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 |
|---|---|---|---|---|---|---|---|---|---|
1985_AHSME_Problems | 16 | 0 | Other | Multiple Choice | If $\usepackage{gensymb} A = 20 \degree$ and $\usepackage{gensymb} B = 25 \degree$, then the value of $\left(1+\tan A\right)\left(1+\tan B\right)$ is
$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2\left(\tan A+\tan B\right) \qquad \mathrm{(E) \ }\text{non... | [
"Noting that $\\usepackage{gensymb} 25 \\degree + 20 \\degree = 45 \\degree$, we apply the angle sum formula \\[\\tan(A+B) = \\frac{\\tan A+\\tan B}{1-\\tan A\\tan B},\\] giving \\begin{align*}1 &= \\tan 45^{\\circ} \\\\ &= \\tan(A+B) \\\\ &= \\frac{\\tan A+\\tan B}{1-\\tan A\\tan B},\\end{align*} so \\[\\tan A + \... | 4 | ./CreativeMath/AHSME/1985_AHSME_Problems/16.json | AHSME |
1985_AHSME_Problems | 6 | 0 | Probability | Multiple Choice | One student in a class of boys and girls is chosen to represent the class. Each student is equally likely to be chosen and the probability that a boy is chosen is $\frac{2}{3}$ of the probability that a girl is chosen. The ratio of the number of boys to the total number of boys and girls is
$\mathrm{(A)\ } \frac{1}{3... | [
"Let the probability that a boy is chosen be $p$, so the probability that a girl is chosen is $1-p$ (as the probabilities must sum to $1$). Therefore \\begin{align*}\\frac{2}{3}(1-p) = p &\\iff 2(1-p) = 3p \\\\ &\\iff 2-2p = 3p \\\\&\\iff 5p = 2 \\\\&\\iff p = \\frac{2}{5}.\\end{align*}\n\n\nNow simply notice that ... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/6.json | AHSME |
1985_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | In some computer languages (such as APL), when there are no parentheses in an algebraic expression, the operations are grouped from right to left. Thus, $a \times b - c$ in such languages means the same as $a(b-c)$ in ordinary algebraic notation. If $a\div b-c+d$ is evaluated in such a language, the result in ordinary ... | [
"The rightmost part of the expression is $c+d$, so $b-c+d$ would be grouped as $b-(c+d)$, and thus the whole expression would be grouped as $a\\div (b-(c+d)) = \\boxed{\\text{(E)} \\ \\frac{a}{b-c-d}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/7.json | AHSME |
1985_AHSME_Problems | 17 | 0 | Geometry | Multiple Choice | Diagonal $DB$ of rectangle $ABCD$ is divided into three segments of length $1$ by parallel lines $L$ and $L'$ that pass through $A$ and $C$ and are perpendicular to $DB$. The area of $ABCD$, rounded to the one decimal place, is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); real x=sqrt(6), y=sqrt(3), a=0.4; pair D=o... | [
"Let $E$ be the point of intersection of $L$ and $\\overline{BD}$. Then, because $AE$ is the altitude to the hypotenuse of right triangle $ABD$, triangles $ADE$ and $BAE$ are similar, giving \\[\\frac{AE}{BE} = \\frac{ED}{EA},\\] and so \\begin{align*}AE &= \\sqrt{BE \\cdot ED} \\\\ &= \\sqrt{(1+1)(1)} \\\\ &= \\sq... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/17.json | AHSME |
1985_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | How many integers $x$ satisfy the equation \[\left(x^2-x-1\right)^{x+2} = 1?\]
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ } 5 \qquad \mathrm{(E) \ }\text{none of these}$
| [
"We recall that for real numbers $a$ and $b$, there are exactly $3$ ways in which we can have $a^b = 1$, namely $a = 1$; $b = 0$ and $a \\neq 0$; or $a = -1$ and $b$ is an even integer.\n\n\nThe first case therefore gives \\begin{align*}x^2-x-1 = 1 &\\iff x^2-x-2 = 0 \\\\&\\iff (x-2)(x+1) = 0 \\\\&\\iff x = 2 \\tex... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/21.json | AHSME |
1985_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | An arbitrary circle can intersect the graph of $y = \sin x$ in
$\mathrm{(A) \ } \text{at most }2\text{ points} \qquad \mathrm{(B) \ }\text{at most }4\text{ points} \qquad \mathrm{(C) \ } \text{at most }6\text{ points} \qquad \mathrm{(D) \ } \text{at most }8\text{ points}$
$\mathrm{(E) \ }\text{more than }16\text{ p... | [
"Consider a circle whose center lies on the positive $y$-axis and which passes through the origin. As the radius of this circle becomes arbitrarily large, its curvature near the $x$-axis becomes almost flat, and so it can intersect the curve $y = \\sin x$ arbitrarily many times (since the $x$-axis itself intersects... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/10.json | AHSME |
1985_AHSME_Problems | 26 | 0 | Number Theory | Multiple Choice | Find the least positive integer $n$ for which $\frac{n-13}{5n+6}$ is a non-zero reducible fraction.
$\mathrm{(A)\ } 45 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \ } 155 \qquad \mathrm{(D) \ } 226 \qquad \mathrm{(E) \ }\text{none of these}$
| [
"For the fraction to be reducible, the greatest common factor of the numerator and the denominator must be greater than $1$. Using the Euclidean algorithm, we compute \\begin{align*}\\gcd\\left(5n+6,n-13\\right) &= \\gcd\\left(5n+6-5(n-13),n-13\\right) \\\\ &= \\gcd\\left(71,n-13\\right).\\end{align*}\nSince $71$ i... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/26.json | AHSME |
1985_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | Let $\left\lfloor x\right\rfloor$ be the greatest integer less than or equal to $x$. Then the number of real solutions to $4x^2-40\left\lfloor x\right\rfloor+51 = 0$ is
$\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ } 3 \qquad \mathrm{(E) \ }4$
| [
"We rearrange the equation as $4x^2 = 40\\left\\lfloor x\\right\\rfloor-51$, where the right-hand side is now clearly an integer, meaning that $4x^2 = n$ for some non-negative integer $n$. Therefore, in the case where $x \\geq 0$, substituting $x = \\frac{\\sqrt{n}}{2}$ gives\n\\[40\\left\\lfloor\\frac{\\sqrt{n}}{2... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/30.json | AHSME |
1985_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | Consider a sequence $x_1,x_2,x_3,\dotsc$ defined by:
\begin{align*}&x_1 = \sqrt[3]{3}, \\ &x_2 = \left(\sqrt[3]{3}\right)^{\sqrt[3]{3}},\end{align*}
and in general
\[x_n = \left(x_{n-1}\right)^{\sqrt[3]{3}} \text{ for } n > 1.\]
What is the smallest value of $n$ for which $x_n$ is an integer?
$\mathrm{(A)\ } 2 \qqu... | [
"Firstly, we will show by induction that \\[x_n = \\left(\\sqrt[3]{3}\\right)^{\\left(\\left(\\sqrt[3]{3}\\right)^{n-1}\\right)}.\\] For the base case, we indeed have \\begin{align*}x_1 &= \\sqrt[3]{3} \\\\ &= \\left(\\sqrt[3]{3}\\right)^1 \\\\ &= \\left(\\sqrt[3]{3}\\right)^{\\left(\\left(\\sqrt[3]{3}\\right)^0\\r... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/27.json | AHSME |
1985_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | If $2x+1=8$, then $4x+1=$
$\mathrm{(A)\ } 15 \qquad \mathrm{(B) \ }16 \qquad \mathrm{(C) \ } 17 \qquad \mathrm{(D) \ } 18 \qquad \mathrm{(E) \ }19$
| [
"We have \\begin{align*}2x+1 = 8 &\\iff 2x = 7 \\\\ &\\iff x = \\frac{7}{2},\\end{align*} so \\begin{align*}4x+1 &= 4\\left(\\frac{7}{2}\\right)+1 \\\\ &= 2(7)+1 \\\\ &= \\boxed{\\text{(A)} \\ 15}.\\end{align*}\n\n\n",
"From $2x = 7$ (as above), we can directly compute \\begin{align*}4x &= 2(2x) \\\\ &= 2(7) \\\\... | 2 | ./CreativeMath/AHSME/1985_AHSME_Problems/1.json | AHSME |
1985_AHSME_Problems | 11 | 0 | Counting | Multiple Choice | How many distinguishable rearrangements of the letters in $CONTEST$ have both the vowels first? (For instance, $OETCNST$ is one such arrangement, but $OTETSNC$ is not.)
$\mathrm{(A)\ } 60 \qquad \mathrm{(B) \ }120 \qquad \mathrm{(C) \ } 240 \qquad \mathrm{(D) \ } 720 \qquad \mathrm{(E) \ }2520$
| [
"We consider the vowels and consonants separately. There are $2$ vowels ($O$ and $E$), giving $2! = 2$ choices for the first two letters; similarly, there are $5$ consonants ($C$, $N$, $S$, and two $T$s), which would give $5! = 120$ possible choices for letters $3$ to $7$, except that since the two $T$s are indisti... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/11.json | AHSME |
1985_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | In an arcade game, the "monster" is the shaded sector of a circle of radius $1$ cm, as shown in the figure. The missing piece (the mouth) has central angle $\usepackage{gensymb} 60\degree$. What is the perimeter of the monster in cm?
[asy] size(100); defaultpen(linewidth(0.7)); filldraw(Arc(origin,1,30,330)--dir(330)... | [
"The two straight sides of the \"monster\" are radii of length $1$ cm, so their total length is $1+1 = 2$ cm. The circular arc forms $\\frac{360^{\\circ}-60^{\\circ}}{360^{\\circ}} = \\frac{5}{6}$ of the entire circle, and a circle with radius $1$ cm has circumference $2\\pi(1) = 2\\pi$ cm, so the length of the arc... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/2.json | AHSME |
1985_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | In $\triangle ABC$, we have $\angle C = 3\angle A$, $a = 27$ and $c = 48$. What is $b$?
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(14,0), C=(10,6); draw(A--B--C--cycle); label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, N); label("$a$", B--C, dir(B--C)*dir(-90)); label("$b$", A--C, dir(C... | [
"Let $\\angle A = x^{\\circ}$, so $\\angle C = 3x^{\\circ}$, and thus $\\angle B = \\left(180-4x\\right)^{\\circ}$. Now let $D$ be a point on side $AB$ such that $\\angle ACD = x^{\\circ}$, so $\\angle BCD = 3x^{\\circ}-x^{\\circ} = 2x^{\\circ}$, which gives \\[\\angle CDB = 180^{\\circ}-2x^{\\circ}-\\left(180-4x\\... | 2 | ./CreativeMath/AHSME/1985_AHSME_Problems/28.json | AHSME |
1985_AHSME_Problems | 12 | 0 | Number Theory | Multiple Choice | Let $p$, $q$ and $r$ be distinct prime numbers, where $1$ is not considered a prime. Which of the following is the smallest positive perfect cube having $n = pq^2r^4$ as a divisor?
$\mathrm{(A)\ } p^8q^8r^8 \qquad \mathrm{(B) }\left(pq^2r^2\right)^3 \qquad \mathrm{(C) } \left(p^2q^2r^2\right)^3 \qquad \mathrm{(D) } \... | [
"A number of the form $p^aq^br^c$ will be a perfect cube precisely when $a$, $b$, and $c$ are multiples of 3. (Clearly, since we are looking for the \\textit{smallest} possible perfect cube, we can assume that it has no other prime factors.) Furthermore, for it to be a multiple of $pq^2r^4$, we must have $a \\geq 1... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/12.json | AHSME |
1985_AHSME_Problems | 24 | 0 | Probability | Multiple Choice | A non-zero digit is chosen in such a way that the probability of choosing digit $d$ is $\log_{10}{(d+1)}-\log_{10}{d}$. The probability that the digit $2$ is chosen is exactly $1/2$ the probability that the digit chosen is in the set
$\mathrm{(A)\ } \{2,3\} \qquad \mathrm{(B) \ }\{3,4\} \qquad \mathrm{(C) \ } \{4,5,... | [
"We have $\\log_{10}{(d+1)}-\\log_{10}{d} = \\log_{10}{\\left(\\frac{d+1}{d}\\right)}$, so the probability of choosing $2$ is $\\log_{10}{\\left(\\frac{3}{2}\\right)}$. The probability that the digit chosen is in the set must therefore be \\begin{align*}2\\log_{10}{\\left(\\frac{3}{2}\\right)} = &\\log_{10}{\\left(... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/24.json | AHSME |
1985_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | The volume of a certain rectangular solid is $8$ cm\textsuperscript{3}, its total surface area is $32$ cm\textsuperscript{2}, and its three dimensions are in geometric progression. The sums of the lengths in cm of all the edges of this solid is
$\mathrm{(A)\ } 28 \qquad \mathrm{(B) \ }32 \qquad \mathrm{(C) \ } 36 \q... | [
"As the dimensions are in geometric progression, let them be $\\frac{b}{r}$, $b$, and $br$ cm, so the volume is $\\left(\\frac{b}{r}\\right)(b)(br) = b^3$, giving $b^3 = 8$ and thus $b = 2$. The surface area condition now yields \\begin{align*}2\\left(\\frac{2}{r}\\right)(2)+2(2)(2r)+2(2r)\\left(\\frac{2}{r}\\right... | 2 | ./CreativeMath/AHSME/1985_AHSME_Problems/25.json | AHSME |
1985_AHSME_Problems | 13 | 0 | Geometry | Multiple Choice | Pegs are put in a board $1$ unit apart both horizontally and vertically. A rubber band is stretched over $4$ pegs as shown in the figure, forming a quadrilateral. Its area in square units is
[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0... | [
"[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7)); draw((0,0)--(4,0)--(4,3)--(0,3)--cycle); label(\"$A$\",(0,3),NW); label(\"$B$\",(4,3),NE); label(\"$C$\",(4,0),SE); label(\"$D$\",(0,0),SW); label(\"$E$\",(1,3),N); label(\"$F$\",(4,... | 2 | ./CreativeMath/AHSME/1985_AHSME_Problems/13.json | AHSME |
1985_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | In their base $10$ representations, the integer $a$ consists of a sequence of $1985$ eights and the integer $b$ consists of a sequence of $1985$ fives. What is the sum of the digits of the base $10$ representation of the integer $9ab$?
$\mathrm{(A)\ } 15880 \qquad \mathrm{(B) \ }17856 \qquad \mathrm{(C) \ } 17865 \q... | [
"By the formula for the sum of a geometric series, \\begin{align*}a &= 8 \\cdot 10^0 + 8 \\cdot 10^1 + \\dotsb + 8 \\cdot 10^{1984} \\\\ &= \\frac{8\\left(10^{1985}-1\\right)}{10-1} \\\\ &= \\frac{8\\left(10^{1985}-1\\right)}{9},\\end{align*} and similarly \\[b = \\frac{5\\left(10^{1985}-1\\right)}{9},\\] so \\begi... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/29.json | AHSME |
1985_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | In right $\triangle ABC$ with legs $5$ and $12$, arcs of circles are drawn, one with center $A$ and radius $12$, the other with center $B$ and radius $5$. They intersect the hypotenuse in $M$ and $N$. Then $MN$ has length
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(12,7), C=(12,0), M=12*dir(A--B)... | [
"Firstly, the Pythagorean theorem gives \\begin{align*}AB &=\\sqrt{AC^2+BC^2} \\\\ &= \\sqrt{12^2+5^2} \\\\ & =\\sqrt{144+25} \\\\ &=\\sqrt{169} \\\\ &= 13.\\end{align*} Also, $AM = AC = 12$ and $BN = BC = 5$ since they are both radii of the respective circles. Thus $MB = AB-AM = 13-12 = 1$, and so $MN = BN-BM = 5-... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/3.json | AHSME |
1985_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | Let $a,a',b,b'$ be real numbers with $a$ and $a'$ nonzero. The solution to $ax+b=0$ is less than the solution to $a'x+b'=0$ if and only if
$\mathrm{(A)\ } a'b < ab' \qquad \mathrm{(B) \ }ab' < a'b \qquad \mathrm{(C) \ } ab < a'b' \qquad \mathrm{(D) \ } \frac{b}{a} < \frac{b'}{a'} \qquad \mathrm{(E) \ }\frac{b'}{a... | [
"The solution to $ax+b=0$ is $x = \\frac{-b}{a}$, while that to $a'x+b'=0$ is $x = \\frac{-b'}{a'}$. The first solution is less than the second precisely if \\[\\frac{-b}{a} < \\frac{-b'}{a'},\\] and multiplying this inequality by $-1$ reversees the inequality sign, yielding $\\boxed{\\text{(E)} \\ \\frac{b'}{a'} <... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/8.json | AHSME |
1985_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | In a circle with center $O$, $AD$ is a diameter, $ABC$ is a chord, $BO = 5$ and $\angle ABO = \ \stackrel{\frown}{CD} \ = 60^{\circ}$. Then the length of $BC$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair O=origin, A=dir(35), C=dir(155), D=dir(215), B=intersectionpoint(dir(125)--O, A--C); draw(C--A--D^^B--O^... | [
"Since $\\angle CAD$ is an angle inscribed in a $60{^\\circ}$ arc, we obtain $\\angle CAD =\\frac{60^{\\circ}}{2} = 30^{\\circ}$, so $\\triangle ABO$ is a $30^{\\circ}$-$60^{\\circ}$-$90^{\\circ}$ right triangle. This gives $AO = BO\\sqrt{3} = 5\\sqrt{3}$ and $AB = 2BO = 10$, and now since $AD$ is a diameter, $AD =... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/22.json | AHSME |
1985_AHSME_Problems | 18 | 0 | Number Theory | Multiple Choice | Six bags of marbles contain $18, 19, 21, 23, 25$ and $34$ marbles, respectively. One bag contains chipped marbles only. The other $5$ bags contain no chipped marbles. Jane takes three of the bags and George takes two of the others. Only the bag of chipped marbles remains. If Jane gets twice as many marbles as George, h... | [
"Let George's bags contain a total of $x$ marbles, so Jane's bag contains $2x$ marbles. This means the total number of non-chipped marbles is $3x \\equiv 0 \\pmod{3}$, while the total number of marbles is $18+19+21+23+25+34 = 140 \\equiv 2 \\pmod{3}$, so the number of chipped marbles must also be congruent to $2-0 ... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/18.json | AHSME |
1985_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | A large bag of coins contains pennies, dimes and quarters. There are twice as many dimes as pennies and three times as many quarters as dimes. An amount of money which could be in the bag is
$\mathrm{(A)\ } $306 \qquad \mathrm{(B) \ } $333 \qquad \mathrm{(C)\ } $342 \qquad \mathrm{(D) \ } $348 \qquad \mathrm{(E) \... | [
"If there are $x$ pennies in the bag, then there are $2x$ dimes and $3(2x) = 6x$ quarters. Since pennies are $$0.01$, dimes are $$0.10$, and quarters are $$0.25$, the total amount of money in the bag is \\[$ \\left(0.01x+(0.10)(2x)+(0.25)(6x)\\right) = $1.71x.\\] Therefore, the possible amounts of money are precise... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/4.json | AHSME |
1985_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | If $a$ and $b$ are positive numbers such that $a^b = b^a$ and $b = 9a$, then the value of $a$ is
$\mathrm{(A)\ } 9 \qquad \mathrm{(B) \ }\frac{1}{9} \qquad \mathrm{(C) \ } \sqrt[9]{9} \qquad \mathrm{(D) \ } \sqrt[3]{9} \qquad \mathrm{(E) \ }\sqrt[4]{3}$
| [
"Substituting $b = 9a$ into $a^b = b^a$ gives \\begin{align*}a^{9a} = \\left(9a\\right)^a &\\iff \\left(a^9\\right)^a = \\left(9a\\right)^a \\qquad \\text{(using the identity } \\left(x^y\\right)^z = x^{yz}\\text{ for } x > 0\\text{)} \\\\ &\\iff a^9 = 9a \\qquad \\text{(taking the } a\\text{th root of both sides, ... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/15.json | AHSME |
1985_AHSME_Problems | 5 | 0 | Arithmetic | Multiple Choice | Which terms must be removed from the sum
\[\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\]
if the sum of the remaining terms is to equal $1$?
$\mathrm{(A)\ } \frac{1}{4}\text{ and }\frac{1}{8} \qquad \mathrm{(B) \ }\frac{1}{4}\text{ and }\frac{1}{12} \qquad \mathrm{(C) \ } \frac{1}{8}... | [
"We compute the entire sum as \\begin{align*}\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{6}+\\frac{1}{8}+\\frac{1}{10}+\\frac{1}{12} &= \\frac{60+30+20+15+12+10}{120} \\\\ &= \\frac{147}{120} \\\\ &= \\frac{49}{40},\\end{align*}\nso the two terms to be removed must sum to $\\frac{49}{40}-1 = \\frac{9}{40}$. That is, $\\fra... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/5.json | AHSME |
1985_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | Consider the graphs of $y = Ax^2$ and $y^2+3 = x^2+4y$, where $A$ is a positive constant and $x$ and $y$ are real variables. In how many points do the two graphs intersect?
$\mathrm{(A) \ }\text{exactly }4 \qquad \mathrm{(B) \ }\text{exactly }2 \qquad$
$\mathrm{(C) \ }\text{at least }1,\text{ but the number varies... | [
"Substituting $y = Ax^2$ into the equation $y^2+3 = x^2+4y$ gives \\begin{align*}\\left(Ax^2\\right)+3 = x^2+4\\left(Ax^2\\right) &\\iff A^2x^4+3 = x^2+4Ax^2 \\\\ &\\iff A^2x^4-\\left(4A+1\\right)x^2+3 = 0 \\\\ &\\iff x^2 = \\frac{4A+1 \\pm \\sqrt{4A^2+8A+1}}{2A^2} \\\\ &\\text{(using the quadratic formula)}.\\end{... | 2 | ./CreativeMath/AHSME/1985_AHSME_Problems/19.json | AHSME |
1985_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | If \[x = \frac{-1+i\sqrt{3}}{2} \qquad\text{and}\qquad y=\frac{-1-i\sqrt{3}}{2},\] where $i^2 = -1$, then which of the following is not correct?
$\mathrm{(A)\ } x^5+y^5 = -1 \qquad \mathrm{(B) \ }x^7+y^7 = -1 \qquad \mathrm{(C) \ } x^9+y^9 = -1 \qquad$
$\mathrm{(D) \ } x^{11}+y^{11} = -1 \qquad \mathrm{(E) \ }x^... | [
"We can write \\begin{align*}&x = \\cos\\left(\\frac{2\\pi}{3}\\right)+i\\sin\\left(\\frac{2\\pi}{3}\\right) = e^{\\frac{2\\pi}{3}i}, \\text{ and} \\\\ &y = \\cos\\left(-\\frac{2\\pi}{3}\\right)+i\\sin\\left(-\\frac{2\\pi}{3}\\right) = e^{-\\frac{2\\pi}{3}i},\\end{align*} which gives \\begin{align*}&x^k = e^{\\frac... | 2 | ./CreativeMath/AHSME/1985_AHSME_Problems/23.json | AHSME |
1985_AHSME_Problems | 9 | 0 | Number Theory | Multiple Choice | The odd positive integers $1, 3, 5, 7, \ldots$, are arranged into five columns continuing with the pattern shown on the right. Counting from the left, the column in which $1985$ appears is the
[asy] int i,j; for(i=0; i<4; i=i+1) { label(string(16*i+1), (2*1,-2*i)); label(string(16*i+3), (2*2,-2*i)); label(string(16*i... | [
"Considering each integer modulo $16$ gives the following pattern:\n[asy] int i,j; for(i=0; i<4; i=i+1) { label(string(2*i+1), (2*i,-2*1)); label(string(15-2*i), (2*(i-1),-2*1.35)); label(string(2*i+1), (2*i,-2*1.7)); label(string(15-2*i), (2*(i-1),-2*2.05)); } [/asy]\n\n\nWe therefore observe that all numbers cong... | 1 | ./CreativeMath/AHSME/1985_AHSME_Problems/9.json | AHSME |
1998_AHSME_Problems | 20 | 0 | Other | Multiple Choice | Three cards, each with a positive integer written on it, are lying face-down on a table. Casey, Stacy, and Tracy are told that
\begin{description}
\item (a) the numbers are all different,
\item (b) they sum to $13$, and
\item (c) they are in increasing order, left to right.
\end{description}
First, Casey looks at t... | [
"Initially, there are the following possibilities for the numbers on the cards: $(1,2,10)$, $(1,3,9)$, $(1,4,8)$, $(1,5,7)$, $(2,3,8)$, $(2,4,7)$, $(2,5,6)$, and $(3,4,6)$.\n\n\nIf Casey saw the number $3$, she would have known the other two numbers. As she does not, we eliminated the possibility $(3,4,6)$.\n\n\nAt... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/20.json | AHSME |
1998_AHSME_Problems | 6 | 0 | Number Theory | Multiple Choice | If $1998$ is written as a product of two positive integers whose difference is as small as possible, then the difference is
$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }17 \qquad \mathrm{(D) \ }47 \qquad \mathrm{(E) \ } 93$
| [
"If we want the difference of the two factors to be as small as possible, then the two numbers must be as close to $\\sqrt{1998}$ as possible.\n\n\nSince $45^2 = 2025$, the factors should be as close to $44$ or $45$ as possible.\n\n\nBreaking down $1998$ into its prime factors gives $1998 = 2\\cdot 3^3 \\cdot 37$. ... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/6.json | AHSME |
1998_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | If $N > 1$, then $\sqrt[3]{N\sqrt[3]{N\sqrt[3]{N}}} =$
$\mathrm{(A) \ } N^{\frac 1{27}} \qquad \mathrm{(B) \ } N^{\frac 1{9}} \qquad \mathrm{(C) \ } N^{\frac 1{3}} \qquad \mathrm{(D) \ } N^{\frac {13}{27}} \qquad \mathrm{(E) \ } N$
| [
"The key identities are $\\sqrt[3]{x^n} = x^{\\frac{n}{3}}$ and $x \\cdot x^{\\frac{a}{b}} = x^{1 + \\frac{a}{b}}$\n\n\n$\\sqrt[3]{N\\sqrt[3]{N\\sqrt[3]{N}}}$\n\n\n$\\sqrt[3]{N\\sqrt[3]{N\\cdot N^{\\frac{1}{3}}}}$\n\n\n$\\sqrt[3]{N\\sqrt[3]{N^{\\frac{4}{3}}}}$\n\n\n$\\sqrt[3]{N\\cdot{N^{\\frac{4}{9}}}}$\n\n\n$\\sqr... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/7.json | AHSME |
1998_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | Let $f(x)$ be a function with the two properties:
\begin{description}
\item (a) for any two real numbers $x$ and $y$, $f(x+y) = x + f(y)$, and
\item (b) $f(0) = 2$.
\end{description}
What is the value of $f(1998)$?
$\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 2 \qquad\mathrm{(C)}\ 1996 \qquad\mathrm{(D)}\ 1998 \qquad\m... | [
"\\[f(1998 + 0) = 1998 + f(0) = 2000 \\Rightarrow \\mathrm{(E)}\\]\n\n\nThe function $f(x) = x+2$ satisfies these properties. \n\n\n"
] | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/17.json | AHSME |
1998_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | In an $h$-meter race, Sunny is exactly $d$ meters ahead of Windy when Sunny finishes the race. The next time they race, Sunny sportingly starts $d$ meters behind Windy, who is at the starting line. Both runners run at the same constant speed as they did in the first race. How many meters ahead is Sunny when Sunny finis... | [
"Let $s$ and $w$ be the speeds of Sunny and Windy. From the first race we know that $\\frac sw = \\frac h{h-d}$. In the second race, Sunny's track length is $h+d$. She will finish this track in $\\frac{h+d}s$. In this time, Windy will run the distance $w\\cdot \\frac{h+d}s = \\frac{(h+d)(h-d)}h$. This is less than ... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/21.json | AHSME |
1998_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | A large square is divided into a small square surrounded by four congruent rectangles as shown. The perimter of each of the congruent rectangles is $14$. What is the area of the large square?
\begin{center}
[asy]pathpen = black+linewidth(0.7); D((0,0)--(7,0)--(7,7)--(0,7)--cycle); D((1,0)--(1,6)); D((0,6)--(6,6)); D(... | [
"Let the length of the longer side be $x$, and the length of the shorter side be $y$. We are given that $2x+2y=14\\implies x+y=7$. However, note that $x+y$ is also the length of a side of the larger square. Thus the area of the larger square is $(x+y)^2=7^2=\\boxed{49\\text{ (A)}}$.\n\n\n",
"Expand the small s... | 2 | ./CreativeMath/AHSME/1998_AHSME_Problems/10.json | AHSME |
1998_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | In quadrilateral $ABCD$, it is given that $\angle A = 120^{\circ}$, angles $B$ and $D$ are right angles, $AB = 13$, and $AD = 46$. Then $AC=$
$\mathrm{(A)}\ 60 \qquad\mathrm{(B)}\ 62 \qquad\mathrm{(C)}\ 64 \qquad\mathrm{(D)}\ 65 \qquad\mathrm{(E)}\ 72$
| [
"Let the extensions of $\\overline{DA}$ and $\\overline{CB}$ be at $E$. Since $\\angle BAD = 120^{\\circ}$, $\\angle BAE = 60^{\\circ}$ and $\\triangle ABE$ is a $30-60-90$ triangle. Also, $\\triangle ABE \\sim \\triangle CDE$, so $\\triangle CDE$ is also a $30-60-90$ triangle.\n\n\n\\begin{center}\n[asy] size(200)... | 2 | ./CreativeMath/AHSME/1998_AHSME_Problems/26.json | AHSME |
1998_AHSME_Problems | 30 | 0 | Number Theory | Multiple Choice | For each positive integer $n$, let
\begin{center}
$a_n = \frac{(n+9)!}{(n-1)!}$\end{center}
Let $k$ denote the smallest positive integer for which the rightmost nonzero digit of $a_k$ is odd. The rightmost nonzero digit of $a_k$ is
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D... | [
"We have $a_n = n(n+1)\\dots (n+9)$.\n\n\nThe value $a_n$ can be written as $2^{x_n} 5^{y_n} r_n$, where $r_n$ is not divisible by 2 and 5. The number of trailing zeroes is $z_n = \\min(x_n,y_n)$. The last non-zero digit is the last digit of $2^{x_n-z_n} 5^{y_n-z_n} r_n$.\n\n\nClearly, the last non-zero digit is ev... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/30.json | AHSME |
1998_AHSME_Problems | 1 | 0 | Geometry | Multiple Choice | Each of the sides of five congruent rectangles is labeled with an integer. In rectangle A, $w = 4, x = 1, y = 6, z = 9$. In rectangle B, $w = 1, x = 0, y = 3, z = 6$. In rectangle C, $w = 3, x = 8, y = 5, z = 2$. In rectangle D, $w = 7, x = 5, y = 4, z = 8$. In rectangle E, $w = 9, x = 2, y = 7, z = 0$. These five rect... | [
"Looking at the list of $w$ and $y$ values that must match up left-to-right, we have $(w,y) = A(4,6), B(1,3), C(3,5), D(7,4), E(9,7)$. Looking for digits that only appear once, we see that $9$, $6$, $1$, and $5$ cannot match up to other digits, and thus must appear on the ends. $1$ and $9$ only appear on the left... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/1.json | AHSME |
1998_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | Let $R$ be a rectangle. How many circles in the plane of $R$ have a diameter both of whose endpoints are vertices of $R$?
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6$
| [
"There are $6$ pairs of vertices of $R$. However, both diagonals determine the same circle, therefore the answer is $\\boxed{5}$.\n\n\n\\begin{center}\n[asy] size(200); defaultpen(0.8); pair A=(0,0), B=(5,0), C=(5,2), D=(0,2); draw ( A--B--C--D--cycle ); draw( circle( (A+B)/2, length((A-B)/2) ), red ); draw( circ... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/11.json | AHSME |
1998_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | Letters $A,B,C,$ and $D$ represent four different digits selected from $0,1,2,\ldots ,9.$ If $(A+B)/(C+D)$ is an integer that is as large as possible, what is the value of $A+B$?
$\mathrm{(A) \ }13 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ } 15\qquad \mathrm{(D) \ }16 \qquad \mathrm{(E) \ } 17$
| [
"If we want $\\frac{A+B}{C+D}$ to be as large as possible, we want to try to maximize the numerator $A+B$ and minimize the denominator $C+D$. Picking $A=9$ and $B=8$ will maximize the numerator, and picking $C=0$ and $D=1$ will minimize the denominator. \n\n\nChecking to make sure the fraction is an integer, $\\f... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/2.json | AHSME |
1998_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | In triangle $ABC$, angle $C$ is a right angle and $CB > CA$. Point $D$ is located on $\overline{BC}$ so that angle $CAD$ is twice angle $DAB$. If $AC/AD = 2/3$, then $CD/BD = m/n$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
$\mathrm{(A) \ }10 \qquad \mathrm{(B) \ }14 \qquad \mathrm{(C) \ }1... | [
"Let $\\theta = \\angle DAB$, so $2\\theta = \\angle CAD$ and $3 \\theta = \\angle CAB$. Then, it is given that $\\cos 2\\theta = \\frac{AC}{AD} = \\frac{2}{3}$ and\n\n\n\n\n\n\n\\begin{center}\n$\\frac{BD}{CD} = \\frac{AC(\\tan 3\\theta - \\tan 2\\theta)}{AC \\tan 2\\theta} = \\frac{\\tan 3\\theta}{\\tan 2\\theta}... | 4 | ./CreativeMath/AHSME/1998_AHSME_Problems/28.json | AHSME |
1998_AHSME_Problems | 12 | 0 | Number Theory | Multiple Choice | How many different prime numbers are factors of $N$ if
\begin{center}
$\log_2 ( \log_3 ( \log_5 (\log_ 7 N))) = 11?$\end{center}
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }3 \qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ }7$
| [
"Re-writing as exponents, we have $\\log_3 ( \\log_5 (\\log_ 7 N)) = 2^{11}$, and so forth, such that $N = 7^{5^{3^{2^{11}}}}$, which only has $7$ as a prime factor $\\mathbf{(A)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/12.json | AHSME |
1998_AHSME_Problems | 24 | 0 | Counting | Multiple Choice | Call a $7$-digit telephone number $d_1d_2d_3-d_4d_5d_6d_7$ \textit{memorable} if the prefix sequence $d_1d_2d_3$ is exactly the same as either of the sequences $d_4d_5d_6$ or $d_5d_6d_7$ (possibly both). Assuming that each $d_i$ can be any of the ten decimal digits $0,1,2, \ldots, 9$, the number of different memorable ... | [
"In this problem, we only need to consider the digits $\\overline{d_4d_5d_6d_7}$. Each possibility of $\\overline{d_4d_5d_6d_7}$ gives $2$ possibilities for $\\overline{d_1d_2d_3}$, which are $\\overline{d_1d_2d_3}=\\overline{d_4d_5d_6}$ and $\\overline{d_1d_2d_3}=\\overline{d_5d_6d_7}$ with the exception of the ca... | 2 | ./CreativeMath/AHSME/1998_AHSME_Problems/24.json | AHSME |
1998_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | A piece of graph paper is folded once so that $(0,2)$ is matched with $(4,0)$, and $(7,3)$ is matched with $(m,n)$. Find $m+n$.
$\mathrm{(A) \ }6.7 \qquad \mathrm{(B) \ }6.8 \qquad \mathrm{(C) \ }6.9 \qquad \mathrm{(D) \ }7.0 \qquad \mathrm{(E) \ }8.0$
| [
"The line of the fold is the perpendicular bisector of the segment that connects $(0,2)$ and $(4,0)$. \nThe point $(m,n)$ is the image of the point $(7,3)$ according to this axis.\nThe situation looks as follows.\n\n\n\\begin{center}\n[asy] size(200); defaultpen(0.8); pair A=(0,2), B=(4,0), C=(7,3); pair u=(1,2), ... | 2 | ./CreativeMath/AHSME/1998_AHSME_Problems/25.json | AHSME |
1998_AHSME_Problems | 13 | 0 | Number Theory | Multiple Choice | Walter rolls four standard six-sided dice and finds that the product of the numbers of the upper faces is $144$. Which of he following could \textbf{not } be the sum of the upper four faces?
$\mathrm{(A) \ }14 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad \mathrm{(E) \ }18$
| [
"We have $144 = 2^4 3^2$.\n\n\nAs the numbers on the dice are less than $9$, the two $3$s must come from different dice. This leaves us with three cases: $(6,6,a,b)$, \n$(6,3,a,b)$, and $(3,3,a,b)$.\n\n\nIn the first case we have $ab=2^2$, leading to the solutions $(6,6,4,1)$ and $(6,6,2,2)$.\n\n\nIn the second cas... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/13.json | AHSME |
1998_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | A point $(x,y)$ in the plane is called a lattice point if both $x$ and $y$ are integers. The area of the largest square that contains exactly three lattice points in its interior is closest to
$\mathrm{(A) \ } 4.0 \qquad \mathrm{(B) \ } 4.2 \qquad \mathrm{(C) \ } 4.5 \qquad \mathrm{(D) \ } 5.0 \qquad \mathrm{(E) \ } ... | [
"[asy] real e = 0.1; dot((0,-1)); dot((1,-1)); dot((-1,0)); dot((0,0)); dot((1,0)); dot((2,0)); dot((-1,1)); dot((0,1)); dot((1,1)); dot((0,2)); dot((-1,-1)); dot((2,2)); dot((1,2)); dot((2,1)); dot((2,-1)); dot((-1,2)); draw((0.8, -1.4+e)--(1.8-e, 0.6)--(-0.2, 1.6-e)--(-1.2+e, -0.4)--cycle); [/asy]\nThe best squa... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/29.json | AHSME |
1998_AHSME_Problems | 3 | 0 | Arithmetic | Multiple Choice | If $\texttt{a,b,}$ and $\texttt{c}$ are digits for which
\begin{center}
$\begin{tabular}{rr}&\ \texttt{7 a 2}\\ -& \texttt{4 8 b} \\ \hline &\ \texttt{c 7 3} \end{tabular}$\end{center}
then $\texttt{a+b+c =}$
$\mathrm{(A) \ }14 \qquad \mathrm{(B) \ }15 \qquad \mathrm{(C) \ }16 \qquad \mathrm{(D) \ }17 \qquad ... | [
"Working from right to left, we see that $2 - b = 3$. Clearly if $b$ is a single digit integer, this cannot be possible. Therefore, there must be some borrowing from $a$. Borrow $1$ from the digit $a$, and you get $12 - b = 3$, giving $b = 9$.\n\n\nSince $1$ was borrowed from $a$, we have from the tens column $(... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/3.json | AHSME |
1998_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | A square with sides of length $1$ is divided into two congruent trapezoids and a pentagon, which have equal areas, by joining the center of the square with points on three of the sides, as shown. Find $x$, the length of the longer parallel side of each trapezoid.
\begin{center}
[asy] pointpen = black; pathpen = blac... | [
"\\begin{center}\n[asy] pointpen = black; pathpen = black; D(unitsquare); D((0,0)); D((1,0)); D((1,1)); D((0,1)); D(D((.5,.5))--D((1,.5))); D(D((.17,1))--(.5,.5)--D((.5,1)));D(D((1-.17,1))--(.5,.5)--D((.17,0))); D((.17,1)--(.17,0));D((1-.17,1)--(1-.17,.5));D((0,.5)--(.5,.5)); MP(\"x\",(.58,1),N); MP(\"I\",(.17/2,.2... | 3 | ./CreativeMath/AHSME/1998_AHSME_Problems/8.json | AHSME |
1998_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | What is the value of the expression
\[\frac{1}{\log_2 100!} + \frac{1}{\log_3 100!} + \frac{1}{\log_4 100!} + \cdots + \frac{1}{\log_{100} 100!}?\]
$\mathrm{(A)}\ 0.01 \qquad\mathrm{(B)}\ 0.1 \qquad\mathrm{(C)}\ 1 \qquad\mathrm{(D)}\ 2 \qquad\mathrm{(E)}\ 10$
\subsubsection{Solution 1}
By the change-of-base formul... | [
"By the change-of-base formula, \n\\[\\log_{k} 100! = \\frac{\\log 100!}{\\log k}\\] \nThus (because if $x=y$, then $\\frac{1}{x}=\\frac{1}{y}$)\n\\[\\frac{1}{\\log_k 100!} = \\frac{\\log k}{\\log 100!}\\]\nThus the sum is \n\\[\\left(\\frac{1}{\\log 100!}\\right)(\\log 2 + \\log 3 + \\cdots + \\log 100) = \\frac{... | 2 | ./CreativeMath/AHSME/1998_AHSME_Problems/22.json | AHSME |
1998_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | A right circular cone of volume $A$, a right circular cylinder of volume $M$, and a sphere of volume $C$ all have the same radius, and the common height of the cone and the cylinder is equal to the diameter of the sphere. Then
$\mathrm{(A) \ } A-M+C = 0 \qquad \mathrm{(B) \ } A+M=C \qquad \mathrm{(C) \ } 2A = M+C$
... | [
"Using the radius $r$ the three volumes can be computed as follows:\n\n\n$A = \\frac 13 (\\pi r^2) \\cdot 2r$\n\n\n$M = (\\pi r^2) \\cdot 2r$\n\n\n$C = \\frac 43 \\pi r^3$\n\n\nClearly, $M = A+C \\Longrightarrow$ the correct answer is $\\mathrm{(A)}$.\n\n\nThe other linear combinations are obviously non-zero, and t... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/18.json | AHSME |
1998_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | Define $[a,b,c]$ to mean $\frac {a+b}c$, where $c \neq 0$. What is the value of
\begin{center}
$\left[[60,30,90],[2,1,3],[10,5,15]\right]?$\end{center}
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }0.5 \qquad \mathrm{(C) \ }1 \qquad \mathrm{(D) \ }1.5 \qquad \mathrm{(E) \ }2$
| [
"Note that $[ta,tb,tc] = \\frac{ta+tb}{tc} = \\frac{t(a+b)}{tc} = \\frac{a+b}{c} = [a,b,c]$. \n\n\nThus $[60,30,90] = [2,1,3] = [10,5,15] = \\frac{2+1}{3} = 1$, and $[1,1,1] = \\frac{1+1}{1} = 2 \\Longrightarrow \\mathbf{(E)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/4.json | AHSME |
1998_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | A parabola has vertex of $(4,-5)$ and has two $x-$intercepts, one positive, and one negative. If this parabola is the graph of $y = ax^2 + bx + c,$ which of $a,b,$ and $c$ must be positive?
$\mathrm{(A) \ } \text{only}\ a \qquad \mathrm{(B) \ } \text{only}\ b \qquad \mathrm{(C) \ } \text{only}\ c \qquad \mathrm{(D) \... | [
"The vertex of the parabola is at $(4,-5)$. Since there are two x-intercepts, it must open upwards. If it opened downard, there would be no roots. Thus, $a > 0$.\n\n\nThe x-coordinate of the vertex is $\\frac{-b}{2a}$. Since $a$ is positive, and the x-intercept is positive, the value $-b$ must be positive too, ... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/14.json | AHSME |
1998_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | A regular hexagon and an equilateral triangle have equal areas. What is the ratio of the length of a side of the triangle to the length of a side of the hexagon?
$\mathrm{(A) \ }\sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }\sqrt{6} \qquad \mathrm{(D) \ }3 \qquad \mathrm{(E) \ }6$
| [
"$A_{\\triangle} = \\frac{s_t^2\\sqrt{3}}{4}$\n\n\n$A_{hex} = \\frac{6s_h^2\\sqrt{3}}{4}$ since a regular hexagon is just six equilateral triangles.\n\n\nSetting the areas equal, we get:\n\n\n$s_t^2 = 6s_h^2$\n\n\n$\\left(\\frac{s_t}{s_h}\\right)^2 = 6$\n\n\n$\\frac{s_t}{s_h} = \\sqrt{6}$, and the answer is $\\boxe... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/15.json | AHSME |
1998_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | If $2^{1998}-2^{1997}-2^{1996}+2^{1995} = k \cdot 2^{1995},$ what is the value of $k$?
$\mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ } 5$
| [
"$2^{1998} - 2^{1997} - 2^{1996} + 2^{1995}$\n\n\nFactor out $2^{1995}$:\n\n\n$2^{1995}(2^{3} - 2^{2} - 2^{1} + 1)$\n\n\nSimplify:\n\n\n$2^{1995}\\cdot (8 - 4 - 2 + 1)$\n\n\n$2^{1995}\\cdot (3)$\n\n\nBy comparing the answer with the original equation, $k=3$, and the answer is $\\text{(C)}.$\n\n\n",
"Divide both s... | 2 | ./CreativeMath/AHSME/1998_AHSME_Problems/5.json | AHSME |
1998_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | How many triangles have area $10$ and vertices at $(-5,0),(5,0)$ and $(5\cos \theta, 5\sin \theta)$ for some angle $\theta$?
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$
| [
"The triangle can be seen as having the base on the $x$ axis and height $|5\\sin\\theta|$. The length of the base is $10$, thus the height must be $2$. The equation $|\\sin\\theta| = \\frac 25$ has $\\boxed{4}$ solutions, one in each quadrant.\n\n\n\\begin{center}\n[asy] size(250); defaultpen(0.8); pair A=(-5,0), ... | 2 | ./CreativeMath/AHSME/1998_AHSME_Problems/19.json | AHSME |
1998_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | The graphs of $x^2 + y^2 = 4 + 12x + 6y$ and $x^2 + y^2 = k + 4x + 12y$ intersect when $k$ satisfies $a \le k \le b$, and for no other values of $k$. Find $b-a$.
$\mathrm{(A) \ }5 \qquad \mathrm{(B) \ }68 \qquad \mathrm{(C) \ }104 \qquad \mathrm{(D) \ }140 \qquad \mathrm{(E) \ }144$
| [
"Both sets of points are quite obviously circles. To show this, we can rewrite each of them in the form $(x-x_0)^2 + (y-y_0)^2 = r^2$.\n\n\nThe first curve becomes $(x-6)^2 + (y-3)^2 = 7^2$, which is a circle centered at $(6,3)$ with radius $7$.\n\n\nThe second curve becomes $(x-2)^2 + (y-6)^2 = 40+k$, which is a c... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/23.json | AHSME |
1998_AHSME_Problems | 9 | 0 | Arithmetic | Multiple Choice | A speaker talked for sixty minutes to a full auditorium. Twenty percent of the audience heard the entire talk and ten percent slept through the entire talk. Half of the remainder heard one third of the talk and the other half heard two thirds of the talk. What was the average number of minutes of the talk heard by memb... | [
"Assume that there are $100$ people in the audience.\n\n\n$20$ people heard $60$ minutes of the talk, for a total of $20\\cdot 60 = 1200$ minutes heard.\n\n\n$10$ people heard $0$ minutes.\n\n\n$\\frac{70}{2} = 35$ people heard $20$ minutes of the talk, for a total of $35\\cdot 20 = 700$ minutes.\n\n\n$35$ people h... | 1 | ./CreativeMath/AHSME/1998_AHSME_Problems/9.json | AHSME |
1967_AHSME_Problems | 20 | 0 | Geometry | Multiple Choice | A circle is inscribed in a square of side $m$, then a square is inscribed in that circle, then a circle is inscribed in the latter square, and so on. If $S_n$ is the sum of the areas of the first $n$ circles so inscribed, then, as $n$ grows beyond all bounds, $S_n$ approaches:
$\textbf{(A)}\ \frac{\pi m^2}{2}\qquad... | [
"All answers correctly grow as $m^2$, so we let $m=1$.\n\n\nThe radius of the first circle is $\\frac{1}{2}$, so its area is $\\frac{\\pi}{4}$.\n\n\nThe diagonal of the second square is the diameter of the first circle, which is $1$. Therefore, the side length of the square is $\\frac{\\sqrt{2}}{2}$.\n\n\nNow we n... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/20.json | AHSME |
1967_AHSME_Problems | 36 | 0 | Algebra | Multiple Choice | Given a geometric progression of five terms, each a positive integer less than $100$. The sum of the five terms is $211$. If $S$ is the sum of those terms in the progression which are squares of integers, then $S$ is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 91\qquad \textbf{(C)}\ 133\qquad \textbf{(D)}\ 195\qquad \tex... | [
"Let the first term be $a$ and the common ratio be $r$, and WLOG let $r \\ge 1$. The five terms are $a, ar, ar^2, ar^3, ar^4$, and the sum is $a(1 + r + r^2 + r^3 + r^4)$. Clearly $r$ must be rational for all terms to be integers. If $r$ were an integer, it could not be $1$, since $a$ would equal $\\frac{211}{1 ... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/36.json | AHSME |
1967_AHSME_Problems | 16 | 0 | Number Theory | Multiple Choice | Let the product $(12)(15)(16)$, each factor written in base $b$, equals $3146$ in base $b$. Let $s=12+15+16$, each term expressed in base $b$. Then $s$, in base $b$, is
$\textbf{(A)}\ 43\qquad \textbf{(B)}\ 44\qquad \textbf{(C)}\ 45\qquad \textbf{(D)}\ 46\qquad \textbf{(E)}\ 47$
| [
"Converting everything into base $10$, we have $(b + 2)(b+5)(b+6) = 3b^3 + b^2 + 4b + 6$. Looking ahead, the constant term of the polynomial will be $2*5*6 -6 = 54$. By the Rational Root Theorem, the only possible integer roots are $1, 2, 3, 6, 9, 18, 27, 54$. Bases $1, 2, 3, 6$ do not have a $6$ as a digit. Tes... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/16.json | AHSME |
1967_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | If $f(x)=4^x$ then $f(x+1)-f(x)$ equals:
$\text{(A)}\ 4\qquad\text{(B)}\ f(x)\qquad\text{(C)}\ 2f(x)\qquad\text{(D)}\ 3f(x)\qquad\text{(E)}\ 4f(x)$
| [
"The desired expression is equal to $4^{x+1} - 4^{x}$\nUsing the fact that $4^{x+1}$=$4^{x}*4$, we see that the answer is\n$3*4^{x}$\n$\\fbox{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/6.json | AHSME |
1967_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | If $\frac{a}{b}<-\frac{c}{d}$ where $a$, $b$, $c$, and $d$ are real numbers and $bd \not= 0$, then:
$\text{(A)}\ a \; \text{must be negative} \qquad \text{(B)}\ a \; \text{must be positive} \qquad$
$\text{(C)}\ a \; \text{must not be zero} \qquad \text{(D)}\ a \; \text{can be negative or zero, but not positive } \\ \... | [
"Notice that $b$ and $d$ are irrelevant to the problem. Also notice that the negative sign is irrelevant since $c$ can be any real number. So the question is equivalent to asking what values $a$ can take on given the inequality $a<c$. The answer, is of course, that $a$ could be anything, or $\\boxed{\\textbf{(E)... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/7.json | AHSME |
1967_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | If $r_1$ and $r_2$ are the distinct real roots of $x^2+px+8=0$, then it must follow that:
$\textbf{(A)}\ |r_1+r_2|>4\sqrt{2}\qquad \textbf{(B)}\ |r_1|>3 \; \text{or} \; |r_2| >3 \\ \textbf{(C)}\ |r_1|>2 \; \text{and} \; |r_2|>2\qquad \textbf{(D)}\ r_1<0 \; \text{and} \; r_2<0\qquad \textbf{(E)}\ |r_1+r_2|<4\sqrt{2}$
... | [
"We are given that the roots are real, so the discriminant is positive, which means $p^2 - 4(8)(1) > 0$. This leads to $|p| > 4\\sqrt{2}$. By Vieta, the sum of the roots is $-p$, so we have $|-(r_1 + r_2)| \\ge 4\\sqrt{2}$, or $|r_1 + r_2| > 4\\sqrt{2}$, which is option $\\fbox{A}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/17.json | AHSME |
1967_AHSME_Problems | 40 | 0 | Geometry | Multiple Choice | Located inside equilateral triangle $ABC$ is a point $P$ such that $PA=8$, $PB=6$, and $PC=10$. To the nearest integer the area of triangle $ABC$ is:
$\textbf{(A)}\ 159\qquad \textbf{(B)}\ 131\qquad \textbf{(C)}\ 95\qquad \textbf{(D)}\ 79\qquad \textbf{(E)}\ 50$
| [
"[asy] draw((0,10)--(8.66,-5)--(-8.66,-5)--cycle); label(\"$A$\",(0,10),N); label(\"$B$\",(-9.5,-5.2),N); label(\"$C$\",(9.5,-5.2),N); dot((-3,0)); label(\"$P$\",(-3,-2),N); draw((-3,0)--(0,10)); draw((-3,0)--(-8.66,-5)); draw((-3,0)--(8.66,-5)); dot((-9,7.5)); label(\"$P'$\",(-9.2,7.5),N); draw((-9,7.5)--(0,10))... | 4 | ./CreativeMath/AHSME/1967_AHSME_Problems/40.json | AHSME |
1967_AHSME_Problems | 37 | 0 | Geometry | Multiple Choice | Segments $AD=10$, $BE=6$, $CF=24$ are drawn from the vertices of triangle $ABC$, each perpendicular to a straight line $RS$, not intersecting the triangle. Points $D$, $E$, $F$ are the intersection points of $RS$ with the perpendiculars. If $x$ is the length of the perpendicular segment $GH$ drawn to $RS$ from the in... | [
"$\\fbox{A}$\n\n\nWLOG let $RS$ be the x-axis, or at least horizontal. The three lengths represent the y-coordinates of points $A,B,C$. As $G$ is by definition the average of $A,B,C$, it's coordinates are the average of the coordinates of $A,B,$ and $C$. Hence, the y-coordinate of $G$, which is also the distance fr... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/37.json | AHSME |
1967_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | In right triangle $ABC$ the hypotenuse $\overline{AB}=5$ and leg $\overline{AC}=3$. The bisector of angle $A$ meets the opposite side in $A_1$. A second right triangle $PQR$ is then constructed with hypotenuse $\overline{PQ}=A_1B$ and leg $\overline{PR}=A_1C$. If the bisector of angle $P$ meets the opposite side in ... | [
"$\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/21.json | AHSME |
1967_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | If $\frac{a}{10^x-1}+\frac{b}{10^x+2}=\frac{2 \cdot 10^x+3}{(10^x-1)(10^x+2)}$ is an identity for positive rational values of $x$, then the value of $a-b$ is:
$\textbf{(A)}\ \frac{4}{3} \qquad \textbf{(B)}\ \frac{5}{3} \qquad \textbf{(C)}\ 2 \qquad \textbf{(D)}\ \frac{11}{4} \qquad \textbf{(E)}\ 3$
| [
"Given the equation:\n\\[\\frac{a}{10^x-1}+\\frac{b}{10^x+2}=\\frac{2 \\cdot 10^x+3}{(10^x-1)(10^x+2)}\\]\n\n\nLet's simplify by letting $y = 10^x$. The equation becomes:\n\\[\\frac{a}{y-1}+\\frac{b}{y+2}=\\frac{2y+3}{(y-1)(y+2)}\\]\n\n\nMultiplying each term by the common denominator $(y-1)(y+2)$, we obtain:\n\\[a... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/10.json | AHSME |
1967_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | If one uses only the tabular information $10^3=1000$, $10^4=10,000$, $2^{10}=1024$, $2^{11}=2048$, $2^{12}=4096$, $2^{13}=8192$, then the strongest statement one can make for $\log_{10}{2}$ is that it lies between:
$\textbf{(A)}\ \frac{3}{10} \; \text{and} \; \frac{4}{11}\qquad \textbf{(B)}\ \frac{3}{10} \; \text{and... | [
"Since $1024$ is greater than $1000$.\n\n\n$\\log (1024) > 3$\n\n\n$10 * \\log (2) > 3$\n\n\nand $\\log (2) > 3/10$.\n\n\n\n\nSimilarly, $8192 < 10000$, so $\\log (8192) < 4$\n\n\n$13 * \\log (2) < 4$\n\n\nand $\\log (2) < 4/13$\n\n\n\n\nTherefore $3/10 < \\log 2 < 4/13$ \nso the answer is $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/26.json | AHSME |
1967_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | A dealer bought $n$ radios for $d$ dollars, $d$ a positive integer. He contributed two radios to a community bazaar at half their cost. The rest he sold at a profit of $8 on each radio sold. If the overall profit was $72, then the least possible value of $n$ for the given information is:
$\textbf{(A)}\ 18\qquad \t... | [
"The first $2$ radios are sold for $\\frac{d}{2}$ dollars each, for income of $2 \\cdot \\frac{d}{2}$, or simply $d$.\n\n\nThe remaining $n-2$ radios are sold for $d+8$ dollars each, for income of $(n-2)(d+8)$, which expands to $nd +8n -2d - 16$.\n\n\nThe amount invested in buying $n$ radios for $d$ dollars is $nd$... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/30.json | AHSME |
1967_AHSME_Problems | 31 | 0 | Algebra | Multiple Choice | Let $D=a^2+b^2+c^2$, where $a$, $b$, are consecutive integers and $c=ab$. Then $\sqrt{D}$ is:
$\textbf{(A)}\ \text{always an even integer}\qquad \textbf{(B)}\ \text{sometimes an odd integer, sometimes not}\\ \textbf{(C)}\ \text{always an odd integer}\qquad \textbf{(D)}\ \text{sometimes rational, sometimes not}\\ \te... | [
"Let $a=x, b=x+1, c = x(x+1)$. Then $D = x^2 + (x+1)^2 + x^2(x+1)^2$, which simplifies to $x^4 + 2x^3 + 3x^2 + 2x + 1$.\n\n\nFrom the options, we want to test if $D$ is always a perfect square. Because the polynomial expression for $D$ is quartic, if it is a perfect square, it would be the square of a quadratic e... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/31.json | AHSME |
1967_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | Two candles of the same length are made of different materials so that one burns out completely at a uniform rate in $3$ hours and the other in $4$ hours. At what time P.M. should the candles be lighted so that, at 4 P.M., one stub is twice the length of the other?
$\textbf{(A) 1:24}\qquad \textbf{(B) 1:28}\qquad \t... | [
"If the candles both have length $\\ell$, then the candle that burns in $3$ hours has a stub of $\\ell$ at $0$ minutes, and a stub of $0$ at $180$ minutes. Since the candle burns at a constant rate (i.e. linearly), the stub length of this candle $t$ minutes after being lit is $f(t) = \\frac{\\ell}{180}(180 - t)$, ... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/27.json | AHSME |
1967_AHSME_Problems | 1 | 0 | Number Theory | Multiple Choice | The three-digit number $2a3$ is added to the number $326$ to give the three-digit number $5b9$. If $5b9$ is divisible by 9, then $a+b$ equals
$\text{(A)}\ 2\qquad\text{(B)}\ 4\qquad\text{(C)}\ 6\qquad\text{(D)}\ 8\qquad\text{(E)}\ 9$
| [
"If $5b9$ is divisible by $9$, this must mean that $5 + b + 9$ is a multiple of $9$. So, \\[5 + b + 9 = 9, 18, 27, 36...\\]\n\n\nBecause $5 + 9 = 14$ and $b$ is in between 0 and 9, \n\n\n\\[5 + b + 9 = 18\\]\n\\[b = 4\\]\n\n\nThe question states that\n\\[2a3 + 326 = 549\\]\nso \n\\[2a3 = 549 - 326\\]\n\\[2a3 = 223\... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/1.json | AHSME |
1967_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | If the perimeter of rectangle $ABCD$ is $20$ inches, the least value of diagonal $\overline{AC}$, in inches, is:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ \sqrt{50}\qquad \textbf{(C)}\ 10\qquad \textbf{(D)}\ \sqrt{200}\qquad \textbf{(E)}\ \text{none of these}$
| [
"For rectangle $ABCD$ with perimeter 20, the diagonal $AC$ is given by:\n\\[AC = \\sqrt{l^2 + w^2}\\]\nTo minimize $AC$, $l$ and $w$ should be equal (i.e., the rectangle is a square). Thus, $l = w = 5$.\nSo, the minimum $AC$ is:\n\\[AC = \\sqrt{5^2 + 5^2} = \\boxed{\\textbf{(B) } \\sqrt{50}}\\]\n~ proloto\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/11.json | AHSME |
1967_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | An equivalent of the expression
$\left(\frac{x^2+1}{x}\right)\left(\frac{y^2+1}{y}\right)+\left(\frac{x^2-1}{y}\right)\left(\frac{y^2-1}{x}\right)$, $xy \not= 0$,
is:
$\text{(A)}\ 1\qquad\text{(B)}\ 2xy\qquad\text{(C)}\ 2x^2y^2+2\qquad\text{(D)}\ 2xy+\frac{2}{xy}\qquad\text{(E)}\ \frac{2x}{y}+\frac{2y}{x}$
| [
"\\[\\left(\\frac{x^2+1}{x}\\right)\\left(\\frac{y^2+1}{y}\\right)+\\left(\\frac{x^2-1}{y}\\right)\\left(\\frac{y^2-1}{x}\\right)\\]\n\\[\\frac{(x^2+1)(y^2+1)}{xy} + \\frac{(x^2-1)(y^2-1)}{xy}\\]\n\\[\\frac{(x^2y^2+x^2+y^2+1) + (x^2y^2-x^2-y^2+1)}{xy}\\]\n\\[\\frac{x^2y^2+x^2+y^2+1 + x^2y^2-x^2-y^2+1}{xy}\\]\n\\[\\... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/2.json | AHSME |
1967_AHSME_Problems | 28 | 0 | Other | Multiple Choice | Given the two hypotheses: $\text{I}$ Some Mems are not Ens and $\text{II}$ No Ens are Veens. If "some" means "at least one", we can conclude that:
$\textbf{(A)}\ \text{Some Mems are not Veens}\qquad \textbf{(B)}\ \text{Some Vees are not Mems}\\ \textbf{(C)}\ \text{No Mem is a Vee}\qquad \textbf{(D)}\ \text{Some Mems... | [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/28.json | AHSME |
1967_AHSME_Problems | 32 | 0 | Geometry | Multiple Choice | In quadrilateral $ABCD$ with diagonals $AC$ and $BD$, intersecting at $O$, $BO=4$, $OD = 6$, $AO=8$, $OC=3$, and $AB=6$. The length of $AD$ is:
$\textbf{(A)}\ 9\qquad \textbf{(B)}\ 10\qquad \textbf{(C)}\ 6\sqrt{3}\qquad \textbf{(D)}\ 8\sqrt{2}\qquad \textbf{(E)}\ \sqrt{166}$
| [
"After drawing the diagram, we see that we actually have a lot of lengths to work with. Considering triangle ABD, we know values of $AB, BD(BD = BO + OD)$, but we want to find the value of $AD$. We can apply stewart's theorem now, letting $m = 4, n = 6, AD = X, AB = 6$, and we have $10 \\cdot 6 \\cdot 4 + 8 \\cdot ... | 3 | ./CreativeMath/AHSME/1967_AHSME_Problems/32.json | AHSME |
1967_AHSME_Problems | 24 | 0 | Number Theory | Multiple Choice | The number of solution-pairs in the positive integers of the equation $3x+5y=501$ is:
$\textbf{(A)}\ 33\qquad \textbf{(B)}\ 34\qquad \textbf{(C)}\ 35\qquad \textbf{(D)}\ 100\qquad \textbf{(E)}\ \text{none of these}$
| [
"We have $y = \\frac{501 - 3x}{5}$. Thus, $501 - 3x$ must be a positive multiple of $5$. If $x = 2$, we find our first positive multiple of $5$. From there, we note that $x = 2 + 5k$ will always return a multiple of $5$ for $501 - 3x$. Our first solution happens at $k=0$. \n\n\n\n\nWe now want to find the smal... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/24.json | AHSME |
1967_AHSME_Problems | 25 | 0 | Number Theory | Multiple Choice | For every odd number $p>1$ we have:
$\textbf{(A)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-2\qquad \textbf{(B)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p\\ \textbf{(C)}\ (p-1)^{\frac{1}{2}(p-1)} \; \text{is divisible by} \; p\qquad \textbf{(D)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{i... | [
"Given that $p$ is odd, $p-1$ must be even, therefore ${{\\frac{1}{2}}(p-1)}$ must be an integer, which will be denoted as n.\n\\[(p-1)^n-1\\]\nBy sum and difference of powers\n\\[=((p-1)-1)((p-1)^{n-1}+\\cdots+1^{n-1})\\]\n\\[=(p-2)((p-1)^{n-1}+\\cdots+1^{n-1})\\]\n$p-2$ divide $(p-1)^{{\\frac{1}{2}}(p-1)}$$\\fbox... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/25.json | AHSME |
1967_AHSME_Problems | 33 | 0 | Geometry | Multiple Choice | [asy] fill(circle((4,0),4),grey); fill((0,0)--(8,0)--(8,-4)--(0,-4)--cycle,white); fill(circle((7,0),1),white); fill(circle((3,0),3),white); draw((0,0)--(8,0),black+linewidth(1)); draw((6,0)--(6,sqrt(12)),black+linewidth(1)); MP("A", (0,0), W); MP("B", (8,0), E); MP("C", (6,0), S); MP("D",(6,sqrt(12)), N); [/asy]
I... | [
"To make the problem much simpler while staying in the constraints of the problem, position point $C$ halfway between $A$ and $B$. Then, call $\\overline{AC} = \\overline{BC}=r$ . The area of the shaded region is then \\[\\frac{ \\pi r^2 - \\pi (r/2)^2 - \\pi (r/2)^2}{2}=\\frac{\\pi r^2}{4}\\]\nBecause $\\overline{... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/33.json | AHSME |
1967_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | $\overline{AB}$ is a diameter of a circle. Tangents $\overline{AD}$ and $\overline{BC}$ are drawn so that $\overline{AC}$ and $\overline{BD}$ intersect in a point on the circle. If $\overline{AD}=a$ and $\overline{BC}=b$, $a \not= b$, the diameter of the circle is:
$\textbf{(A)}\ |a-b|\qquad \textbf{(B)}\ \frac{1}{... | [
"$\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/29.json | AHSME |
1967_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | The side of an equilateral triangle is $s$. A circle is inscribed in the triangle and a square is inscribed in the circle. The area of the square is:
$\text{(A)}\ \frac{s^2}{24}\qquad\text{(B)}\ \frac{s^2}{6}\qquad\text{(C)}\ \frac{s^2\sqrt{2}}{6}\qquad\text{(D)}\ \frac{s^2\sqrt{3}}{6}\qquad\text{(E)}\ \frac{s^2}{3... | [
"The radius of our circle is $\\frac{\\sqrt{3}}{6}s$. This means the square has side length $\\frac{\\sqrt{6}}{6}s$ which has area of $\\frac{s^2}{6}=\\fbox{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/3.json | AHSME |
1967_AHSME_Problems | 34 | 0 | Geometry | Multiple Choice | Points $D$, $E$, $F$ are taken respectively on sides $AB$, $BC$, and $CA$ of triangle $ABC$ so that $AD:DB=BE:CE=CF:FA=1:n$. The ratio of the area of triangle $DEF$ to that of triangle $ABC$ is:
$\textbf{(A)}\ \frac{n^2-n+1}{(n+1)^2}\qquad \textbf{(B)}\ \frac{1}{(n+1)^2}\qquad \textbf{(C)}\ \frac{2n^2}{(n+1)^2}\qqua... | [
"WLOG, let's assume that $\\triangle ABC$ is equilateral. Therefore, $[ABC]=\\frac{(1+n)^2\\sqrt3}{4}$ and $[DBE]=[ADF]=[EFC]=n \\cdot sin(60)/2$. Then $[DEF]=\\frac{(n^2-n+1)\\sqrt3}{4}$. Finding the ratio yields $\\fbox{A}$. -Dark_Lord\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/34.json | AHSME |
1967_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | To $m$ ounces of a $m\%$ solution of acid, $x$ ounces of water are added to yield a $(m-10)\%$ solution. If $m>25$, then $x$ is
$\textbf{(A)}\ \frac{10m}{m-10} \qquad \textbf{(B)}\ \frac{5m}{m-10} \qquad \textbf{(C)}\ \frac{m}{m-10} \qquad \textbf{(D)}\ \frac{5m}{m-20} \\ \textbf{(E)}\ \text{not determined by the g... | [
"$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/8.json | AHSME |
1967_AHSME_Problems | 22 | 0 | Number Theory | Multiple Choice | For natural numbers, when $P$ is divided by $D$, the quotient is $Q$ and the remainder is $R$. When $Q$ is divided by $D'$, the quotient is $Q'$ and the remainder is $R'$. Then, when $P$ is divided by $DD'$, the remainder is:
$\textbf{(A)}\ R+R'D\qquad \textbf{(B)}\ R'+RD\qquad \textbf{(C)}\ RR'\qquad \textbf{(D)}\... | [
"We are given $P = QD + R$ and $Q = D'Q' + R'$.\n\n\nPlugging the second equation into the first yields:\n\n\n$P = (D'Q' + R')D + R$\n\n\n\n\n$P = (DD')Q' + (R'D + R)$\n\n\nIf we divide $P$ by $DD'$, the quotient would be $Q'$, and the remainder would be $R'D + R$, which is option $\\fbox{A}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/22.json | AHSME |
1967_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | If $x^2-5x+6<0$ and $P=x^2+5x+6$ then
$\textbf{(A)}\ P \; \text{can take any real value} \qquad \textbf{(B)}\ 20<P<30\\ \textbf{(C)}\ 0<P<20 \qquad \textbf{(D)}\ P<0 \qquad \textbf{(E)}\ P>30$
| [
"We are given that $x^2 - 5x + 6 < 0$, which, when factored, gives $(x - 2)(x-3) < 0$. This has a solution of $2<x<3$, because the original quadratic is $\\cup$-shaped, and thus dips below the x-axis between the roots.\n\n\nSince $x^2 + 5x + 6$ has a vertex minimum at $x = -\\frac{5}{2}$, so it is increasing on th... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/18.json | AHSME |
1967_AHSME_Problems | 38 | 0 | Algebra | Multiple Choice | Given a set $S$ consisting of two undefined elements "pib" and "maa", and the four postulates: $P_1$: Every pib is a collection of maas, $P_2$: Any two distinct pibs have one and only one maa in common, $P_3$: Every maa belongs to two and only two pibs, $P_4$: There are exactly four pibs. Consider the three theorems: ... | [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/38.json | AHSME |
1967_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | Given $\frac{\log{a}}{p}=\frac{\log{b}}{q}=\frac{\log{c}}{r}=\log{x}$, all logarithms to the same base and $x \not= 1$. If $\frac{b^2}{ac}=x^y$, then $y$ is:
$\text{(A)}\ \frac{q^2}{p+r}\qquad\text{(B)}\ \frac{p+r}{2q}\qquad\text{(C)}\ 2q-p-r\qquad\text{(D)}\ 2q-pr\qquad\text{(E)}\ q^2-pr$
| [
"We are given:\n\\[\\frac{b^2}{ac} = x^y\\]\n\n\nTaking the logarithm on both sides:\n\\[\\log{\\left(\\frac{b^2}{ac}\\right)} = \\log{x^y}\\]\n\n\nUsing the properties of logarithms:\n\\[2\\log{b} - \\log{a} - \\log{c} = y \\log{x}\\]\n\n\nSubstituting the values given in the problem statement:\n\\[2q \\log{x} - p... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/4.json | AHSME |
1967_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | Let $f(t)=\frac{t}{1-t}$, $t \not= 1$. If $y=f(x)$, then $x$ can be expressed as
$\textbf{(A)}\ f\left(\frac{1}{y}\right)\qquad \textbf{(B)}\ -f(y)\qquad \textbf{(C)}\ -f(-y)\qquad \textbf{(D)}\ f(-y)\qquad \textbf{(E)}\ f(y)$
| [
"Since we know that $y=f(x)$, we can solve for $y$ in terms of $x$. This gives us\n\n\n$y=\\frac{x}{1-x}$\n\n\n$\\Rightarrow y(1-x)=x$\n\n\n$\\Rightarrow y-yx=x$\n\n\n$\\Rightarrow y=yx+x$\n\n\n$\\Rightarrow y=x(y+1)$\n\n\n$\\Rightarrow x=\\frac{y}{y+1}$\n\n\nTherefore, we want to find the function with $y$ that ou... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/14.json | AHSME |
1967_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | The difference in the areas of two similar triangles is $18$ square feet, and the ratio of the larger area to the smaller is the square of an integer. The area of the smaller triange, in square feet, is an integer, and one of its sides is $3$ feet. The corresponding side of the larger triangle, in feet, is:
$\textb... | [
"If the areas of the two triangles are $A_1$ and $A_2$ with $A_1 > A_2$, we are given that $A_1 - A_2 = 18$ and $\\frac{A_1}{A_2} = k^2$\n\n\nPlugging in the second equation into the first leads to $(k^2 - 1)A_2 = 18$. \n\n\nIf $A_2$ is an integer, it can only be a factor of $18$ - namely $1, 2, 3, 6, 9, 18$. \n\... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/15.json | AHSME |
1967_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | A triangle is circumscribed about a circle of radius $r$ inches. If the perimeter of the triangle is $P$ inches and the area is $K$ square inches, then $\frac{P}{K}$ is:
$\text{(A)}\text{ independent of the value of} \; r\qquad\text{(B)}\ \frac{\sqrt{2}}{r}\qquad\text{(C)}\ \frac{2}{\sqrt{r}}\qquad\text{(D)}\ \frac{... | [
"The area $K$ of the triangle can be expressed in terms of its inradius $r$ and its semiperimeter $s$ as:\n\\[K = r \\times s = r \\times \\frac{P}{2}\\]\n\n\nSo, $\\frac{P}{K} = \\boxed{\\textbf{(C) } \\frac{2}{r}}$.\n\n\n~ proloto\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/5.json | AHSME |
1967_AHSME_Problems | 39 | 0 | Algebra | Multiple Choice | Given the sets of consecutive integers $\{1\}$,$\{2, 3\}$,$\{4,5,6\}$,$\{7,8,9,10\}$,$\; \cdots \;$, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let $S_n$ be the sum of the elements in the nth set. The... | [
"The last element of the 21st set is $\\frac{21\\times 22}{2}=231$, and hence the first element of the 21st set is $211$.\nSo $S_{21}=\\frac{231\\times 232}{2}-\\frac{210\\times 211}{2}=4641$, hence our answer is $\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/39.json | AHSME |
1967_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | The area of a rectangle remains unchanged when it is made $2 \frac{1}{2}$ inches longer and $\frac{2}{3}$ inch narrower, or when it is made $2 \frac{1}{2}$ inches shorter and $\frac{4}{3}$ inch wider. Its area, in square inches, is:
$\textbf{(A)}\ 30\qquad \textbf{(B)}\ \frac{80}{3}\qquad \textbf{(C)}\ 24\qquad \tex... | [
"We are given $xy = (x+\\frac{5}{2})(y-\\frac{2}{3}) = (x - \\frac{5}{2})(y + \\frac{4}{3})$\n\n\nFOILing each side gives:\n\n\n$xy = xy - \\frac{2}{3}x + \\frac{5}{2}y - \\frac{5}{3} = xy + \\frac{4}{3}x - \\frac{5}{2}y - \\frac{10}{3}$\n\n\nTaking the last two parts and moving everything to the left gives:\n\n\n$... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/19.json | AHSME |
1967_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | If $x$ is real and positive and grows beyond all bounds, then $\log_3{(6x-5)}-\log_3{(2x+1)}$ approaches:
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 3\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ \text{no finite number}$
| [
"Since $\\log_b x - \\log_b y = \\log_b \\frac{x}{y}$, the expression is equal to $\\log_3 \\frac{6x - 5}{2x + 1}$.\n\n\nThe expression $\\frac{6x - 5}{2x + 1}$ is equal to $3 - \\frac{8}{2x + 1}$. As $x$ gets large, the second term approaches $0$, and thus $\\frac{6x - 5}{2x + 1}$ approaches $3$. Thus, the expre... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/23.json | AHSME |
1967_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | Let $K$, in square units, be the area of a trapezoid such that the shorter base, the altitude, and the longer base, in that order, are in arithmetic progression. Then:
$\textbf{(A)}\ K \; \text{must be an integer} \qquad \textbf{(B)}\ K \; \text{must be a rational fraction} \\ \textbf{(C)}\ K \; \text{must be an irr... | [
"From the problem we can set the altitude equal to $a$, the shorter base equal to $a-d$, and the longer base equal to $a+d$. By the formula for the area of a trapezoid, we have $K=a^2$. However, since $a$ can equal any real number $(3, 2.7, \\pi)$, none of the statements $A, B, C, D$ need to be true, so the answer ... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/9.json | AHSME |
1967_AHSME_Problems | 35 | 0 | Algebra | Multiple Choice | The roots of $64x^3-144x^2+92x-15=0$ are in arithmetic progression. The difference between the largest and smallest roots is:
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ \frac{1}{2}\qquad \textbf{(D)}\ \frac{3}{8}\qquad \textbf{(E)}\ \frac{1}{4}$
| [
"By Vieta, the sum of the roots is $-\\frac{-144}{64} = \\frac{9}{4}$. Because the roots are in arithmetic progression, the middle root is the average of the other two roots, and is also the average of all three roots. Therefore, $\\frac{\\frac{9}{4}}{3} = \\frac{3}{4}$ is the middle root.\n\n\nThe other two root... | 1 | ./CreativeMath/AHSME/1967_AHSME_Problems/35.json | AHSME |
1972_AHSME_Problems | 20 | 0 | Other | Multiple Choice | If $\tan x=\dfrac{2ab}{a^2-b^2}$ where $a>b>0$ and $0^\circ <x<90^\circ$, then $\sin x$ is equal to
$\textbf{(A) }\frac{a}{b}\qquad \textbf{(B) }\frac{b}{a}\qquad \textbf{(C) }\frac{\sqrt{a^2-b^2}}{2a}\qquad \textbf{(D) }\frac{\sqrt{a^2-b^2}}{2ab}\qquad \textbf{(E) }\frac{2ab}{a^2+b^2}$
Solution
| [
"We start by letting $\\tan x = \\frac{\\sin x}{\\cos x}$ so that our equation is now: \\[\\frac{\\sin x}{\\cos x} = \\frac{2ab}{a^2-b^2}\\] Multiplying through and rearranging gives us the equation: \\[\\cos x = \\frac{a^2-b^2}{2ab} * \\sin x\\] We now apply the Pythagorean identity $\\sin ^2 x + \\cos ^2 x =1$, u... | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/20.json | AHSME |
1972_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | There are two positive numbers that may be inserted between $3$ and $9$ such that the first three are in geometric progression
while the last three are in arithmetic progression. The sum of those two positive numbers is
$\textbf{(A) }13\textstyle\frac{1}{2}\qquad \textbf{(B) }11\frac{1}{4}\qquad \textbf{(C) }10\frac... | [
"Let $a$ be first and $b$ be second. We can then get equations based on our knowledge: $b-a = 9-b$ and $b/a = a/3$. We then get $b = (9+a)/2$ from our first equation and we substitute that into the second to get $\\frac{9+a}{2a} = a/3$ which simplifies to $2a^2-3a-27=0$ which be $(2a-9)(a+3) = 0$ and $a=9/2$. Then ... | 2 | ./CreativeMath/AHSME/1972_AHSME_Problems/16.json | AHSME |
1972_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | If $3^{2x}+9=10\left(3^{x}\right)$, then the value of $(x^2+1)$ is
$\textbf{(A) }1\text{ only}\qquad \textbf{(B) }5\text{ only}\qquad \textbf{(C) }1\text{ or }5\qquad \textbf{(D) }2\qquad \textbf{(E) }10$
| [
"We can rearrange to get $3^{2x}-10\\left(3^{x}\\right)+9=0$. Factor or use the quadratic formula to find that $3^{x}= 9\\text{ or }1$, which makes $x = 2\\text{ or }0$ and our answer \\[\\boxed{\\textbf{(C) }1\\text{ or }5}.\\]\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/6.json | AHSME |
1972_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | If $yz:zx:xy=1:2:3$, then $\dfrac{x}{yz}:\dfrac{y}{zx}$ is equal to
$\textbf{(A) }3:2\qquad \textbf{(B) }1:2\qquad \textbf{(C) }1:4\qquad \textbf{(D) }2:1\qquad \textbf{(E) }4:1$
| [
"We want to find\n\\[\\dfrac{x}{yz}:\\dfrac{y}{zx} = \\dfrac{x}{yz} \\cdot \\dfrac{zx}{y} = \\dfrac{x^2}{y^2}.\\]\n\n\nWe are given $yz:zx=1:2,$ and dividing both sides by $z$ gives $y:x = 1:2.$ Therefore,\n\\[\\dfrac{x^2}{y^2} = \\dfrac{1}{\\dfrac{1}{2^2}} = 4.\\]\n\n\nThe answer is $\\textbf{(E)}.$\n\n\n-edited b... | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/7.json | AHSME |
1972_AHSME_Problems | 17 | 0 | Probability | Multiple Choice | A piece of string is cut in two at a point selected at random. The probability that the longer piece is at least x times as large as the shorter piece is
$\textbf{(A) }\frac{1}{2}\qquad \textbf{(B) }\frac{2}{x}\qquad \textbf{(C) }\frac{1}{x+1}\qquad \textbf{(D) }\frac{1}{x}\qquad \textbf{(E) }\frac{2}{x+1}$
| [
"The string could be represented by a line plot segment on the interval $(0,x+1)$. The interval $(0,1)$ is a solution as well as $(x,x+1)$. The probability is therefore $\\boxed{\\textbf{(E) }\\frac{2}{x+1}}.$ ~lopkiloinm\n\n\n"
] | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/17.json | AHSME |
1972_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | For $x$ real, the inequality $1\le |x-2|\le 7$ is equivalent to
$\textbf{(A) }x\le 1\text{ or }x\ge 3\qquad \textbf{(B) }1\le x\le 3\qquad \textbf{(C) }-5\le x\le 9\qquad \\ \textbf{(D) }-5\le x\le 1\text{ or }3\le x\le 9\qquad \textbf{(E) }-6\le x\le 1\text{ or }3\le x\le 10$
| [
"We can split the inequality into two smaller inequalities and solve them individually.\n\\[|x-2| \\ge 1 \\quad \\rightarrow \\quad x \\ge 3 \\quad \\text{and} \\quad x \\le 1\\]\n\\[|x-2| \\le 7 \\quad \\rightarrow \\quad x \\le 9 \\quad \\text{and} \\quad x \\ge -5\\]\n\n\nCombining these inequalities, we get $x ... | 1 | ./CreativeMath/AHSME/1972_AHSME_Problems/10.json | AHSME |
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