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 |
|---|---|---|---|---|---|---|---|---|---|
1983_AHSME_Problems | 3 | 0 | Number Theory | Multiple Choice | Three primes $p,q$, and $r$ satisfy $p+q = r$ and $1 < p < q$. Then $p$ equals
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 7\qquad \textbf{(D)}\ 13\qquad \textbf{(E)}\ 17$
| [
"We are given that $p,q$ and $r$ are primes. In order for $p$ and $q$ to sum to another prime, either $p$ or $q$ has to be even, because the sum of two odd numbers would be even, and the only even prime is $2$ (but $p + q = 2$ would have, as the only solution in positive integers, $p = q = 1$, and $1$ is not prime)... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/3.json | AHSME |
1983_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | Let $f(x) = \frac{x+1}{x-1}$. Then for $x^2 \neq 1, f(-x)$ is
$\textbf{(A)}\ \frac{1}{f(x)}\qquad \textbf{(B)}\ -f(x)\qquad \textbf{(C)}\ \frac{1}{f(-x)}\qquad \textbf{(D)}\ -f(-x)\qquad \textbf{(E)}\ f(x)$
| [
"We find $f(-x) = \\frac{-x+1}{-x-1} = \\frac{x-1}{x+1} = \\frac{1}{f(x)}$, so the answer is $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/8.json | AHSME |
1983_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | Consider the two functions $f(x) = x^2+2bx+1$ and $g(x) = 2a(x+b)$, where the variable $x$ and the constants $a$ and $b$ are real numbers.
Each such pair of constants $a$ and $b$ may be considered as a point $(a,b)$ in an $ab$-plane.
Let $S$ be the set of such points $(a,b)$ for which the graphs of $y = f(x)$ and $y ... | [
"We must describe geometrically those $(a,b)$ for which the equation $x^2+2bx+1=2a(x+b)$, i.e. $x^2+2(b-a)x+(1-2ab)=0$, has no solutions (equivalent to the graphs not intersecting). By considering the discriminant of this quadratic equation, there are no solutions if and only if $\\left(2(b-a)\\right)^2 - 4(1)(1-2a... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/22.json | AHSME |
1983_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | Let $f$ be a polynomial function such that, for all real $x$,
$f(x^2 + 1) = x^4 + 5x^2 + 3$.
For all real $x, f(x^2-1)$ is
$\textbf{(A)}\ x^4+5x^2+1\qquad \textbf{(B)}\ x^4+x^2-3\qquad \textbf{(C)}\ x^4-5x^2+1\qquad \textbf{(D)}\ x^4+x^2+3\qquad \textbf{(E)}\ \text{none of these}$
| [
"Let $y = x^2 + 1$. Then $x^2 = y - 1$, so we can write the given equation as \n\\begin{align*}f(y) &= x^4 + 5x^2 + 3 \\\\ &= (x^2)^2 + 5x^2 + 3 \\\\ &= (y - 1)^2 + 5(y - 1) + 3 \\\\ &= y^2 - 2y + 1 + 5y - 5 + 3 \\\\ &= y^2 + 3y - 1.\\end{align*}\nThen substituting $x^2 - 1$ for $y$, we get\n\\begin{align*}f(x^2 - ... | 2 | ./CreativeMath/AHSME/1983_AHSME_Problems/18.json | AHSME |
1983_AHSME_Problems | 14 | 0 | Number Theory | Multiple Choice | The units digit of $3^{1001} 7^{1002} 13^{1003}$ is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ 3\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 9$
| [
"First, we notice that $3^0$ is congruent to $1 \\ \\text{(mod 10)}$, $3^1$ is $3 \\ \\text{(mod 10)}$, $3^2$ is $9 \\ \\text{(mod 10)}$, $3^3$ is $7 \\ \\text{(mod 10)}$, $3^4$ is $1 \\ \\text{(mod 10)}$, and so on. This turns out to be a cycle repeating every $4$ terms, so $3^{1001}$ is congruent to $3 \\ \\text{... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/14.json | AHSME |
1983_AHSME_Problems | 15 | 0 | Probability | Multiple Choice | Three balls marked $1,2$ and $3$ are placed in an urn. One ball is drawn, its number is recorded, and then the ball is returned to the urn. This process is repeated and then repeated once more, and each ball is equally likely to be drawn on each occasion. If the sum of the numbers recorded is $6$, what is the probabili... | [
"Since a ball is drawn three times and the sum is $6$, the only ways this can happen are the permutations of $1, 2, 3$ and $2, 2, 2$. Therefore, with there being $3! + 1 = 7$ equally-likely possibilities in total, the probability that the ball numbered $2$ was drawn all three times is $\\boxed{\\textbf{(C)}\\ \\fra... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/15.json | AHSME |
1983_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | Triangle $ABC$ has a right angle at $C$. If $\sin A = \frac{2}{3}$, then $\tan B$ is
$\textbf{(A)}\ \frac{3}{5}\qquad \textbf{(B)}\ \frac{\sqrt 5}{3}\qquad \textbf{(C)}\ \frac{2}{\sqrt 5}\qquad \textbf{(D)}\ \frac{\sqrt{5}}{2}\qquad \textbf{(E)}\ \frac{5}{3}$
| [
"Since $\\sin$ can be seen as \"opposite over hypotenuse\" in a right triangle, we can label the diagram as shown.\n[asy] pair A,B,C; C = (0,0); B = (2,0); A = (0,1.7); draw(A--B--C--A); draw(rightanglemark(B,C,A,8)); label(\"$A$\",A,W); label(\"$B$\",B,SE); label(\"$C$\",C,SW); label(\"$2x$\",(B+C)/2,S); label(\"$... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/5.json | AHSME |
1983_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | Point $D$ is on side $CB$ of triangle $ABC$. If
$\angle{CAD} = \angle{DAB} = 60^\circ\mbox{, }AC = 3\mbox{ and }AB = 6$,
then the length of $AD$ is
$\textbf{(A)} \ 2 \qquad \textbf{(B)} \ 2.5 \qquad \textbf{(C)} \ 3 \qquad \textbf{(D)} \ 3.5 \qquad \textbf{(E)} \ 4$
| [
"Let $AD = y$. Since $AD$ bisects $\\angle{BAC}$, the Angle Bisector Theorem gives $\\frac{DB}{CD} = \\frac{AB}{AC} = 2$, so let $CD = x$ and $DB = 2x$. Applying the Law of Cosines to $\\triangle CAD$ gives $x^2 = 3^2 + y^2 - 3y$, and to $\\triangle DAB$ gives $(2x)^2 = 6^2 + y^2 - 6y$. Subtracting $4$ times the fi... | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/19.json | AHSME |
1983_AHSME_Problems | 9 | 0 | Arithmetic | Multiple Choice | In a certain population the ratio of the number of women to the number of men is $11$ to $10$.
If the average (arithmetic mean) age of the women is $34$ and the average age of the men is $32$,
then the average age of the population is
$\textbf{(A)}\ 32\frac{9}{10}\qquad \textbf{(B)}\ 32\frac{20}{21}\qquad \textbf{(... | [
"Assume, without loss of generality, that there are exactly $11$ women and $10$ men. Then the total age of the women is $34 \\cdot 11 = 374$ and the total age of the men is $32 \\cdot 10 = 320$. Therefore the overall average is $\\frac{374+320}{11+10} = \\boxed{\\textbf{(D)}\\ 33\\frac{1}{21}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1983_AHSME_Problems/9.json | AHSME |
1968_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | The measures of the interior angles of a convex polygon of $n$ sides are in arithmetic progression. If the common difference is $5^{\circ}$ and the largest angle is $160^{\circ}$, then $n$ equals:
$\text{(A) } 9\quad \text{(B) } 10\quad \text{(C) } 12\quad \text{(D) } 16\quad \text{(E) } 32$
| [
"The formula for the sum of the angles in any polygon is $180(n-2)$. Because this particular polygon is convex and has its angles in an arithmetic sequence with its largest angle being $160$, we can find the sum of the angles.\n\n\n$a_{n}=160$\n\n\n$a_{1}=160-5(n-1)$\n\n\nPlugging this into the formula for finding ... | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/20.json | AHSME |
1968_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | If $x$ is such that $\frac{1}{x}<2$ and $\frac{1}{x}>-3$, then:
$\text{(A) } -\frac{1}{3}<x<\frac{1}{2}\quad \text{(B) } -\frac{1}{2}<x<3\quad \text{(C) } x>\frac{1}{2}\quad\\ \text{(D) } x>\frac{1}{2} \text{ or} -\frac{1}{3}<x<0\quad \text{(E) } x>\frac{1}{2} \text{ or } x<-\frac{1}{3}$
| [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/16.json | AHSME |
1968_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | Let side $AD$ of convex quadrilateral $ABCD$ be extended through $D$, and let side $BC$ be extended through $C$, to meet in point $E.$ Let $S$ be the degree-sum of angles $CDE$ and $DCE$, and let $S'$ represent the degree-sum of angles $BAD$ and $ABC.$ If $r=S/S'$, then:
$\text{(A) } r=1 \text{ sometimes, } r>1 \text... | [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/6.json | AHSME |
1968_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | Let $O$ be the intersection point of medians $AP$ and $CQ$ of triangle $ABC.$ if $OQ$ is 3 inches, then $OP$, in inches, is:
$\text{(A) } 3\quad \text{(B) } \frac{9}{2}\quad \text{(C) } 6\quad \text{(D) } 9\quad \text{(E) } \text{undetermined}$
| [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/7.json | AHSME |
1968_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | Let $f(n)=\frac{x_1+x_2+\cdots +x_n}{n}$, where $n$ is a positive integer. If $x_k=(-1)^k, k=1,2,\cdots ,n$, the set of possible values of $f(n)$ is:
$\text{(A) } \{0\}\quad \text{(B) } \{\frac{1}{n}\}\quad \text{(C) } \{0,-\frac{1}{n}\}\quad \text{(D) } \{0,\frac{1}{n}\}\quad \text{(E) } \{1,\frac{1}{n}\}$
| [
"$\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/17.json | AHSME |
1968_AHSME_Problems | 21 | 0 | Number Theory | Multiple Choice | If $S=1!+2!+3!+\cdots +99!$, then the units' digit in the value of S is:
$\text{(A) } 9\quad \text{(B) } 8\quad \text{(C) } 5\quad \text{(D) } 3\quad \text{(E) } 0$
| [
"Note that every factorial after $5!$ has a unit digit of $0$, meaning that we can disregard them. Thus, we only need to find the units digit of $1! + 2! + 3! + 4!$, and as $1! + 2! + 3! + 4! \\equiv 3$ mod $10$, which means that the unit digit is $3$, we have our answer of $\\fbox{D}$ as desired.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/21.json | AHSME |
1968_AHSME_Problems | 10 | 0 | Other | Multiple Choice | Assume that, for a certain school, it is true that
I: Some students are not honest.
II: All fraternity members are honest.
A necessary conclusion is:
$\text{(A) Some students are fraternity members.} \quad\\ \text{(B) Some fraternity member are not students.} \quad\\ \text{(C) Some students are not fraternity mem... | [
"$\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/10.json | AHSME |
1968_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | Let $S=2+4+6+\cdots +2N$, where $N$ is the smallest positive integer such that $S>1,000,000$. Then the sum of the digits of $N$ is:
$\text{(A) } 27\quad \text{(B) } 12\quad \text{(C) } 6\quad \text{(D) } 2\quad \text{(E) } 1$
| [
"Note that $S = 2(1+2+3+...+N) = N(N +1)$. It follows that $N = 1000$, so the sum of the digits of $N$ is $\\fbox{E}$. \n\n\n\n\nFrostFox\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/26.json | AHSME |
1968_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | Convex polygons $P_1$ and $P_2$ are drawn in the same plane with $n_1$ and $n_2$ sides, respectively, $n_1\le n_2$. If $P_1$ and $P_2$ do not have any line segment in common, then the maximum number of intersections of $P_1$ and $P_2$ is:
$\text{(A) } 2n_1\quad \text{(B) } 2n_2\quad \text{(C) } n_1n_2\quad \text{(D) ... | [
"Notice how $P_2$ can pass through each line segment of $P_1$ at most twice. To have more than two intersections, the line passing through $P_1$ would have a zigzag shape which is impossible for convex polygons. Therefore, the intersections does not depend on $P_2$ and the answer is $\\fbox{A}$\n\n\n",
"Try to ge... | 2 | ./CreativeMath/AHSME/1968_AHSME_Problems/30.json | AHSME |
1968_AHSME_Problems | 31 | 0 | Geometry | Multiple Choice | [asy] draw((0,0)--(10,20*sqrt(3)/2)--(20,0)--cycle,black+linewidth(.75)); draw((20,0)--(20,12)--(32,12)--(32,0)--cycle,black+linewidth(.75)); draw((32,0)--(37,10*sqrt(3)/2)--(42,0)--cycle,black+linewidth(.75)); MP("I",(10,0),N);MP("II",(26,0),N);MP("III",(37,0),N); MP("A",(0,0),S);MP("B",(20,0),S);MP("C",(32,0),S);MP("... | [
"$\\fbox{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/31.json | AHSME |
1968_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | Let $S_n=1-2+3-4+\cdots +(-1)^{n-1}n$, where $n=1,2,\cdots$. Then $S_{17}+S_{33}+S_{50}$ equals:
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } -1\quad \text{(E) } -2$
| [
"If $n$ is even, $S_{n}$ is negative $n/2$. If $n$ is odd, then $S_{n}$ is $(n+1)/2$.\n(These can be found using simple calculations.)\nTherefore, we know $S_{17}+S_{33}+S_{50}$ =$9+17-25$, which is $\\fbox{B}$.\n\n\n\n\nSolution By FrostFox\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/27.json | AHSME |
1968_AHSME_Problems | 1 | 0 | Geometry | Multiple Choice | Let $P$ units be the increase in circumference of a circle resulting from an increase in $\pi$ units in the diameter. Then $P$ equals:
$\text{(A) } \frac{1}{\pi}\quad\text{(B) } \pi\quad\text{(C) } \frac{\pi^2}{2}\quad\text{(D) } \pi^2\quad\text{(E) } 2\pi$
| [
"Let $d$ be the diameter of the original circle. If $d$ is increased by $\\pi$, then the new circumference is $\\pi d + \\pi^2$. The difference in circumference is therefore $\\pi d + \\pi^2 - \\pi d = \\pi^2$\n\n\nTherefore, the answer is $\\fbox{D}$\n\n\nSolution by VivekA\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/1.json | AHSME |
1968_AHSME_Problems | 11 | 0 | Geometry | Multiple Choice | If an arc of $60^{\circ}$ on circle $I$ has the same length as an arc of $45^{\circ}$ on circle $II$, the ratio of the area of circle $I$ to that of circle $II$ is:
$\text{(A) } 16:9\quad \text{(B) } 9:16\quad \text{(C) } 4:3\quad \text{(D) } 3:4\quad \text{(E) } \text{none of these}$
| [
"$\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/11.json | AHSME |
1968_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | The real value of $x$ such that $64^{x-1}$ divided by $4^{x-1}$ equals $256^{2x}$ is:
$\text{(A) } -\frac{2}{3}\quad\text{(B) } -\frac{1}{3}\quad\text{(C) } 0\quad\text{(D) } \frac{1}{4}\quad\text{(E) } \frac{3}{8}$
| [
"$\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/2.json | AHSME |
1968_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | If the arithmetic mean of $a$ and $b$ is double their geometric mean, with $a>b>0$, then a possible value for the ratio $a/b$, to the nearest integer, is:
$\text{(A) } 5\quad \text{(B) } 8\quad \text{(C) } 11\quad \text{(D) } 14\quad \text{(E) none of these}$
| [
"$\\fbox{D}$\n\n\n\n\n$\\frac{a+b}{2}=2\\cdot\\sqrt{ab}$\n\n\n$\\frac{a}{b} +1=4\\cdot\\sqrt{\\frac{a}{b}}$\n\n\nsetting $x=\\sqrt{\\frac{a}{b}}$ we get a quadratic equation is$x^2+1=4x$ with solutions $x=\\frac{4\\pm \\sqrt{16-4}}{2}$\n\n\n$x^2=\\frac{a}{b}=(4+3)+4\\sqrt{3}=13.8=14$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/28.json | AHSME |
1968_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | A circle passes through the vertices of a triangle with side-lengths $7\tfrac{1}{2},10,12\tfrac{1}{2}.$ The radius of the circle is:
$\text{(A) } \frac{15}{4}\quad \text{(B) } 5\quad \text{(C) } \frac{25}{4}\quad \text{(D) } \frac{35}{4}\quad \text{(E) } \frac{15\sqrt{2}}{2}$
| [
"The triangle that goes through all the vertices of the triangle is the circumcircle of the triangle.\n$(7\\frac{1}{2})^{2}+10^{2}=(12\\frac{1}{2})^{2}$, so the triangle is a right triangle.The radius of a circumcircle \nof a right triangle is half the hypotenuse. $\\frac{1}{2}\\cdot \\frac{25}{2}=\\frac{25}{4}\\im... | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/12.json | AHSME |
1968_AHSME_Problems | 32 | 0 | Geometry | Multiple Choice | $A$ and $B$ move uniformly along two straight paths intersecting at right angles in point $O$. When $A$ is at $O$, $B$ is $500$ yards short of $O$. In two minutes they are equidistant from $O$, and in $8$ minutes more they are again equidistant from $O$. Then the ratio of $A$'s speed to $B$'s speed is:
$\text{(A) } 4... | [
"$\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/32.json | AHSME |
1968_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | A painting $18$" X $24$" is to be placed into a wooden frame with the longer dimension vertical. The wood at the top and bottom is twice as wide as the wood on the sides. If the frame area equals that of the painting itself, the ratio of the smaller to the larger dimension of the framed painting is:
$\text{(A) } 1:3\... | [
"Let the width of the frame on the sides to be $x$.\n\n\nThen, the width of the frame on the top and bottom is $2x$.\n\n\nThe area of the frame is then $x\\cdot 2x-18\\cdot24$\n\n\nSetting the area of the frame equal to the area of the picture,\n\\[(2x+18)(4x+24)-18\\cdot24 = 18\\cdot24\\]\nSolving,\n\\[8x^2+120x+4... | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/24.json | AHSME |
1968_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | Ace runs with constant speed and Flash runs $x$ times as fast, $x>1$. Flash gives Ace a head start of $y$ yards, and, at a given signal, they start off in the same direction. Then the number of yards Flash must run to catch Ace is:
$\text{(A) } xy\quad \text{(B) } \frac{y}{x+y}\quad \text{(C) } \frac{xy}{x-1}\quad \t... | [
"$\\fbox{C}$\n\n\n",
"Let $k$ denotes the distance Ace needs to run after the $y$ yard. Since the distance they run with same amount of time is proportional to their speed, we have\n\\[\\frac{1}{x}=\\frac{k}{y+k}\\]\n\\[k=\\frac{y}{x-1}\\]\nThus the total distance ran by Flash is\n\\[y+k=y+\\frac{y}{x-1}=\\frac{x... | 2 | ./CreativeMath/AHSME/1968_AHSME_Problems/25.json | AHSME |
1968_AHSME_Problems | 33 | 0 | Number Theory | Multiple Choice | A number $N$ has three digits when expressed in base $7$. When $N$ is expressed in base $9$ the digits are reversed. Then the middle digit is:
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$
| [
"Call the number $\\overline{abc}$ in base 7.\n\n\nThen, $49a+7b+c=81c+9b+a$. (Breaking down the number in base-form).\n\n\nAfter combining like terms and moving the variables around,\n$48a=2b+80c$,$b=40c-24a=8(5c-2a)$. This shows that $b$ is a multiple of 8 (we only have to find the middle digit under \\textit{one... | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/33.json | AHSME |
1968_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | If $m$ and $n$ are the roots of $x^2+mx+n=0 ,m \ne 0,n \ne 0$, then the sum of the roots is:
$\text{(A) } -\frac{1}{2}\quad \text{(B) } -1\quad \text{(C) } \frac{1}{2}\quad \text{(D) } 1\quad \text{(E) } \text{undetermined}$
| [
"By Vieta's Theorem, $mn = n$ and $-(m + n) = m$. Dividing the first equation by $n$ gives $m = 1$. Multiplying the 2nd by -1 gives $m + n = -m$. The RHS is -1, so the answer is $\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/13.json | AHSME |
1968_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | Given the three numbers $x,y=x^x,z=x^{x^x}$ with $.9<x<1.0$. Arranged in order of increasing magnitude, they are:
$\text{(A) } x,z,y\quad \text{(B) } x,y,z\quad \text{(C) } y,x,z\quad \text{(D) } y,z,x\quad \text{(E) } z,x,y$
| [
"$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/29.json | AHSME |
1968_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | A straight line passing through the point $(0,4)$ is perpendicular to the line $x-3y-7=0$. Its equation is:
$\text{(A) } y+3x-4=0\quad \text{(B) } y+3x+4=0\quad \text{(C) } y-3x-4=0\quad \\ \text{(D) } 3y+x-12=0\quad \text{(E) } 3y-x-12=0$
| [
"$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/3.json | AHSME |
1968_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | With $400$ members voting the House of Representatives defeated a bill. A re-vote, with the same members voting, resulted in the passage of the bill by twice the margin by which it was originally defeated. The number voting for the bill on the revote was $\frac{12}{11}$ of the number voting against it originally. How m... | [
"$\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/34.json | AHSME |
1968_AHSME_Problems | 8 | 0 | Arithmetic | Multiple Choice | A positive number is mistakenly divided by $6$ instead of being multiplied by $6.$ Based on the correct answer, the error thus committed, to the nearest percent, is :
$\text{(A) } 100\quad \text{(B) } 97\quad \text{(C) } 83\quad \text{(D) } 17\quad \text{(E) } 3$
| [
"$\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/8.json | AHSME |
1968_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | A segment of length $1$ is divided into four segments. Then there exists a quadrilateral with the four segments as sides if and only if each segment is:
$\text{(A) equal to } \frac{1}{4}\quad\\ \text{(B) equal to or greater than } \frac{1}{8} \text{ and less than }\frac{1}{2}\quad\\ \text{(C) greater than } \frac{1}{... | [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/22.json | AHSME |
1968_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | Side $AB$ of triangle $ABC$ has length 8 inches. Line $DEF$ is drawn parallel to $AB$ so that $D$ is on segment $AC$, and $E$ is on segment $BC$. Line $AE$ extended bisects angle $FEC$. If $DE$ has length $5$ inches, then the length of $CE$, in inches, is:
$\text{(A) } \frac{51}{4}\quad \text{(B) } 13\quad \text{(C) ... | [
"Draw a line passing through $A$ and parallel to $BC$. Let $\\angle FEC = 2n$. By alternate-interior-angles or whatever, $\\angle BAE = n$, so $BAE$ is an isosceles triangle, and it follows that $BE = 8$. $\\triangle ABC \\sim \\triangle DEC$. Let $CE = x$. We have\n\\[\\frac{8}{8+x} = \\frac{5}{x} \\Rightarrow 40+... | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/18.json | AHSME |
1968_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | Define an operation $\star$ for positive real numbers as $a\star b=\frac{ab}{a+b}$. Then $4 \star (4 \star 4)$ equals:
$\text{(A) } \frac{3}{4}\quad \text{(B) } 1\quad \text{(C) } \frac{4}{3}\quad \text{(D) } 2\quad \text{(E )} \frac{16}{3}$
| [
"$4 \\star 4 \\star 4 = 4 \\star 2 = \\frac{4}{3}$\n\n\nTherefore, our answer is $\\fbox{C}$\n\n\nSolution by VivekA\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/4.json | AHSME |
1968_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | If $x$ and $y$ are non-zero numbers such that $x=1+\frac{1}{y}$ and $y=1+\frac{1}{x}$, then $y$ equals
$\text{(A) } x-1\quad \text{(B) } 1-x\quad \text{(C) } 1+x\quad \text{(D) } -x\quad \text{(E) } x$
| [
"We see after multiplying the first equation by $y$, that \n\n\n$xy=y+1.$\n\n\nSimilarly, we see that after multiplying the second equation by $x$, we get that \n\n\n$xy=x+1.$\n\n\nThus $x+1=y+1 \\implies x=y$, giving us our final answer of $\\fbox{E}.$\n\n\n~SirAppel\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/14.json | AHSME |
1968_AHSME_Problems | 15 | 0 | Number Theory | Multiple Choice | Let $P$ be the product of any three consecutive positive odd integers. The largest integer dividing all such $P$ is:
$\text{(A) } 15\quad \text{(B) } 6\quad \text{(C) } 5\quad \text{(D) } 3\quad \text{(E) } 1$
| [
"Product $P$ can be written as $2n-1$,$2n+1$,$2n+3$. Because $P$ is defined as a \"3 consecutive odd integer\" product impies that $P$ must be divisible by at least 3. A is ruled out because factors of 5 only arise every 5 terms, if we were to take the 3 terms in the middle of the factors of 5 we wouldn't have a fa... | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/15.json | AHSME |
1968_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | If $f(n)=\tfrac{1}{3} n(n+1)(n+2)$, then $f(r)-f(r-1)$ equals:
$\text{(A) } r(r+1)\quad \text{(B) } (r+1)(r+2)\quad \text{(C) } \tfrac{1}{3} r(r+1)\quad \\ \text{(D) } \tfrac{1}{3} (r+1)(r+2)\quad \text{(E )} \tfrac{1}{3} r(r+1)(2r+1)$
| [
"$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/5.json | AHSME |
1968_AHSME_Problems | 19 | 0 | Counting | Multiple Choice | Let $n$ be the number of ways $10$ dollars can be changed into dimes and quarters, with at least one of each coin being used. Then $n$ equals:
$\text{(A) } 40\quad \text{(B) } 38\quad \text{(C) } 21\quad \text{(D) } 20\quad \text{(E) } 19$
| [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/19.json | AHSME |
1968_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | If all the logarithms are real numbers, the equality
$log(x+3)+log(x-1)=log(x^2-2x-3)$
is satisfied for:
$\text{(A) all real values of }x \quad\\ \text{(B) no real values of } x\quad\\ \text{(C) all real values of } x \text{ except } x=0\quad\\ \text{(D) no real values of } x \text{ except } x=0\quad\\ \text{(E) all ... | [
"$\\fbox{B}$\n\n\n",
"From the given we have\n\\[\\log(x+3)+\\log(x-1)=\\log(x^2-2x-3)\\]\n\\[\\log(x^2+2x-3)=\\log(x^2-2x-3)\\]\n\\[x^2+2x-3=x^2-2x-3\\]\n\\[x=0\\]\nHowever substituing into $\\log(x-1)$ gets a negative argument, which is impossible $\\boxed{D}$.\n\n\n~ Nafer\n\n\n"
] | 2 | ./CreativeMath/AHSME/1968_AHSME_Problems/23.json | AHSME |
1968_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | The sum of the real values of $x$ satisfying the equality $|x+2|=2|x-2|$ is:
$\text{(A) } \frac{1}{3}\quad \text{(B) } \frac{2}{3}\quad \text{(C) } 6\quad \text{(D) } 6\tfrac{1}{3}\quad \text{(E) } 6\tfrac{2}{3}$
| [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/9.json | AHSME |
1968_AHSME_Problems | 35 | 0 | Geometry | Multiple Choice | [asy] draw(arc((0,0),10, 0, 180),black+linewidth(.75)); draw((-10,0)--(10,0),black+linewidth(.75)); draw((-sqrt(96),2)--(sqrt(96),2),black+linewidth(.75)); draw((-8,6)--(8,6),black+linewidth(.75)); draw((0,0)--(0,10),black+linewidth(.75)); draw((-8,6)--(-8,2),black+linewidth(.75)); draw((8,6)--(8,2),black+linewidth(.75... | [
"Let $OG = a - 2h$, where $h = JH = HG$. Since the areas of rectangle $EHGL$ and trapezoid $EHGC$ are both half of rectangle $LMFE$ and trapezoid $EFDC$, respectively, the ratios between their areas will remain the same, so let us consider rectangle $EHGL$ and trapezoid $EHGC$.\n\n\nDraw radii $OC$ and $OE$, both o... | 1 | ./CreativeMath/AHSME/1968_AHSME_Problems/35.json | AHSME |
1982_AHSME_Problems | 20 | 0 | Number Theory | Multiple Choice | The number of pairs of positive integers $(x,y)$ which satisfy the equation $x^2+y^2=x^3$ is
$\text {(A)} 0 \qquad \text {(B)} 1 \qquad \text {(C)} 2 \qquad \text {(D)} \text{not finite} \qquad \text {(E)} \text{none of these}$
| [
"Rearrange the equation to $y^2=x^3-x^2=x^2(x-1)$. This equation is satisfied whenever $x-1$ is a perfect square. There are infinite possible values of $x$, and thus the answer is $\\boxed{D: \\text{Not Finite}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/20.json | AHSME |
1982_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | A wooden cube has edges of length $3$ meters. Square holes, of side one meter, centered in each face are cut through to the opposite face. The edges of the holes are parallel to the edges of the cube. The entire surface area including the inside, in square meters, is
$\text {(A)} 54 \qquad \text {(B)} 72 \qquad \te... | [
"Each exterior unit square which is removed exposes 4 faces of the unit interior squares, so the entire surface area in square meters is $6 \\cdot 3^2 - 6 + 24=72.$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/16.json | AHSME |
1982_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | The sum of all but one of the interior angles of a convex polygon equals $2570^\circ$. The remaining angle is
$\text{(A)} \ 90^\circ \qquad \text{(B)} \ 105^\circ \qquad \text{(C)} \ 120^\circ \qquad \text{(D)}\ 130^\circ\qquad \text{(E)}\ 144^\circ$
| [
"Note that the sum of the interior angles of a convex polygon of $n$ sides is $180(n-2)^\\circ$, and each interior angle belongs to $[0, 180^\\circ)$. Therefore, we must have $n - 2 = \\lfloor \\frac{2570^\\circ}{180^\\circ} \\rfloor = 15$. Then the missing angle must be $180*15^\\circ - 2570^\\circ = 130^\\circ$, ... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/6.json | AHSME |
1982_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | If the operation $x \star y$ is defined by $x \star y = (x+1)(y+1) - 1$, then which one of the following is FALSE?
$\text{(A)} \ x \star y = y\star x \text{ for all real } x,y. \\ \text{(B)} \ x \star (y + z) = ( x \star y ) + (x \star z) \text{ for all real } x,y, \text{ and } z.\\ \text{(C)} \ (x-1) \star (x+1) =... | [
"(A) is true because multiplication is commutative.\n\n\n(B) is false because we have $x\\star (y + z) = (x+1)(y+z+1) - 1$ and \n$x\\star y + x\\star z = (x+1)(y+1) - 1 + (x+1)(z+1) - 1$\n$(x+1)(y+z+2) - 2$\n$(x+1)(y+z+1) - (x+1) - 2$\n$[(x+1)(y+z+1) - 1] + x,$\nwhich does not match with the previous expression.\n\... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/7.json | AHSME |
1982_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | How many real numbers $x$ satisfy the equation $3^{2x+2}-3^{x+3}-3^{x}+3=0$?
$\text {(A)} 0 \qquad \text {(B)} 1 \qquad \text {(C)} 2 \qquad \text {(D)} 3 \qquad \text {(E)} 4$
| [
"Let $a = 3^x$. Then the preceding equation can be expressed as the quadratic, \\[9a^2-28a+3 = 0\\] Solving the quadratic yields the roots $3$ and $1/9$. Setting these equal to $3^x$, we can immediately see that there are $\\boxed{2}$ real values of $x$ that satisfy the equation.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/17.json | AHSME |
1982_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | In the adjoining figure, the triangle $ABC$ is a right triangle with $\angle BCA=90^\circ$. Median $CM$ is perpendicular to median $BN$,
and side $BC=s$. The length of $BN$ is
[asy] size(200); defaultpen(linewidth(0.7)+fontsize(10));real r=54.72; pair B=origin, C=dir(r), A=intersectionpoint(B--(9,0), C--C+4*dir(r-90... | [
"Suppose that $P$ is the intersection of $\\overline{CM}$ and $\\overline{BN}.$ Let $BN=x.$ By the properties of centroids, we have $BP=\\frac23 x.$\n\n\nNote that $\\triangle BPC\\sim\\triangle BCN$ by AA. From the ratio of similitude $\\frac{BP}{BC}=\\frac{BC}{BN},$ we get\n\\begin{align*} BP\\cdot BN &= BC^2 \\\... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/21.json | AHSME |
1982_AHSME_Problems | 10 | 0 | Other | Multiple Choice | Since $BO$ and $CO$ are angle bisectors of angles $B$ and $C$ respectively, $\angle MBO = \angle OBC$ and similarly $\angle NCO = \angle OCB$. Because $MN$ and $BC$ are parallel, $\angle OBC = \angle MOB$ and $\angle NOC = \angle OCB$ by corresponding angles. This relation makes $\triangle MOB$ and $\triangle NOC$ isos... | [
"Since $BO$ and $CO$ are angle bisectors of angles $B$ and $C$ respectively, $\\angle MBO = \\angle OBC$ and similarly $\\angle NCO = \\angle OCB$. Because $MN$ and $BC$ are parallel, $\\angle OBC = \\angle MOB$ and $\\angle NOC = \\angle OCB$ by corresponding angles. This relation makes $\\triangle MOB$ and $\\tri... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/10.json | AHSME |
1982_AHSME_Problems | 26 | 0 | Number Theory | Multiple Choice | If the base $8$ representation of a perfect square is $ab3c$, where $a\ne 0$, then $c$ equals
$\text{(A)} 0\qquad \text{(B)}1 \qquad \text{(C)} 3\qquad \text{(D)} 4\qquad \text{(E)} \text{not uniquely determined}$
| [
"A perfect square will be $(8k+r)^2=64k^2+16kr+r^2\\equiv r^2\\pmod{16}$ where $r=0,1,...,7$.\n\n\nNotice that $r^2\\equiv 1,4,9,0 \\pmod{16}$.\n\n\nNow $ab3c$ in base 8 is $a8^3+b8^2+3(8)+c\\equiv 8+c\\pmod{16}$. It being a perfect square means $8+c\\equiv 1,4,9,0 \\pmod{16}$. That means that c can only be 1 so th... | 2 | ./CreativeMath/AHSME/1982_AHSME_Problems/26.json | AHSME |
1982_AHSME_Problems | 30 | 0 | Algebra | Multiple Choice | Find the units digit of the decimal expansion of \[\left(15 + \sqrt{220}\right)^{19} + \left(15 + \sqrt{220}\right)^{82}.\]
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ \text{none of these}$
| [
"Let $A=15+\\sqrt{220}$ and $B=15-\\sqrt{220}.$ Note that $A^{19}+B^{19}$ and $A^{82}+B^{82}$ are both integers: When we expand (Binomial Theorem) and combine like terms for each expression, the rational terms are added and the irrational terms are canceled.\n\n\nWe have\n\\begin{align*} A^{19}+B^{19} &= \\left[\\b... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/30.json | AHSME |
1982_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | Suppose $z=a+bi$ is a solution of the polynomial equation \[c_4z^4+ic_3z^3+c_2z^2+ic_1z+c_0=0,\] where $c_0, c_1, c_2, c_3, a,$ and $b$ are real constants and $i^2=-1.$ Which of the following must also be a solution?
$\textbf{(A)}\ -a-bi\qquad \textbf{(B)}\ a-bi\qquad \textbf{(C)}\ -a+bi\qquad \textbf{(D)}\ b+ai ... | [
"Let $z=wi,$ so the given polynomial equation becomes\n\\begin{align*} c_4(wi)^4+ic_3(wi)^3+c_2(wi)^2+ic_1(wi)+c_0&=0 \\\\ c_4w^4+c_3w^3-c_2w^2-c_1w+c_0&=0. &&(\\bigstar) \\end{align*}\nNote that $(\\bigstar)$ is a polynomial equation in $w$ with real coefficients.\n\n\nWe are given that $w=\\frac{a+bi}{i}=b-ai$ is... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/27.json | AHSME |
1982_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | When the polynomial $x^3-2$ is divided by the polynomial $x^2-2$, the remainder is
$\text{(A)} \ 2 \qquad \text{(B)} \ -2 \qquad \text{(C)} \ -2x-2 \qquad \text{(D)} \ 2x+2 \qquad \text{(E)} \ 2x-2$
| [
"Working out $\\frac{x^3-2}{x^2-2}$ using polynomial long division, we get $x + \\frac{2x-2}{x^2-2}$. \nThus the answer is $2x -2$, for choice $\\boxed{(E)}$.\n\n\n",
"Substituting an easy value, like $3$, you get $\\frac{25}{7}$, with a remainder of 4. Then you just plug in x to get $E$\n\n\n-dragoon\n\n\n"
] | 2 | ./CreativeMath/AHSME/1982_AHSME_Problems/1.json | AHSME |
1982_AHSME_Problems | 11 | 0 | Other | Multiple Choice | All of the digits of the numbers must be 0 to 9. If the first and last digits are x and y, we have $x-y=2$ and $0<x<9$ or $y-x=2,$ and $0<y<9.$ Substituting we have $0<x+2<9,$ and $0<y+2<9.$ Thus $0<x<7$ and $0<y<7,$ which yields 16 pairs (x, y) such that the absolute value of the difference between the x and y is $2.$... | [
"All of the digits of the numbers must be 0 to 9. If the first and last digits are x and y, we have $x-y=2$ and $0<x<9$ or $y-x=2,$ and $0<y<9.$ Substituting we have $0<x+2<9,$ and $0<y+2<9.$ Thus $0<x<7$ and $0<y<7,$ which yields 16 pairs (x, y) such that the absolute value of the difference between the x and y is... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/11.json | AHSME |
1982_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | If a number eight times as large as $x$ is increased by two, then one fourth of the result equals
$\text{(A)} \ 2x + \frac{1}{2} \qquad \text{(B)} \ x + \frac{1}{2} \qquad \text{(C)} \ 2x+2 \qquad \text{(D)}\ 2x+4 \qquad \text{(E)}\ 2x+16$
| [
"The number $8$ times as large as $x$ will be $8x$, and $8x$ increased by two will give $8x+2$. Hence finally, the answer is $\\frac{1}{4}(8x+2) = \\boxed{\\text{(A)}\\ 2x + \\frac{1}{2}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/2.json | AHSME |
1982_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | A set of consecutive positive integers beginning with $1$ is written on a blackboard. One number is erased. The average (arithmetic mean) of the remaining numbers is $35\frac{7}{17}$. What number was erased?
$\textbf{(A)}\ 6\qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ 8 \qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\... | [
"Suppose that there are $n$ positive integers in the set initially, so their sum is $\\frac{n(n+1)}{2}$ by arithmetic series. The average of the remaining numbers is minimized when $n$ is erased, and is maximized when $1$ is erased.\n\n\nIt is clear that $n>1.$ We write and solve a compound inequality for $n:$\n\\b... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/28.json | AHSME |
1982_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | Let $f(x) = ax^7+bx^3+cx-5$, where $a,b$ and $c$ are constants. If $f(-7) = 7$, then $f(7)$ equals
$\text {(A)} -17 \qquad \text {(B)} -7 \qquad \text {(C)} 14 \qquad \text {(D)} 21\qquad \text {(E)} \text{not uniquely determined}$
| [
"$f(x)$ is an odd function shifted down 5 units. Thus, it can be written as $f(x)=g(x)-5$ where $g(x)=ax^7+bx^3+cx$. Thus: $f(-7)=g(-7)-5=7$ and $g(-7)=12$. Using this and the fact $g(x)$ is odd, we can evaluate $f(7)$, which is:\n\n\n\\[f(7) = g(7)-5 = -g(-7)-5 = -12-5 = -17\\]\n\n\nTherefore, the answer is $\\box... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/12.json | AHSME |
1982_AHSME_Problems | 25 | 0 | Probability | Multiple Choice | The adjacent map is part of a city: the small rectangles are blocks, and the paths in between are streets.
Each morning, a student walks from intersection $A$ to intersection $B$, always walking along streets shown,
and always going east or south. For variety, at each intersection where he has a choice, he chooses wi... | [
"[asy] defaultpen(linewidth(0.7)+fontsize(8)); size(250); path p=origin--(5,0)--(5,3)--(0,3)--cycle; path q=(5,19)--(6,19)--(6,20)--(5,20)--cycle; int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<6; j=j+1) { draw(shift(6*i, 4*j)*p); }} clip((4,2)--(25,2)--(25,21)--(4,21)--cycle); fill(q^^shift(18,-16)*q^^shift(18,-12)*q,... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/25.json | AHSME |
1982_AHSME_Problems | 29 | 0 | Algebra | Multiple Choice | Let $x,y,$ and $z$ be three positive real numbers whose sum is $1.$ If no one of these numbers is more than twice any other,
then the minimum possible value of the product $xyz$ is
$\textbf{(A)}\ \frac{1}{32}\qquad \textbf{(B)}\ \frac{1}{36}\qquad \textbf{(C)}\ \frac{4}{125}\qquad \textbf{(D)}\ \frac{1}{127}\qquad ... | [
"Suppose that the product $xyz$ is minimized at $(x,y,z)=(x_0,y_0,z_0).$ Without the loss of generality, let $x_0 \\leq y_0 \\leq z_0$ and fix $y=y_0.$\n\n\nTo minimize $xy_0z,$ we minimize $xz.$ Note that $x+z=1-y_0.$ By a corollary of the AM-GM Inequality \\textit{\\textbf{(If two nonnegative numbers have a const... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/29.json | AHSME |
1982_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | Evaluate $(x^x)^{(x^x)}$ at $x = 2$.
$\text{(A)} \ 16 \qquad \text{(B)} \ 64 \qquad \text{(C)} \ 256 \qquad \text{(D)} \ 1024 \qquad \text{(E)} \ 65,536$
| [
"Plugging in $2$ as $x$ gives $4^4$, which is merely $\\boxed{\\text{(C)} \\ 256 }$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/3.json | AHSME |
1982_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | In a narrow alley of width $w$ a ladder of length $a$ is placed with its foot at point $P$ between the walls.
Resting against one wall at $Q$, the distance $k$ above the ground makes a $45^\circ$ angle with the ground.
Resting against the other wall at $R$, a distance $h$ above the ground, the ladder makes a $75^\cir... | [
"[asy] import olympiad; size(200); pair T,P,Q,M,L,R; T=(0,0); P=(2.5,0); Q=(10,7.5); M=(0,7.5); L=(10,0); R=(0,10); draw(R--T--L--Q); draw(P--Q--R--cycle); draw(Q--M); label(\"$P$\", P, S); dot(P); label(\"$Q$\", Q, E); dot(Q); label(\"$R$\", R, W); dot(R); label(\"$S$\", M, W); dot... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/22.json | AHSME |
1982_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | In the adjoining figure of a rectangular solid, $\angle DHG=45^\circ$ and $\angle FHB=60^\circ$. Find the cosine of $\angle BHD$.
[asy] import three;defaultpen(linewidth(0.7)+fontsize(10)); currentprojection=orthographic(1/3+1/10,1-1/10,1/3); real r=sqrt(3); triple A=(0,0,r), B=(0,r,r), C=(1,r,r), D=(1,0,r), E=O, F=(... | [
"WLOG, let $CD=1$. \n\n\nLooking at square GHDC, we see that $\\angle DHC=45$, which implies that $DC=CH=1$ and $DH=\\sqrt{2}$\n\n\nTaking each cross-section one at a time, we look at square DHFB. We obviously know that CHB is a $30$ degree angle, giving $BH=\\frac{2\\sqrt{3}}{3}$, and $BC=\\frac{\\sqrt{3}}{3}$.\n\... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/18.json | AHSME |
1982_AHSME_Problems | 14 | 0 | Geometry | Multiple Choice | In the adjoining figure, points $B$ and $C$ lie on line segment $AD$, and $AB, BC$, and $CD$ are diameters of circle $O, N$, and $P$, respectively. Circles $O, N$, and $P$ all have radius $15$ and the line $AG$ is tangent to circle $P$ at $G$. If $AG$ intersects circle $N$ at points $E$ and $F$, then chord $EF$ has le... | [
"Drop a perpendicular line from $N$ to $AG$ at point $H$. $AN=45$, and since $\\triangle{AGP}$ is similar to $\\triangle{AHN}$. $NH=9$. $NE=NF=15$ so by the Pythagorean Theorem, $EH=HF=12$. Thus $EF=\\boxed{24.}$ Answer is then $\\boxed{C}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/14.json | AHSME |
1982_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | Let $[z]$ denote the greatest integer not exceeding $z$. Let $x$ and $y$ satisfy the simultaneous equations
\begin{align*} y&=2[x]+3 \\ y&=3[x-2]+5. \end{align*}
If $x$ is not an integer, then $x+y$ is
$\text {(A) } \text{ an integer} \qquad \text {(B) } \text{ between 4 and 5} \qquad \text{(C) }\text{ betwee... | [
"We simply ignore the floor of $x$. Then, we have $y$ = $2x + 3$ = $3(x-2)+5$. Solving for $3x - 1 = 2x + 3$, we get $x = 4$. For the floor of $x$, we have $x$ is between $4$ and $5$. Plugging in $8$ + $3$ = $11$ for $y$, we have $y = 11$. We have $11 + 4.x$ = $\\boxed {(D)}$\n\n\n~Arcticturn\n\n\n",
"Since $x$ i... | 2 | ./CreativeMath/AHSME/1982_AHSME_Problems/15.json | AHSME |
1982_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | Two positive numbers $x$ and $y$ are in the ratio $a: b$ where $0 < a < b$. If $x+y = c$, then the smaller of $x$ and $y$ is
$\text{(A)} \ \frac{ac}{b} \qquad \text{(B)} \ \frac{bc-ac}{b} \qquad \text{(C)} \ \frac{ac}{a+b} \qquad \text{(D)}\ \frac{bc}{a+b}\qquad \text{(E)}\ \frac{ac}{b-a}$
| [
"We can write 2 equations.\n\n\n$\\frac{x}{y}=\\frac{a}{b}$\n\n\nand\n\n\n$x+y=c$\n\n\nSolving for $x$ and $y$ in terms of $a, b, c$ we get :\n\n\n$x=\\frac{ac}{a+b}$ and $y=\\frac{bc}{a+b}$\n\n\nSince we know $a$ is less than $b$ and $\\frac{x}{y}=\\frac{bc}{a+b}$, the smaller of $x$ and $y$ must be $x$. Therefore... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/5.json | AHSME |
1982_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | Let $f(x)=|x-2|+|x-4|-|2x-6|$ for $2 \leq x\leq 8$. The sum of the largest and smallest values of $f(x)$ is
$\textbf {(A)}\ 1 \qquad \textbf {(B)}\ 2 \qquad \textbf {(C)}\ 4 \qquad \textbf {(D)}\ 6 \qquad \textbf {(E)}\ \text{none of these}$
| [
"Note that at $x=2,3,4,$ one of the three absolute values is equal to $0.$\n\n\nWithout using absolute values, we rewrite $f(x)$ as a piecewise function:\n\\[f(x) = \\begin{cases} (x-2)+(4-x)-(6-2x) & \\mathrm{if} \\ 2\\leq x<3 \\\\ (x-2)+(4-x)-(2x-6) & \\mathrm{if} \\ 3\\leq x<4 \\\\ (x-2)+(x-4)-(2x-6) & \\mathrm{... | 1 | ./CreativeMath/AHSME/1982_AHSME_Problems/19.json | AHSME |
1982_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | The lengths of the sides of a triangle are consecutive integers, and the largest angle is twice the smallest angle.
The cosine of the smallest angle is
$\textbf{(A)}\ \frac{3}{4}\qquad \textbf{(B)}\ \frac{7}{10}\qquad \textbf{(C)}\ \frac{2}{3}\qquad \textbf{(D)}\ \frac{9}{14}\qquad \textbf{(E)}\ \text{none of these... | [
"In $\\triangle ABC,$ let $a=n,b=n+1,c=n+2,$ and $\\angle A=\\theta$ for some positive integer $n.$ We are given that $\\angle C=2\\theta,$ and we need $\\cos\\theta.$\n\n\nWe apply the Law of Cosines to solve for $\\cos\\theta:$ \\[\\cos\\theta=\\frac{b^2+c^2-a^2}{2bc}=\\frac{n+5}{2(n+2)}.\\]\nWe apply the Law of ... | 2 | ./CreativeMath/AHSME/1982_AHSME_Problems/23.json | AHSME |
1997_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | Which one of the following integers can be expressed as the sum of $100$ consecutive positive integers?
$\textbf{(A)}\ 1,\!627,\!384,\!950\qquad\textbf{(B)}\ 2,\!345,\!678,\!910\qquad\textbf{(C)}\ 3,\!579,\!111,\!300\qquad\textbf{(D)}\ 4,\!692,\!581,\!470\qquad\textbf{(E)}\ 5,\!815,\!937,\!260$
| [
"Notice how the sum of 100 consecutive integers is $(x-49)+(x-48)+(x-47)...+x+...(x+47)+(x+48)+(x+49)+(x+50)$. \n\n\nCancelling out the constants give us $100x + 50$. \n\n\nLooking over at the list of possible values, we quickly realise that the only possible solution is $\\boxed{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/20.json | AHSME |
1997_AHSME_Problems | 16 | 0 | Counting | Multiple Choice | The three row sums and the three column sums of the array
\[\left[\begin{matrix}4 & 9 & 2\\ 8 & 1 & 6\\ 3 & 5 & 7\end{matrix}\right]\]
are the same. What is the least number of entries that must be altered to make all six sums different from one another?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3... | [
"If you change $3$ numbers, then you either change one number in each column and row (ie sudoku-style):\n\n\n\\[\\left[\\begin{matrix}* & 9 & 2\\\\ 8 & * & 6\\\\ 3 & 5 & *\\end{matrix}\\right]\\]\n\n\nOr you leave at least one row and one column unchanged:\n\n\n\\[\\left[\\begin{matrix}* & 9 & 2\\\\ * & * & 6\\\\ 3... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/16.json | AHSME |
1997_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | Consider the sequence
$1,-2,3,-4,5,-6,\ldots,$
whose $n$th term is $(-1)^{n+1}\cdot n$. What is the average of the first $200$ terms of the sequence?
$\textbf{(A)}-\!1\qquad\textbf{(B)}-\!0.5\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 0.5\qquad\textbf{(E)}\ 1$
| [
"The average of a list is the sum of all numbers divided by the size of the list.\n\n\nThe sum of the list can be found by adding the numbers in pairs: $(1 + -2) + (3 + -4) + ... + (199 + -200)$\n\n\nThe sum of each pair is $-1$, and there are $100$ pairs, so the total sum is $-100$.\n\n\nThere are $200$ numbers o... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/6.json | AHSME |
1997_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | The sum of seven integers is $-1$. What is the maximum number of the seven integers that can be larger than $13$?
$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 6\qquad\textbf{(E)}\ 7$
| [
"If the first six integers are $14$, the last number can be $(-14\\cdot 6) - 1 = -85$. The sum of all seven integers will be $-1$.\n\n\nHowever, if all seven integers are over $13$, the smallest possible sum is $14\\cdot 7 = 98$.\n\n\nThus, the answer is $6$, which is option $\\boxed{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/7.json | AHSME |
1997_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | A line $x=k$ intersects the graph of $y=\log_5 x$ and the graph of $y=\log_5 (x + 4)$. The distance between the points of intersection is $0.5$. Given that $k = a + \sqrt{b}$, where $a$ and $b$ are integers, what is $a+b$?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E... | [
"Since the line $x=k$ is vertical, we are only concerned with vertical distance.\n\n\nIn other words, we want to find the value of $k$ for which the distance $|\\log_5 x - \\log_5 (x+4)| = \\frac{1}{2}$\n\n\nSince $\\log_5 x$ is a strictly increasing function, we have:\n\n\n$\\log_5 (x + 4) - \\log_5 x = \\frac{1}{... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/17.json | AHSME |
1997_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | For any positive integer $n$, let
$f(n) =\left\{\begin{matrix}\log_{8}{n}, &\text{if }\log_{8}{n}\text{ is rational,}\\ 0, &\text{otherwise.}\end{matrix}\right.$
What is $\sum_{n = 1}^{1997}{f(n)}$?
$\textbf{(A)}\ \log_{8}{2047}\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ \frac{55}{3}\qquad\textbf{(D)}\ \frac{58}{3}... | [
"For positive integers $n$, $\\log_8 n$ is rational if and only if $n = 2^k$ for an integer $k$. That's because $\\log_8 2^k = k\\log_8 2 = \\frac{k}{3}$.\n\n\nSo we actually want to find $\\sum_{k=0}^{10} \\log_8 2^k$, since $2^{11}$ will be over $1997$.\n\n\nUsing log properties, we get $\\sum_{k=0}^{10} k \\log... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/21.json | AHSME |
1997_AHSME_Problems | 10 | 0 | Probability | Multiple Choice | Two six-sided dice are fair in the sense that each face is equally likely to turn up. However, one of the dice has the $4$ replaced by $3$ and the other die has the $3$ replaced by $4$ . When these dice are rolled, what is the probability that the sum is an odd number?
$\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \f... | [
"On the first die, the chance of an odd number is $\\frac{4}{6} = \\frac{2}{3}$, and the chance of an even number is $\\frac{1}{3}$.\n\n\nOn the second die, the chance of an odd number is $\\frac{1}{3}$, and the chance of an even number is $\\frac{2}{3}$.\n\n\nTo get an odd sum, we need exactly one even and one odd... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/10.json | AHSME |
1997_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$, angle $APB$ is twice angle $ACB$, and $\overline{AC}$ intersects $\overline{BP}$ at point $D$. If $PB = 3$ and $PD= 2$, then $AD\cdot CD =$
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,... | [
"The product of two lengths with a common point brings to mind the Power of a Point Theorem.\n\n\nSince $PA = PB$, we can make a circle with radius $PA$ that is centered on $P$, and both $A$ and $B$ will be on that circle. Since $\\angle APB = \\widehat {AB} = 2 \\angle ACB$, we can see that point $C$ will also li... | 2 | ./CreativeMath/AHSME/1997_AHSME_Problems/26.json | AHSME |
1997_AHSME_Problems | 30 | 0 | Counting | Multiple Choice | For positive integers $n$, denote $D(n)$ by the number of pairs of different adjacent digits in the binary (base two) representation of $n$. For example, $D(3) = D(11_{2}) = 0$, $D(21) = D(10101_{2}) = 4$, and $D(97) = D(1100001_{2}) = 2$. For how many positive integers less than or equal $97$ to does $D(n) = 2$?
$\t... | [
"If $D(n)$ is even, then the binary expansion of $n$ will both begin and end with a $1$, because all positive binary numbers begin with a $1$, and if you switch digits twice, you will have a $1$ at the end. Thus, we are only concerned with the $49$ odd numbers between $1$ and $98$ inclusive.\n\n\nAll of these odd ... | 3 | ./CreativeMath/AHSME/1997_AHSME_Problems/30.json | AHSME |
1997_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | Consider those functions $f$ that satisfy $f(x+4)+f(x-4) = f(x)$ for all real $x$. Any such function is periodic, and there is a least common positive period $p$ for all of them. Find $p$.
$\textbf{(A)}\ 8\qquad\textbf{(B)}\ 12\qquad\textbf{(C)}\ 16\qquad\textbf{(D)}\ 24\qquad\textbf{(E)}\ 32$
| [
"Recall that $p$ is the fundamental period of function $f$ iff $p$ is the smallest positive $p$ such that $f(x) = f(x + p)$ for all $x$.\n\n\n\n\nIn this case, we know that $f(x+ 4) + f(x - 4) = f(x)$. Plugging in $x+4$ in for $x$ to get the next equation in the recursion, we also get $f(x + 8) + f(x) = f(x + 4)$.... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/27.json | AHSME |
1997_AHSME_Problems | 1 | 0 | Arithmetic | Multiple Choice | If $\texttt{a}$ and $\texttt{b}$ are digits for which
$\begin{array}{ccc}& 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{array}$
then $\texttt{a+b =}$
$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 9 \qquad \mathrm{(E) \ }12$
| [
"From the units digit calculation, we see that the units digit of $a\\times 3$ is $9$. Since $0 \\le a \\le 9$ and $a$ is an integer, the only value of $a$ that works is is $a=3$. As a double-check, that does work, since $23 \\times 3 = 69$, which is the first line of the multiplication.\n\n\nThe second line of t... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/1.json | AHSME |
1997_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | In the sixth, seventh, eighth, and ninth basketball games of the season, a player scored $23$,$14$, $11$, and $20$ points, respectively. Her points-per-game average was higher after nine games than it was after the first five games. If her average after ten games was greater than $18$, what is the least number of point... | [
"The sum of the scores for games $6$ through $9$ is $68$. The average in these four games is $\\frac{68}{4} = 17$.\n\n\nThe total points in all ten games is greater than $10\\cdot 18 = 180$. Thus, it must be at least $181$.\n\n\nThere are at least $181 - 68 = 113$ points in the other six games: games $1-5$ and g... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/11.json | AHSME |
1997_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | The adjacent sides of the decagon shown meet at right angles. What is its perimeter?
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; dot(origin);dot((12,0));dot((12,1));dot((9,1));dot((9,7));dot((7,7));dot((7,10));dot((3,10));dot((3,8));dot((0,8)); draw(origin--(12,0)--(12,1)--(9,1)--(9,7)--(7,7)--(7,10)--(3,10)--(3... | [
"The three unlabelled vertical sides have the same sum as the two labelled vertical sides, which is $10$.\n\n\nThe four unlabelled horizontal sides have the same sum as the one large horizontal side, which is $12$.\n\n\nThus, the perimeter is $2(12+10) = 44$, which is option $\\boxed{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/2.json | AHSME |
1997_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | How many ordered triples of integers $(a,b,c)$ satisfy $|a+b|+c = 19$ and $ab+|c| = 97$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$
| [
"WLOG, let $a \\ge 0$, and let $a \\ge b$. We can say this because if we have one solution $(a,b) = (a_0, b_0)$ with $a_0 \\ge 0$ and $a_0 > b_0$, we really have the four solutions $(a_0, b_0), (-a_0, -b_0), (b_0, a_0), (-b_0, -a_0)$ by the symmetry of the original problem. \n\n\nFurthermore, we assert that these... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/28.json | AHSME |
1997_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | If $m$ and $b$ are real numbers and $mb>0$, then the line whose equation is $y=mx+b$ \textit{cannot} contain the point
$\textbf{(A)}\ (0,1997)\qquad\textbf{(B)}\ (0,-1997)\qquad\textbf{(C)}\ (19,97)\qquad\textbf{(D)}\ (19,-97)\qquad\textbf{(E)}\ (1997,0)$
| [
"Geometrically, $b$ is the y-intercept, and $m$ is the slope.\n\n\nSince $mb>0$, then either we have a positive y-intercept and a positive slope, or a negative y-intercept and a negative slope.\n\n\nLines with a positive y-intercept and positive slope can go through quadrants I, II, and III. They cannot go through... | 3 | ./CreativeMath/AHSME/1997_AHSME_Problems/12.json | AHSME |
1997_AHSME_Problems | 24 | 0 | Counting | Multiple Choice | A rising number, such as $34689$, is a positive integer each digit of which is larger than each of the digits to its left. There are $\binom{9}{5} = 126$ five-digit rising numbers. When these numbers are arranged from smallest to largest, the $97^{\text{th}}$ number in the list does not contain the digit
$\textbf{(A)... | [
"The list starts with $12345$. There are $\\binom{8}{4} = 70$ four-digit rising numbers that do not begin with $1$, and thus also $70$ five digit rising numbers that do begin with $1$ that are formed by simply putting a $1$ before the four digit number.\n\n\nThus, the $71^{\\text{st}}$ number is $23456$. There ar... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/24.json | AHSME |
1997_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | Let $ABCD$ be a parallelogram and let $\overrightarrow{AA^\prime}$, $\overrightarrow{BB^\prime}$, $\overrightarrow{CC^\prime}$, and $\overrightarrow{DD^\prime}$ be parallel rays in space on the same side of the plane determined by $ABCD$. If $AA^{\prime} = 10$, $BB^{\prime}= 8$, $CC^\prime = 18$, and $DD^\prime = 22$ a... | [
"Let $ABCD$ be a unit square with $A(0,0,0)$, $B(0,1,0)$, $C(1,1,0)$, and $D(1,0,0)$. Assume that the rays go in the +z direction. In this case, $A^\\prime(0,0,10)$, $B^\\prime(0,1,8)$, $C^\\prime(1,1,18)$, and $D^\\prime(1,0,22)$. Finding the midpoints of $A^\\prime C^\\prime$ and $B^\\prime D^\\prime$ gives $... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/25.json | AHSME |
1997_AHSME_Problems | 13 | 0 | Number Theory | Multiple Choice | How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of is a perfect square?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$
| [
"Let $N = 10t + u$, where $t$ is the tens digit and $u$ is the units digit.\n\n\nThe condition of the problem is that $10t + u + 10u + t$ is a perfect square.\n\n\nSimplifying and factoring, we want $11(t+u)$ to be a perfect square.\n\n\nThus, $t+u$ must at least be a multiple of $11$, and since $t$ and $u$ are dig... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/13.json | AHSME |
1997_AHSME_Problems | 29 | 0 | Number Theory | Multiple Choice | Call a positive real number special if it has a decimal representation that consists entirely of digits $0$ and $7$. For example, $\frac{700}{99}= 7.\overline{07}= 7.070707\cdots$ and $77.007$ are special numbers. What is the smallest $n$ such that $1$ can be written as a sum of $n$ special numbers?
$\textbf{(A)}\ 7\... | [
"Define a super-special number to be a number whose decimal expansion only consists of $0$'s and $1$'s. The problem is equivalent to finding the number of super-special numbers necessary to add up to $\\frac{1}{7}=0.142857142857\\hdots$. This can be done in $8$ numbers if we take\n\\[0.111111\\hdots, 0.011111\\hdot... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/29.json | AHSME |
1997_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | If $x$, $y$, and $z$ are real numbers such that
\begin{center}
$(x-3)^2 + (y-4)^2 + (z-5)^2 = 0$,\end{center}
then $x + y + z =$
$\mathrm{(A)\ } -12 \qquad \mathrm{(B) \ }0 \qquad \mathrm{(C) \ } 8 \qquad \mathrm{(D) \ } 12 \qquad \mathrm{(E) \ }50$
| [
"If the sum of three squared expressions is zero, then each expression itself must be zero, since $a^2 \\ge 0$ with the equality iff $a=0$.\n\n\nIn this case, $x-3=0$, $y-4=0$, and $z-5=0$. Adding the three equations and moving the constant to the right gives $x + y + z = 12$, and the answer is $\\boxed{D}$.\n\n\n... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/3.json | AHSME |
1997_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | Mientka Publishing Company prices its bestseller Where's Walter? as follows:
$C(n) =\left\{\begin{matrix}12n, &\text{if }1\le n\le 24\\ 11n, &\text{if }25\le n\le 48\\ 10n, &\text{if }49\le n\end{matrix}\right.$
where $n$ is the number of books ordered, and $C(n)$ is the cost in dollars of $n$ books. Notice that $... | [
"Clearly, the areas of concern are where the piecewise function shifts value. \n\n\nSince $C(25) = 11\\cdot 25 = 275$, we want to find the least value of $n$ for which $C(n) > 275$.\n\n\nIf $n \\le 24$, then $C(n) = 12n$, so for $C(n) > 275$, $12n > 275$, which is equivalent to $n > 22.91$. Thus, both $n=23$ and ... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/8.json | AHSME |
1997_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | Ashley, Betty, Carlos, Dick, and Elgin went shopping. Each had a whole number of dollars to spend, and together they had $56$ dollars. The absolute difference between the amounts Ashley and Betty had to spend was $19$ dollars. The absolute difference between the amounts Betty and Carlos had was $7$ dollars, between Car... | [
"Working backwards, if $6 \\le E \\le 10$, then $6 \\pm 11 \\le A \\le 10 \\pm 11$. Since $A$ is a positive integer, $17 \\le A \\le 21$. \n\n\n\n\nSince $17 \\le A \\le 21$, we know that $17 \\pm 19 \\le B \\le 21 \\pm 19$. But if $B=36$, which is the smallest possible \"plus\" value, then $E + A + B = 6 + 17 + ... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/22.json | AHSME |
1997_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | A list of integers has mode $32$ and mean $22$. The smallest number in the list is $10$. The median $m$ of the list is a member of the list. If the list member $m$ were replaced by $m+10$, the mean and median of the new list would be $24$ and $m+10$, respectively. If were $m$ instead replaced by $m-8$, the median of th... | [
"Let there be $n$ integers on the list. The list of $n$ integers has mean $22$, so the sum of the integers is $22n$.\n\n\nReplacing $m$ with $m+10$ will increase the sum of the list from $22n$ to $22n + 10$.\n\n\nThe new mean of the list is $24$, so the new sum of the list is also $24n$.\n\n\nThus, we get $22n + 1... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/18.json | AHSME |
1997_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | If $a$ is $50\%$ larger than $c$, and $b$ is $25\%$ larger than $c$, then $a$ is what percent larger than $b$?
$\mathrm{(A)\ } 20\% \qquad \mathrm{(B) \ }25\% \qquad \mathrm{(C) \ } 50\% \qquad \mathrm{(D) \ } 100\% \qquad \mathrm{(E) \ }200\%$
| [
"Translating each sentence into an equation, $a = 1.5c$ and $b = 1.25c$.\n\n\nWe want a relationship between $a$ and $b$. Dividing the second equation into the first will cancel the $c$, so we try that and get:\n\n\n$\\frac{a}{b} = \\frac{1.5}{1.25}$\n\n\n$\\frac{a}{b} = \\frac{150}{125}$\n\n\n$\\frac{a}{b} = \\fr... | 2 | ./CreativeMath/AHSME/1997_AHSME_Problems/4.json | AHSME |
1997_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | The number of geese in a flock increases so that the difference between the populations in year $n+2$ and year $n$ is directly proportional to the population in year $n+1$. If the populations in the years $1994$, $1995$, and $1997$ were $39$, $60$, and $123$, respectively, then the population in $1996$ was
$\textbf{(... | [
"Let $x$ be the population in $1996$, and let $k$ be the constant of proportionality.\n\n\nIf $n=1994$, then the difference in population between $1996$ and $1994$ is directly proportional to the population in $1995$.\n\n\nTranslating this sentence, $(x - 39) = k(60)$\n\n\nSimilarly, letting $n=1995$ gives the sent... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/14.json | AHSME |
1997_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | Medians $BD$ and $CE$ of triangle $ABC$ are perpendicular, $BD=8$, and $CE=12$. The area of triangle $ABC$ is
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot... | [
"[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair F = midpoint(B--C); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G);dot(F); label(\"$A$\",A,S);label(\"$B$\",B,N);label(\"$C$\"... | 2 | ./CreativeMath/AHSME/1997_AHSME_Problems/15.json | AHSME |
1997_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | A rectangle with perimeter $176$ is divided into five congruent rectangles as shown in the diagram. What is the perimeter of one of the five congruent rectangles?
[asy] defaultpen(linewidth(.8pt)); draw(origin--(0,3)--(4,3)--(4,0)--cycle); draw((0,1)--(4,1)); draw((2,0)--midpoint((0,1)--(4,1))); real r = 4/3; draw((r,... | [
"Let $l$ represent the length of one of the smaller rectangles, and let $w$ represent the width of one of the smaller rectangles, with $w < l$.\n\n\nFrom the large rectangle, we see that the top has length $3w$, the right has length $l + w$, the bottom has length $2l$, and the left has length $l + 2$.\n\n\nSince th... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/5.json | AHSME |
1997_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | A circle with center $O$ is tangent to the coordinate axes and to the hypotenuse of the $30^\circ$-$60^\circ$-$90^\circ$ triangle $ABC$ as shown, where $AB=1$. To the nearest hundredth, what is the radius of the circle?
[asy] defaultpen(linewidth(.8pt)); dotfactor=3; pair A = origin; pair B = (1,0); pair C = (0,sqr... | [
"[asy] defaultpen(linewidth(.8pt)); dotfactor=3; pair A = origin; pair B = (1,0); pair C = (0,sqrt(3)); pair O = (2.33,2.33); pair D = (2.33,0); pair E = (0, 2.33); pair F = (0.35, 1.1); dot(A);dot(B);dot(C);dot(O);dot(D);dot(E);dot(F); label(\"$A$\",A,SW);label(\"$B$\",B,SE);label(\"$C$\",C,W);label(\"$O$\",O,NW);... | 2 | ./CreativeMath/AHSME/1997_AHSME_Problems/19.json | AHSME |
1997_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | [asy] size(6cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); draw((-1,1)--(2,1)); draw((-1,0)--(1,0)); draw((-1,1)--(-1,0)); draw((0,-1)--(0,3)); draw((1,2)--(1,0)); draw((-1,1)--(1,1)); draw((0,2)--(1,2)); draw((0,3)--(1,2)); draw((0,-1)--(2,1)); draw((0,-1)--((0,-1) + sqrt(2)*dir(-15))); draw(((0,-1) + sqrt(2)*dir(-1... | [
"Since there are three squares, the figure may appear to be based upon a unit cube.\n\n\nSince there is an equilateral triangle of side length $\\sqrt{2}$, this triangle has the length of a facial (non-spatial) diagonal of a unit cube.\n\n\nSince there are three isosceles right triangles that are half the area of a... | 1 | ./CreativeMath/AHSME/1997_AHSME_Problems/23.json | AHSME |
1997_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | In the figure, $ABCD$ is a $2 \times 2$ square, $E$ is the midpoint of $\overline{AD}$, and $F$ is on $\overline{BE}$. If $\overline{CF}$ is perpendicular to $\overline{BE}$, then the area of quadrilateral $CDEF$ is
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = (0,2); pair B = origin; pair C = (2,0); pair... | [
"Since $\\angle EBA = \\angle FCB$ and $\\angle FBC = \\angle AEB$, we have $\\triangle ABE \\sim \\triangle FCB$.\n\n\n$\\frac{AB}{FC} = \\frac{BE}{CB} = \\frac{EA}{BF}$\n\n\n$\\frac{2}{FC} = \\frac{\\sqrt{5}}{2} = \\frac{1}{BF}$\n\n\nFrom those two equations, we find that $CF = \\frac{4}{\\sqrt{5}}$ and $BF = \\f... | 2 | ./CreativeMath/AHSME/1997_AHSME_Problems/9.json | AHSME |
1975_AHSME_Problems | 20 | 0 | Geometry | Multiple Choice | In the adjoining figure triangle $ABC$ is such that $AB = 4$ and $AC = 8$. IF $M$ is the midpoint of $BC$ and $AM = 3$, what is the length of $BC$?
[asy] draw((-4,0)--(4,0)--(-1,4)--cycle); draw((-1, 4)--(0, 0.00001)); label("B", (-4,0), S); label("C", (4,0), S); label("A", (-1, 4), N); label("M", (0, 0.0001), S); [/... | [
"Let $BM=CM=x$. Then, by Stewart's Theorem, we have\n\\[2x^3+18x=16x+64x\\]\n\\[\\implies x^2+9=40\\]\n\\[\\implies x=\\sqrt{31}\\implies BC=\\boxed{2\\sqrt{31}}.\\]\nThe answer is $\\boxed{B}.$\n-brainiacmaniac31\n\n\n"
] | 1 | ./CreativeMath/AHSME/1975_AHSME_Problems/20.json | AHSME |
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