=10$ a... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/12.json | AHSME |
1989_AHSME_Problems | 24 | 0 | Counting | Multiple Choice | Five people are sitting at a round table. Let $f\geq 0$ be the number of people sitting next to at least 1 female and $m\geq0$ be the number of people sitting next to at least one male. The number of possible ordered pairs $(f,m)$ is
$\mathrm{(A) \ 7 } \qquad \mathrm{(B) \ 8 } \qquad \mathrm{(C) \ 9 } \qquad \mathrm{... | [
"Suppose there are more men than women; then there are between zero and two women.\n\n\nIf there are no women, the pair is $(0,5)$. If there is one woman, the pair is $(2,5)$.\n\n\nIf there are two women, there are two arrangements: one in which they are together, and one in which they are apart, giving the pairs $... | 2 | ./CreativeMath/AHSME/1989_AHSME_Problems/24.json | AHSME |
1989_AHSME_Problems | 25 | 0 | Counting | Multiple Choice | In a certain cross country meet between 2 teams of 5 runners each, a runner who finishes in the $n$th position contributes $n$ to his teams score. The team with the lower score wins. If there are no ties among the runners, how many different winning scores are possible?
(A) 10 (B) 13 (C) 27 (D) 120 (E) 126
| [
"The scores of all ten runners must sum to $55$. So a winning score is anything between $1+2+3+4+5=15$ and $\\lfloor\\tfrac{55}{2}\\rfloor=27$ inclusive. It is easy to check that this range is covered by considering $1+2+3+4+x$, $1+2+3+x+10$ and $1+2+x+9+10$, so the answer is $\\boxed{13}$.\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/25.json | AHSME |
1989_AHSME_Problems | 13 | 0 | Geometry | Multiple Choice | Two strips of width 1 overlap at an angle of $\alpha$ as shown. The area of the overlap (shown shaded) is
[asy] pair a = (0,0),b= (6,0),c=(0,1),d=(6,1); transform t = rotate(-45,(3,.5)); pair e = t*a,f=t*b,g=t*c,h=t*d; pair i = intersectionpoint(a--b,e--f),j=intersectionpoint(a--b,g--h),k=intersectionpoint(c--d,e--f)... | [
"[asy] pair a = (0,0),b= (6,0),c=(0,1),d=(6,1); transform t = rotate(-45,(3,.5)); pair e = t*a,f=t*b,g=t*c,h=t*d; pair i = intersectionpoint(a--b,e--f),j=intersectionpoint(a--b,g--h),k=intersectionpoint(c--d,e--f),l=intersectionpoint(c--d,g--h); draw(a--b^^c--d^^e--f^^g--h); filldraw(i--j--l--k--cycle,blue); label(... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/13.json | AHSME |
1989_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | What is the value of the sum $S=\sum_{k=0}^{49}(-1)^k\binom{99}{2k}=\binom{99}{0}-\binom{99}{2}+\binom{99}{4}-\cdots -\binom{99}{98}?$
(A) $-2^{50}$ (B) $-2^{49}$ (C) 0 (D) $2^{49}$ (E) $2^{50}$
| [
"By the Binomial Theorem, $(1+i)^{99}=\\sum_{n=0}^{99}\\binom{99}{j}i^n =$ $\\binom{99}{0}i^0+\\binom{99}{1}i^1+\\binom{99}{2}i^2+\\binom{99}{3}i^3+\\binom{99}{4}i^4+\\cdots +\\binom{99}{98}i^{98}$. \n\n\nUsing the fact that $i^1=i$, $i^2=-1$, $i^3=-i$, $i^4=1$, and $i^{n+4}=i^n$, the sum becomes: \n\n\n$(1+i)^{99}... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/29.json | AHSME |
1989_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | A square is cut into three rectangles along two lines parallel to a side, as shown. If the perimeter of each of the three rectangles is 24, then the area of the original square is
[asy] draw((0,0)--(9,0)--(9,9)--(0,9)--cycle); draw((3,0)--(3,9), dashed); draw((6,0)--(6,9), dashed);[/asy]
$\textrm{(A)}\ 24\qquad\tex... | [
"Let $x$ be the width of a rectangle so that the side of the square is $3x$. Since $2(3x)+2(x)=24$ we have $x=3$. Thus the area of the square is $(3\\cdot3)^2=81$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/3.json | AHSME |
1989_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | For how many integers $n$ between 1 and 100 does $x^2+x-n$ factor into the product of two linear factors with integer coefficients?
$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$
| [
"For $x^2+x-n$ to factor into a product of two linear factors, we must have $x^2+x-n = (x + a)(x + b)$, where $a$ and $b$ are integers.\n\n\nBy expansion of the product of the linear factors and comparison to the original quadratic,\n\n\n$ab = -n$\n\n\n$a + b = 1$.\n\n\nThe only way for this to work if $n$ is a pos... | 2 | ./CreativeMath/AHSME/1989_AHSME_Problems/8.json | AHSME |
1989_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | A child has a set of $96$ distinct blocks. Each block is one of $2$ materials (plastic, wood), $3$ sizes (small, medium, large), $4$ colors (blue, green, red, yellow), and $4$ shapes (circle, hexagon, square, triangle). How many blocks in the set differ from the 'plastic medium red circle' in exactly $2$ ways? (The 'wo... | [
"The process of choosing a block can be represented by a generating function. Each choice we make can match the 'plastic medium red circle' in one of its qualities $(1)$ or differ from it in $k$ different ways $(kx)$. Choosing the material is represented by the factor $(1+1x)$, choosing the size by the factor $(1+2... | 3 | ./CreativeMath/AHSME/1989_AHSME_Problems/22.json | AHSME |
1989_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | The set of all real numbers for which
\[x+\sqrt{x^2+1}-\frac{1}{x+\sqrt{x^2+1}}\]
is a rational number is the set of all
(A) integers $x$ (B) rational $x$ (C) real $x$
(D) $x$ for which $\sqrt{x^2+1}$ is rational
(E) $x$ for which $x+\sqrt{x^2+1}$ is rational
| [
"Rationalizing the denominator of $\\frac{1}{x+\\sqrt{x^2+1}}$, it simplifies: $\\frac{1}{x+\\sqrt{x^2+1}}$ = $\\frac{x-\\sqrt{x^2+1}}{x^2-(x^2+1)}$ = $\\frac{x-\\sqrt{x^2+1}}{-1}$ = $-(x-\\sqrt{x^2+1})$. Substituting this into the original equation, we get $x + \\sqrt{x^2+1} - (-(x-\\sqrt{x^2+1})) = x+\\sqrt{x^2+1... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/18.json | AHSME |
1989_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | In the figure, $ABCD$ is an isosceles trapezoid with side lengths $AD=BC=5$, $AB=4$, and $DC=10$. The point $C$ is on $\overline{DF}$ and $B$ is the midpoint of hypotenuse $\overline{DE}$ in right triangle $DEF$. Then $CF=$
[asy] defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0);... | [
"Drop perpendiculars from $A$ and $B$; then the triangle $DBY$ is similar to $DEF$ but with corresponding sides of half the length.\n[asy] defaultpen(fontsize(10)); pair D=origin, A=(3,4), B=(7,4), C=(10,0), E=(14,8), F=(14,0); draw(B--C--F--E--B--A--D--B^^C--D, linewidth(0.7)); dot(A^^B^^C^^D^^E^^F); pair point=(5... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/4.json | AHSME |
1989_AHSME_Problems | 14 | 0 | Geometry | Multiple Choice | $\cot 10+\tan 5=$
$\mathrm{(A) \csc 5 } \qquad \mathrm{(B) \csc 10 } \qquad \mathrm{(C) \sec 5 } \qquad \mathrm{(D) \sec 10 } \qquad \mathrm{(E) \sin 15 }$
| [
"We have \\[\\cot 10 +\\tan 5=\\frac{\\cos 10}{\\sin 10}+\\frac{\\sin 5}{\\cos 5}=\\frac{\\cos10\\cos5+\\sin10\\sin5}{\\sin10\\cos 5}=\\frac{\\cos(10-5)}{\\sin10\\cos5}=\\frac{\\cos5}{\\sin10\\cos5}=\\csc10\\]\n\n\n"
] | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/14.json | AHSME |
1989_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | In $\triangle ABC$, $AB=5$, $BC=7$, $AC=9$, and $D$ is on $\overline{AC}$ with $BD=5$. Find the ratio of $AD:DC$.
[asy] draw((3,4)--(0,0)--(9,0)--(3,4)--(6,0)); dot((0,0));dot((9,0));dot((3,4));dot((6,0)); label("A", (0,0), W); label("B", (3,4), N); label("C", (9,0), E); label("D", (6,0), S);[/asy]
$\textrm{(A)}\ ... | [
"[asy] draw((3,4)--(0,0)--(9,0)--(3,4)--(6,0)); draw((3,4)--(3,0),dotted); dot((0,0));dot((9,0));dot((3,4));dot((6,0));//dot((3,0)); label(\"A\", (0,0), W); label(\"B\", (3,4), N); label(\"C\", (9,0), E); label(\"D\", (6,0), S); label(\"$h$\", (3,1.5),W); draw(rightanglemark((3,1),(3,0),(4,0),10)); label(\"$x$\",(1... | 2 | ./CreativeMath/AHSME/1989_AHSME_Problems/15.json | AHSME |
1989_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | Toothpicks of equal length are used to build a rectangular grid as shown. If the grid is 20 toothpicks high and 10 toothpicks wide, then the number of toothpicks used is
[asy] real xscl = 1.2; int[] x = {0,1,2,4,5},y={0,1,3,4,5}; for(int a:x){ for(int b:y) { dot((a*xscl,b)); } } for(int a:x) { pair prev = (a,y[0]); f... | [
"There are 21 horizontal lines made of 10 matches, and 11 vertical lines made of 20 matches, and $21\\cdot10+11\\cdot20=\\boxed{430}$.\n\n\nAlternatively, the frame can be dissected into $20\\cdot10$ L shapes, the right-hand border, and the bottom border:\n[asy] path p,q; for(int i=0;i<3;++i) { for(int j=3;j<6;++... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/5.json | AHSME |
1989_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3, 4, and 5. What is the area of the triangle?
$\mathrm{(A) \ 6 } \qquad \mathrm{(B) \frac{18}{\pi^2} } \qquad \mathrm{(C) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(D) \frac{9}{\pi^2}(\sqrt{3}-1) } \qqua... | [
"The three arcs make up the entire circle, so the circumference of the circle is $3+4+5=12$ and the radius is $\\frac{12}{2\\pi}=\\frac{6}{\\pi}$. Also, the lengths of the arcs are proportional to their corresponding central angles. Thus, we can write the values of the arcs as $3\\theta$, $4\\theta$, and $5\\theta$... | 2 | ./CreativeMath/AHSME/1989_AHSME_Problems/19.json | AHSME |
1989_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | A particle moves through the first quadrant as follows. During the first minute it moves from the origin to $(1,0)$. Thereafter, it continues to follow the directions indicated in the figure, going back and forth between the positive x and y axes, moving one unit of distance parallel to an axis in each minute. At which... | [
"Squares of size $1\\times1,\\ 2\\times2,\\ 3\\times3,\\ ...$ are successively enclosed between the path and the axes.\n\n\n[asy] import graph; axialshade((0,0)--(0,3)--(3,3)--(3,0)--cycle,yellow,(-6,-6),white,(3,3)); axialshade((0,0)--(0,2)--(2,2)--(2,0)--cycle,orange,(-3,-3),white,(2,2)); axialshade((0,0)--(0,1)-... | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/23.json | AHSME |
1989_AHSME_Problems | 9 | 0 | Counting | Multiple Choice | Mr. and Mrs. Zeta want to name their baby Zeta so that its monogram (first, middle, and last initials) will be in alphabetical order with no letter repeated. How many such monograms are possible?
$\textrm{(A)}\ 276\qquad\textrm{(B)}\ 300\qquad\textrm{(C)}\ 552\qquad\textrm{(D)}\ 600\qquad\textrm{(E)}\ 15600$
| [
"We see that for any combination of two distinct letters other than Z (as the last name will automatically be Z), there is only one possible way to arrange them in alphabetical order, thus the answer is just $\\dbinom{25}{2}=\\boxed{\\mathrm{(B)\\,300}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1989_AHSME_Problems/9.json | AHSME |
1994_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | Suppose $x,y,z$ is a geometric sequence with common ratio $r$ and $x \neq y$. If $x, 2y, 3z$ is an arithmetic sequence, then $r$ is
$\textbf{(A)}\ \frac{1}{4} \qquad\textbf{(B)}\ \frac{1}{3} \qquad\textbf{(C)}\ \frac{1}{2} \qquad\textbf{(D)}\ 2 \qquad\textbf{(E)}\ 4$
| [
"Let $y=xr, z=xr^2$. Since $x, 2y, 3z$ are an arithmetic sequence, there is a common difference and we have $2xr-x=3xr^2-2xr$. Dividing through by $x$, we get $2r-1=3r^2-2r$ or, rearranging, $(r-1)(3r-1)=0$. Since we are given $x\\neq y\\implies r\\neq 1$, the answer is $\\boxed{\\textbf{(B)}\\ \\frac{1}{3}}$\n\n\n... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/20.json | AHSME |
1994_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | Some marbles in a bag are red and the rest are blue. If one red marble is removed, then one-seventh of the remaining marbles are red. If two blue marbles are removed instead of one red, then one-fifth of the remaining marbles are red. How many marbles were in the bag originally?
$\textbf{(A)}\ 8 \qquad\textbf{(B)}\ 2... | [
"Let $r$ and $b$ be the number of red and blue marbles originally in the bag respectively. After $1$ red marble is removed, there are $r+b-1$ marbles left in the bag and $r-1$ red marbles left. So \\[\\frac{r-1}{r+b-1}=\\frac{1}{7}.\\] When $2$ blue marbles are removed, there are $r$ red marbles and $r+b-2$ total m... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/16.json | AHSME |
1994_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | In the sequence
\[..., a, b, c, d, 0, 1, 1, 2, 3, 5, 8,...\]
each term is the sum of the two terms to its left. Find $a$.
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 3$
| [
"We work backwards to find $a$. \n\n\n\\[d+0=1\\implies d=1\\]\n\\[c+1=0\\implies c=-1\\]\n\\[b+(-1)=1\\implies b=2\\]\n\\[a+2=-1\\implies a=\\boxed{\\textbf{(A)}-3.}\\]\n\n\n--Solution by TheMaskedMagician\n\n\n"
] | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/6.json | AHSME |
1994_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | Squares $ABCD$ and $EFGH$ are congruent, $AB=10$, and $G$ is the center of square $ABCD$. The area of the region in the plane covered by these squares is
[asy] draw((0,0)--(10,0)--(10,10)--(0,10)--cycle); draw((5,5)--(12,-2)--(5,-9)--(-2,-2)--cycle); label("A", (0,0), W); label("B", (10,0), E); label("C", (10,10), NE);... | [
"The area of the entire region in the plane is the area of the figure. However, we cannot simply add the two areas of the squares. We find the area of $\\triangle ABG$ and subtract this from $200$, the total area of the two squares.\n\n\n\n\nSince $G$ is the center of $ABCD$, $BG$ is half of the diagonal of the squ... | 2 | ./CreativeMath/AHSME/1994_AHSME_Problems/7.json | AHSME |
1994_AHSME_Problems | 17 | 0 | Geometry | Multiple Choice | An $8$ by $2\sqrt{2}$ rectangle has the same center as a circle of radius $2$. The area of the region common to both the rectangle and the circle is
$\textbf{(A)}\ 2\pi \qquad\textbf{(B)}\ 2\pi+2 \qquad\textbf{(C)}\ 4\pi-4 \qquad\textbf{(D)}\ 2\pi+4 \qquad\textbf{(E)}\ 4\pi-2$
| [
"[asy] import cse5; import olympiad; real s=2*sqrt(2); pair A=(0,0),B=(0,s),C=(8,s),D=(8,0),O=(4,sqrt(2)),X; D(A--B--C--D--cycle); D(CR(O,2)); pair[] P; P=IPs(CR(O,2),box(A,C)); for(int i=0; i<4; i=i+1) { D(O--P[i],black); } X=foot(O,B,C); D(O--X); D(rightanglemark(O,X,C)); D(O); D(MP(\"O\",O,S));[/asy]\n\n\nWe dra... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/17.json | AHSME |
1994_AHSME_Problems | 21 | 0 | Number Theory | Multiple Choice | Find the number of counter examples to the statement:
\[``\text{If N is an odd positive integer the sum of whose digits is 4 and none of whose digits is 0, then N is prime}."\]
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 4$
| [
"Since the sum of the digits of $N$ is $4$ and none of the digits are $0$, $N$'s digits must the elements of one of the sets $\\{1,1,1,1\\},\\{1,1,2\\}$, $\\{2,2\\}$, $\\{1,3\\}$, or $\\{4\\}$. \n\n\nIn the first case, $N = 1111 = 101 \\cdot 11$ so this is a counter example. \n\n\nIn the second case, $N=112$ is ex... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/21.json | AHSME |
1994_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | For distinct real numbers $x$ and $y$, let $M(x,y)$ be the larger of $x$ and $y$ and let $m(x,y)$ be the smaller of $x$ and $y$. If $a<b<c<d<e$, then
\[M(M(a,m(b,c)),m(d,m(a,e)))=\]
$\textbf{(A)}\ a \qquad\textbf{(B)}\ b \qquad\textbf{(C)}\ c \qquad\textbf{(D)}\ d \qquad\textbf{(E)}\ e$
| [
"We work thorough the equation step by step, simplifying as follows:\n\n\n\\begin{align*}M(M(a,m(b,c)),m(d,m(a,e)))&=M(M(a,b),m(d,a))\\\\&=M(b,a)\\\\&=\\boxed{\\textbf{(B) }b}\\end{align*}\n\n\n--Solution by TheMaskedMagician\n\n\n"
] | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/10.json | AHSME |
1994_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | A regular polygon of $m$ sides is exactly enclosed (no overlaps, no gaps) by $m$ regular polygons of $n$ sides each. (Shown here for $m=4, n=8$.) If $m=10$, what is the value of $n$?
[asy] size(200); defaultpen(linewidth(0.8)); draw(unitsquare); path p=(0,1)--(1,1)--(1+sqrt(2)/2,1+sqrt(2)/2)--(1+sqrt(2)/2,2+sqrt(2)/2)-... | [
"To find the number of sides on the regular polygons that surround the decagon, we can find the interior angles and work from there. Knowing that the measure of the interior angle of any regular polygon is $\\frac{(n-2)*180}{n}$, the measure of the decagon's interior angle is $\\frac{8*180}{10} = 144$ degrees. \n\n... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/26.json | AHSME |
1994_AHSME_Problems | 30 | 0 | Probability | Multiple Choice | When $n$ standard 6-sided dice are rolled, the probability of obtaining a sum of 1994 is greater than zero and is the same as the probability of obtaining a sum of $S$. The smallest possible value of $S$ is
$\textbf{(A)}\ 333 \qquad\textbf{(B)}\ 335 \qquad\textbf{(C)}\ 337 \qquad\textbf{(D)}\ 339 \qquad\textbf{(E)}\ ... | [
"Let $d_i$ be the number on the $i$th die. There is a symmetry where we can replace each die's number with $d_i' = 7-d_i$. Note that applying the symmetry twice we get back to where we started since $7-(7-d_i)=d_i$, so this symmetry is its own inverse. Under this symmetry the sum $R=\\sum_{i=1}^n d_i$ is replace... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/30.json | AHSME |
1994_AHSME_Problems | 27 | 0 | Probability | Multiple Choice | A bag of popping corn contains $\frac{2}{3}$ white kernels and $\frac{1}{3}$ yellow kernels. Only $\frac{1}{2}$ of the white kernels will pop, whereas $\frac{2}{3}$ of the yellow ones will pop. A kernel is selected at random from the bag, and pops when placed in the popper. What is the probability that the kernel selec... | [
"To find the probability that the kernel is white, the probability of $P(\\mathrm{white|popped}) = \\frac{P(\\mathrm{white, popped})}{P(\\mathrm{popped})}$.\n\n\nRunning a bit of calculations $P(\\mathrm{white, popped}) = \\frac{1}{3}$ while $P(\\mathrm{popped}) = \\frac{1}{3} + \\frac{2}{9} = \\frac{5}{9}$. Pluggi... | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/27.json | AHSME |
1994_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | $4^4 \cdot 9^4 \cdot 4^9 \cdot 9^9=$
$\textbf{(A)}\ 13^{13} \qquad\textbf{(B)}\ 13^{36} \qquad\textbf{(C)}\ 36^{13} \qquad\textbf{(D)}\ 36^{36} \qquad\textbf{(E)}\ 1296^{26}$
| [
"Note that $a^x\\times a^y=a^{x+y}$. So $4^4\\cdot 4^9=4^{13}$ and $9^4\\cdot 9^9=9^{13}$. Therefore, $4^{13}\\cdot 9^{13}=(4\\cdot 9)^{13}=\\boxed{\\textbf{(C)}\\ 36^{13}}$.\n\n\n--Solution by TheMaskedMagician\n\n\n"
] | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/1.json | AHSME |
1994_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | A large rectangle is partitioned into four rectangles by two segments parallel to its sides. The areas of three of the resulting rectangles are shown. What is the area of the fourth rectangle?
[asy] draw((0,0)--(10,0)--(10,7)--(0,7)--cycle); draw((0,5)--(10,5)); draw((3,0)--(3,7)); label("6", (1.5,6)); label("?", (1.5,... | [
"[asy] pair A=(0,0),B=(10,0),C=(10,7),D=(0,7),EE=(0,5),F=(10,5),G=(3,0),H=(3,7); path BG=shift(0,-0.5)*(B--G); path BF=shift(0.5,0)*(B--F); path FC=shift(0.5,0)*(F--C); path DH=shift(0,0.5)*(D--H); draw(A--B--C--D--cycle); draw(EE--F); draw(G--H); draw(BG,L=Label(\"$7$\",position=MidPoint,align=(0,-1)),arrow=Arrows... | 2 | ./CreativeMath/AHSME/1994_AHSME_Problems/2.json | AHSME |
1994_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | If $i^2=-1$, then $(i-i^{-1})^{-1}=$
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ -2i \qquad\textbf{(C)}\ 2i \qquad\textbf{(D)}\ -\frac{i}{2} \qquad\textbf{(E)}\ \frac{i}{2}$
| [
"We simplify step by step as follows: \\begin{align*}(i-i^{-1})^{-1}&=\\frac{1}{i-i^{-1}}\\\\&=\\frac{1}{i-\\frac{1}{i}}\\\\&=\\frac{1}{\\left(\\frac{i^2-1}{i}\\right)}\\\\&=\\frac{i}{i^2-1}\\\\&=\\boxed{\\textbf{(D) }-\\frac{i}{2}.}\\end{align*}\n\n\n--Solution by TheMaskedMagician\n\n\n"
] | 1 | ./CreativeMath/AHSME/1994_AHSME_Problems/12.json | AHSME |
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