competition_id
string | problem_id
int64 | difficulty
int64 | category
string | problem_type
string | problem
string | solutions
sequence | solutions_count
int64 | source_file
string | competition
string |
|---|---|---|---|---|---|---|---|---|---|
1959_AHSME_Problems | 48 | 0 | Algebra | Multiple Choice | Given the polynomial $a_0x^n+a_1x^{n-1}+\cdots+a_{n-1}x+a_n$, where $n$ is a positive integer or zero, and $a_0$ is a positive integer. The remaining $a$'s are integers or zero. Set $h=n+a_0+|a_1|+|a_2|+\cdots+|a_n|$. [See example 25 for the meaning of $|x|$.] The number of polynomials with $h=3$ is: $\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 9$
| [
"We perform casework by the value of $n$, the degree of our polynomial $a_0x^n+a_1x^{n-1}+\\cdots+a_{n-1}x+a_n$.\n\n\nCase $n = 0$: In this case we are forced to set $a_0 = 3$. This contributes $1$ possibility.\n\n\nCase $n = 1$: In this case we must have $a_0 + |a_1| = 2$, so our polynomial could be $1 + 1x, 0 + 2x, -1 + 1x$. This contributes $3$ possibilities.\n\n\nCase $n = 2$: In this case we must have $a_0 + |a_1| + |a_2| = 1$. However, because $a_0$ must be positive, it has to be $1$, so our polynomial can only be $0 + 0x + 1x^2$. This contributes $1$ possibility.\n\n\nCase $n\\geq 3$: This case is impossible because $h = n+a_0+|a_1|+|a_2|+\\cdots+|a_n|\\geq n + a_0\\geq 3 + 1 = 4 > 3$, so it contributes $0$ possibilities.\n\n\nAdding the results from all four cases, we find that there are $1 + 3 + 1 + 0 = 5$ possibilities in total, so our answer is $\\boxed{(B)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/48.json | AHSME |
1959_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | The symbol $|a|$ means $+a$ if $a$ is greater than or equal to zero, and $-a$ if a is less than or equal to zero; the symbol $<$ means "less than";
the symbol $>$ means "greater than."
The set of values $x$ satisfying the inequality $|3-x|<4$ consists of all $x$ such that:
$\textbf{(A)}\ x^2<49 \qquad\textbf{(B)}\ x^2>1 \qquad\textbf{(C)}\ 1<x^2<49\qquad\textbf{(D)}\ -1<x<7\qquad\textbf{(E)}\ -7<x<1$
| [
"The equation $|3-x| < 4$ can be solved by splitting it into two inequalities: $3-x<4$ and $3-x<-4$. The solutions to those inequalities are $x<-1$ and $x>7$, respectively. The common interval of those two inequalities is $\\textbf{(D)}\\ -1<x<7$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/25.json | AHSME |
1959_AHSME_Problems | 13 | 0 | Arithmetic | Multiple Choice | The arithmetic mean (average) of a set of $50$ numbers is $38$. If two numbers, namely, $45$ and $55$, are discarded, the mean of the remaining set of numbers is: $\textbf{(A)}\ 36.5 \qquad\textbf{(B)}\ 37\qquad\textbf{(C)}\ 37.2\qquad\textbf{(D)}\ 37.5\qquad\textbf{(E)}\ 37.52$
| [
"Since the arithmetic mean of the $50$ numbers is $38$, their sum must be $50*38 = 1900$. After $45$ and $55$ are discarded, the sum decreases by $45 + 55 = 100$, so it must become $1900 - 100 = 1800$. \nBut this means that the new mean of the remaining $50 - 2 = 48$ numbers must be $\\frac{1800}{48} = 37.5$. Thusly, our answer is $\\boxed{\\textbf{(D)}}$, and we are done.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/13.json | AHSME |
1959_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | If the diagonals of a quadrilateral are perpendicular to each other, the figure would always be included under the general classification:
$\textbf{(A)}\ \text{rhombus} \qquad\textbf{(B)}\ \text{rectangles} \qquad\textbf{(C)}\ \text{square} \qquad\textbf{(D)}\ \text{isosceles trapezoid}\qquad\textbf{(E)}\ \text{none of these}$
| [
"Note that a kite has perpendicular diagonals, but is not a subcategory of any of the existing answer choices. Therefore, the answer is $\\boxed{\\textbf{E}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/3.json | AHSME |
1959_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | Let the roots of $x^2-3x+1=0$ be $r$ and $s$. Then the expression $r^2+s^2$ is: $\textbf{(A)}\ \text{a positive integer} \qquad\textbf{(B)}\ \text{a positive fraction greater than 1}\qquad\textbf{(C)}\ \text{a positive fraction less than 1}\qquad\textbf{(D)}\ \text{an irrational number}\qquad\textbf{(E)}\ \text{an imaginary number}$
| [
"You may recognize that $r^2+s^2$ can be written as $(r+s)^2-2rs$. Then, by Vieta's formulas, \\[r+s=-(-3)=3\\]and \\[rs=1.\\]Therefore, plugging in the values for $r+s$ and $rs$, \\[(r+s)^2-2rs=(3)^2-2(1)=9-2=7.\\]Hence, we can say that the expression $r^2+s^2$ is $\\boxed{\\text{(A)}\\ \\text{a positive integer.}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/34.json | AHSME |
1959_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | The line joining the midpoints of the diagonals of a trapezoid has length $3$. If the longer base is $97,$ then the shorter base is: $\textbf{(A)}\ 94 \qquad\textbf{(B)}\ 92\qquad\textbf{(C)}\ 91\qquad\textbf{(D)}\ 90\qquad\textbf{(E)}\ 89$
| [
"Let x be the length of the shorter base. \n3 = (97 - x)/2 \n\n\n6 = 97 - x \n\n\nx = 91 \n\n\nThus, 91.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/22.json | AHSME |
1959_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | The arithmetic mean (average) of the first $n$ positive integers is:
$\textbf{(A)}\ \frac{n}{2} \qquad\textbf{(B)}\ \frac{n^2}{2}\qquad\textbf{(C)}\ n\qquad\textbf{(D)}\ \frac{n-1}{2}\qquad\textbf{(E)}\ \frac{n+1}{2}$
| [
"The sum of the first $n$ positive integers is the same as the nth triangular number, which can be expressed as $\\frac{(n)(n+1)}{2}$. Since the question is asking for the arithmetic mean, the whole sum is divided by $n$, thus giving us $\\frac {\\frac{(n)(n+1)}{2}}{n}$, which then simplifies to $\\textbf{(E)} \\frac{n+1}{2}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/18.json | AHSME |
1959_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | If $78$ is divided into three parts which are proportional to $1, \frac13, \frac16,$ the middle part is:
$\textbf{(A)}\ 9\frac13 \qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 17\frac13 \qquad\textbf{(D)}\ 18\frac13\qquad\textbf{(E)}\ 26$
| [
"Let the part proportional to $1$ equal $x$. Then, the parts are $x, \\frac{1}{3}x$, and $\\frac{1}{6}x$. The sum of these parts should be $78$, so $\\frac{9}{6}x=1\\frac{1}{2}x=78$. Solving for $x$, $x=52$. The middle part is $\\frac{1}{3}x$, so the answer is $17\\frac{1}{3}$, or $\\boxed{\\textbf{C}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/4.json | AHSME |
1959_AHSME_Problems | 14 | 0 | Number Theory | Multiple Choice | Given the set $S$ whose elements are zero and the even integers, positive and negative. Of the five operations applied to any pair of elements: (1) addition (2) subtraction (3) multiplication (4) division (5) finding the arithmetic mean (average), those elements that only yield elements of $S$ are: $\textbf{(A)}\ \text{all} \qquad\textbf{(B)}\ 1,2,3,4\qquad\textbf{(C)}\ 1,2,3,5\qquad\textbf{(D)}\ 1,2,3\qquad\textbf{(E)}\ 1,3,5$
| [
"The first three listed operations, applied to even integers, all yield even integers trivially. As for the fourth operation (division) and fifth operation (average of pair), we can find pairs of even numbers that are mapped to odd integers, such as $\\frac{2}{2} = 1$, and $\\frac{0+2}{2} = 1$. Therefore, the operations that map even integers to even integers only are operations $1, 2, 3$, so our answer is $\\boxed{\\textbf{(D)}}$ and we are done.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/14.json | AHSME |
1959_AHSME_Problems | 42 | 0 | Number Theory | Multiple Choice | Given three positive integers $a,b,$ and $c$. Their greatest common divisor is $D$; their least common multiple is $m$.
Then, which two of the following statements are true?
$\text{(1)}\ \text{the product MD cannot be less than abc} \qquad \\ \text{(2)}\ \text{the product MD cannot be greater than abc}\qquad \\ \text{(3)}\ \text{MD equals abc if and only if a,b,c are each prime}\qquad \\ \text{(4)}\ \text{MD equals abc if and only if a,b,c are each relatively prime in pairs} \text{ (This means: no two have a common factor greater than 1.)}$
$\textbf{(A)}\ 1,2 \qquad\textbf{(B)}\ 1,3\qquad\textbf{(C)}\ 1,4\qquad\textbf{(D)}\ 2,3\qquad\textbf{(E)}\ 2,4$
| [
"Because $1\\times2\\times4>1\\times4$, 1 is false. Because $1\\times1\\times1=1\\times1$, 3 is false. It follows that the answer is $\\boxed{\\textbf{E}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/42.json | AHSME |
1959_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | In a right triangle the square of the hypotenuse is equal to twice the product of the legs. One of the acute angles of the triangle is: $\textbf{(A)}\ 15^{\circ} \qquad\textbf{(B)}\ 30^{\circ} \qquad\textbf{(C)}\ 45^{\circ} \qquad\textbf{(D)}\ 60^{\circ}\qquad\textbf{(E)}\ 75^{\circ}$
| [
"WLOG, by scaling, that the hypotenuse has length 1. Let $\\theta$ be an angle opposite from some leg. Then the two legs have length $\\sin\\theta$ and $\\cos\\theta$ respectively, so we have $2\\sin\\theta\\cos\\theta = 1^2$. From trigonometry, we know that this equation is true when $\\theta = 45^{\\circ}$, so our answer is $\\boxed{\\textbf{(C)}}$ and we are done.\n\n\n",
"Look at the options. Note that if $\\textbf{(A)}$ is the correct answer, one of the acute angles of the triangle will measure to $15$ degrees. This implies that the other acute angle of the triangle would measure to be $75$ degrees, which would imply that $\\textbf{(E)}$ is another correct answer. However, there is only one correct answer per question, so $\\textbf{(A)}$ can't be a correct answer. Using a similar argument, neither $\\textbf{(B)}$, $\\textbf{(D)}$ , nor $\\textbf{(E)}$ can be a correct answer. Since $\\textbf{(C)}$ is the only answer choice left and there must be one correct answer, the answer must be $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 2 | ./CreativeMath/AHSME/1959_AHSME_Problems/15.json | AHSME |
1959_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | The value of $\left(256\right)^{.16}\left(256\right)^{.09}$ is:
$\textbf{(A)}\ 4 \qquad \\ \textbf{(B)}\ 16\qquad \\ \textbf{(C)}\ 64\qquad \\ \textbf{(D)}\ 256.25\qquad \\ \textbf{(E)}\ -16$
| [
"When we multiply numbers with exponents, we add the exponents together and leave the bases unchanged. We can apply this concept to computate $256^{0.16} \\cdot 256^{0.09}$: \n\\[256^{0.16} \\cdot 256^{0.09} = 256^{0.16+0.09}=256^{0.25}.\\]\nNow we can convert the decimal exponent to a fraction: \\[256^{0.25} = 256^{\\frac{1}{4}}.\\]\nNow, let us convert the expression into radical form. Since $4$ is the denominator of the fractional exponent, it will be the index exponent: \\[256^{\\frac{1}{4}}=\\sqrt[4]{256}.\\]\nSince $256 =16^2=(4^2)^2=4^4$, we can solve for the fourth root of $256$: \\[\\sqrt[4]{256}=\\sqrt[4]{4^4}=4.\\]\nTherefore, $(256)^{.16} \\cdot (256)^{.09}=\\boxed{\\textbf{(A) }\\ 4}.$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/5.json | AHSME |
1959_AHSME_Problems | 39 | 0 | Algebra | Multiple Choice | Let S be the sum of the first nine terms of the sequence $x+a, x^2+2a, x^3+3a, \cdots.$
Then S equals:
$\textbf{(A)}\ \frac{50a+x+x^8}{x+1} \qquad\textbf{(B)}\ 50a-\frac{x+x^{10}}{x-1}\qquad\textbf{(C)}\ \frac{x^9-1}{x+1}+45a\qquad\textbf{(D)}\ \frac{x^{10}-x}{x-1}+45a\qquad\textbf{(E)}\ \frac{x^{11}-x}{x-1}+45a$
| [
"We know $x+x^2\\dots+x^9=x(1+x\\dots+x^8)=x\\frac{x^9-1}{x-1}=\\frac{x^{10}-x}{x-1}$. The only answer with this term is $\\boxed{\\textbf{D}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/39.json | AHSME |
1959_AHSME_Problems | 19 | 0 | Counting | Multiple Choice | With the use of three different weights, namely $1$ lb., $3$ lb., and $9$ lb., how many objects of different weights can be weighed, if the objects is to be weighed and the given weights may be placed in either pan of the scale? $\textbf{(A)}\ 15 \qquad\textbf{(B)}\ 13\qquad\textbf{(C)}\ 11\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 7$
| [
"The heaviest object that could be weighed with this set weighs $1 + 3 + 9 = 13$ lb., and we can weigh any positive integer weight at most that. This means that $13$ different objects could be weighed, so our answer is $\\boxed{\\textbf{(B)}}$ and we are done.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/19.json | AHSME |
1959_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | The set of solutions of the equation $\log_{10}\left( a^2-15a\right)=2$ consists of
$\textbf{(A)}\ \text{two integers } \qquad\textbf{(B)}\ \text{one integer and one fraction}\qquad \textbf{(C)}\ \text{two irrational numbers }\qquad\textbf{(D)}\ \text{two non-real numbers} \qquad\textbf{(E)}\ \text{no numbers, that is, the empty set}$
| [
"Understand that $\\log_{10}\\left(x\\right)=y$ can be expressed as $x = y^2$.\n\n\nBy applying that same logic to $\\log_{10}\\left( a^2-15a\\right)=2$, we get $a^2-15a = 10^2$\n\n\nWe can use factoring methods to bring us to $(a-20)(a+5)=0$, which, as each binomial produces one integer solution, gives us a set of $\\textbf{(A)}$ two integers in the solution set.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/23.json | AHSME |
1959_AHSME_Problems | 9 | 0 | Arithmetic | Multiple Choice | A farmer divides his herd of $n$cows among his four sons so that one son gets one-half the herd, a second son, one-fourth, a third son, one-fifth, and the fourth son, $7$ cows. Then $n$ is: $\textbf{(A)}\ 80 \qquad\textbf{(B)}\ 100\qquad\textbf{(C)}\ 140\qquad\textbf{(D)}\ 180\qquad\textbf{(E)}\ 240$
| [
"The first three sons get $\\frac{1}{2}+\\frac{1}{4}+\\frac{1}{5}=\\frac{19}{20}$ of the herd, so that the fourth son should get $\\frac{1}{20}$ of it. But the fourth son gets $7$ cows, so the size of the herd is $n=\\frac{7}{\\frac{1}{20}} = 140$. Then our answer is $\\boxed{C}$, and we are done.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1959_AHSME_Problems/9.json | AHSME |
1951_AHSME_Problems | 20 | 0 | Algebra | Multiple Choice | When simplified and expressed with negative exponents, the expression $(x + y)^{ - 1}(x^{ - 1} + y^{ - 1})$ is equal to:
$\textbf{(A)}\ x^{ - 2} + 2x^{ - 1}y^{ - 1} + y^{ - 2} \qquad\textbf{(B)}\ x^{ - 2} + 2^{ - 1}x^{ - 1}y^{ - 1} + y^{ - 2} \qquad\textbf{(C)}\ x^{ - 1}y^{ - 1}$
$\textbf{(D)}\ x^{ - 2} + y^{ - 2} \qquad\textbf{(E)}\ \frac {1}{x^{ - 1}y^{ - 1}}$
| [
"Note that $(x + y)^{-1}(x^{-1} + y^{-1}) = \\dfrac{1}{x + y}\\cdot\\left(\\dfrac{1}{x} + \\dfrac{1}{y}\\right) = \\dfrac{1}{x + y}\\cdot\\dfrac{x + y}{xy} = \\dfrac{1}{xy} = x^{-1}y^{-1}$. The answer is $\\textbf{(C)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/20.json | AHSME |
1951_AHSME_Problems | 36 | 0 | Geometry | Multiple Choice | Which of the following methods of proving a geometric figure a locus is not correct?
$\textbf{(A)}\ \text{Every point of the locus satisfies the conditions and every point not on the locus does}\\ \text{not satisfy the conditions.}$
$\textbf{(B)}\ \text{Every point not satisfying the conditions is not on the locus and every point on the locus}\\ \text{does satisfy the conditions.}$
$\textbf{(C)}\ \text{Every point satisfying the conditions is on the locus and every point on the locus satisfies}\\ \text{the conditions.}$
$\textbf{(D)}\ \text{Every point not on the locus does not satisfy the conditions and every point not satisfying}\\ \text{the conditions is not on the locus.}$
$\textbf{(E)}\ \text{Every point satisfying the conditions is on the locus and every point not satisfying the} \\ \text{conditions is not on the locus.}$
| [
"Statement $\\boxed{\\textbf{(B)}}$ is wrong because it does not imply that all points that satisfy the conditions are on the locus.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/36.json | AHSME |
1951_AHSME_Problems | 41 | 0 | Algebra | Multiple Choice | The formula expressing the relationship between $x$ and $y$ in the table is:
\[\begin{tabular}{|c|c|c|c|c|c|}\hline x & 2 & 3 & 4 & 5 & 6\\ \hline y & 0 & 2 & 6 & 12 & 20\\ \hline\end{tabular}\]
$\textbf{(A)}\ y = 2x-4\qquad\textbf{(B)}\ y = x^{2}-3x+2\qquad\textbf{(C)}\ y = x^{3}-3x^{2}+2x$
$\textbf{(D)}\ y = x^{2}-4x\qquad\textbf{(E)}\ y = x^{2}-4$
| [
"Just plug the $x,y$ pair $(6,20)$ into each of the 5 answer choices:\n\n\n(A): $2(6)-4=8\\ne20$\n\n\n(B): $6^2-3(6)+2=20$\n\n\n(C): $6^3-3(6^2)+2(6)=120\\ne20$\n\n\n(D): $6^2-4(6)=12\\ne20$\n\n\n(E): $6^2-4=32\\ne20$\n\n\nThe only one that works is $\\boxed{\\textbf{(B)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/41.json | AHSME |
1951_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | If in applying the quadratic formula to a quadratic equation
\[f(x) \equiv ax^2 + bx + c = 0,\]
it happens that $c = \frac{b^2}{4a}$, then the graph of $y = f(x)$ will certainly:
$\mathrm{(A) \ have\ a\ maximum } \qquad \mathrm{(B) \ have\ a\ minimum} \qquad$ $\mathrm{(C) \ be\ tangent\ to\ the\ x-axis} \qquad$ $\mathrm{(D) \ be\ tangent\ to\ the\ y-axis} \qquad$ $\mathrm{(E) \ lie\ in\ one\ quadrant\ only}$
| [
"The discriminant of the quadratic equation is $b^2 - 4ac = b^2 - 4a\\left(\\frac{b^2}{4a}\\right) = 0$. This indicates that the equation has only one root (applying the quadratic formula, we get $x = \\frac{-b + \\sqrt{0}}{2a} = -b/2a$). Thus it follows that $f(x)$ touches the x-axis exactly once, and hence is tangent to the x-axis $\\Rightarrow \\mathrm{(C)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/16.json | AHSME |
1951_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | The bottom, side, and front areas of a rectangular box are known. The product of these areas
is equal to:
$\textbf{(A)}\ \text{the volume of the box} \qquad\textbf{(B)}\ \text{the square root of the volume} \qquad\textbf{(C)}\ \text{twice the volume}$
$\textbf{(D)}\ \text{the square of the volume} \qquad\textbf{(E)}\ \text{the cube of the volume}$
| [
"Let the length of the edges of this box have lengths $a$, $b$, and $c$. We're given $ab$, $bc$, and $ca$. The product of these values is $a^2b^2c^2$, which is the square of the volume of the box. $\\boxed{\\textbf{(D)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/6.json | AHSME |
1951_AHSME_Problems | 7 | 0 | Arithmetic | Multiple Choice | An error of $.02''$ is made in the measurement of a line $10''$ long, while an error of only $.2''$ is made in a measurement of a line $100''$ long. In comparison with the relative error of the first measurement, the relative error of the second measurement is:
$\textbf{(A)}\ \text{greater by }.18 \qquad\textbf{(B)}\ \text{the same} \qquad\textbf{(C)}\ \text{less}$
$\textbf{(D)}\ 10 \text{ times as great} \qquad\textbf{(E)}\ \text{correctly described by both (A) and (D)}$
| [
"There error percentage of the first measure is $0.2\\%$. The error percentage of the second measure is also $0.2\\%$. Therefore, the answer is $\\textbf{(B)}\\ \\text{the same}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/7.json | AHSME |
1951_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | Indicate in which one of the following equations $y$ is neither directly nor inversely proportional to $x$:
$\textbf{(A)}\ x + y = 0 \qquad\textbf{(B)}\ 3xy = 10 \qquad\textbf{(C)}\ x = 5y \qquad\textbf{(D)}\ 3x + y = 10$
$\textbf{(E)}\ \frac {x}{y} = \sqrt {3}$
| [
"Notice that for any directly or inversely proportional values, it can be expressed as $\\frac{x}{y}=k$ or $xy=k$. Now we try to convert each into its standard form counterpart.\n\n\n$\\textbf{(A)}\\ x + y = 0\\implies \\frac{x}{y}=-1$\n\n\n$\\textbf{(B)}\\ 3xy = 10\\implies xy=\\frac{10}{3}$\n\n\n$\\textbf{(C)}\\ x = 5y\\implies \\frac{x}{y}=5$\n\n\n$\\textbf{(E)}\\ \\frac {x}{y} = \\sqrt {3}\\implies \\frac {x}{y} = \\sqrt {3}$\n\n\nAs we can see, the only equation without a \"standard\" form is $\\textbf{(D)}$, so our answer is $\\boxed{\\textbf{(D)}\\ 3x + y = 10}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/17.json | AHSME |
1951_AHSME_Problems | 40 | 0 | Algebra | Multiple Choice | $\left(\frac{(x+1)^{2}(x^{2}-x+1)^{2}}{(x^{3}+1)^{2}}\right)^{2}\cdot\left(\frac{(x-1)^{2}(x^{2}+x+1)^{2}}{(x^{3}-1)^{2}}\right)^{2}$ equals:
$\textbf{(A)}\ (x+1)^{4}\qquad\textbf{(B)}\ (x^{3}+1)^{4}\qquad\textbf{(C)}\ 1\qquad\textbf{(D)}\ [(x^{3}+1)(x^{3}-1)]^{2}$
$\textbf{(E)}\ [(x^{3}-1)^{2}]^{2}$
| [
"First, note that we can pull the exponents out of every factor, since they are all squared. This results in\n$\\left(\\frac{(x+1)(x^{2}-x+1)}{x^{3}+1}\\right)^{4}\\cdot\\left(\\frac{(x-1)(x^{2}+x+1)}{x^{3}-1}\\right)^{4}$\nNow, multiplying the numerators together gives\n$\\left(\\frac{x^3+1}{x^3+1}\\right)^{4}\\cdot\\left(\\frac{x^3-1}{x^3-1}\\right)^{4}$,\nwhich simplifies to $\\boxed{1\\textbf{ (C)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/40.json | AHSME |
1951_AHSME_Problems | 37 | 0 | Number Theory | Multiple Choice | A number which when divided by $10$ leaves a remainder of $9$, when divided by $9$ leaves a remainder of $8$, by $8$ leaves a remainder of $7$, etc., down to where, when divided by $2$, it leaves a remainder of $1$, is:
$\textbf{(A)}\ 59\qquad\textbf{(B)}\ 419\qquad\textbf{(C)}\ 1259\qquad\textbf{(D)}\ 2519\qquad\textbf{(E)}\ \text{none of these answers}$
| [
"If we add $1$ to the number, it becomes divisible by $10, 9, 8, \\cdots, 2, 1$. The LCM of $1$ through $10$ is $2520$, therefore the number we want to find is $2520-1=\\boxed{\\textbf{(D)}\\ 2519}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/37.json | AHSME |
1951_AHSME_Problems | 21 | 0 | Algebra | Multiple Choice | Given: $x > 0, y > 0, x > y$ and $z\ne 0$. The inequality which is not always correct is:
$\textbf{(A)}\ x + z > y + z \qquad\textbf{(B)}\ x - z > y - z \qquad\textbf{(C)}\ xz > yz$
$\textbf{(D)}\ \frac {x}{z^2} > \frac {y}{z^2} \qquad\textbf{(E)}\ xz^2 > yz^2$
| [
"$\\textbf{(A)}\\ x + z > y + z\\implies x>y$, just subtract $z$ from both sides\n\n\n$\\textbf{(B)}\\ x - z > y - z\\implies x>y$, just add $z$ to both sides\n\n\n$\\textbf{(C)}\\ xz > yz\\implies x>y\\text{ if }x>0$, so that means that our desired answer is $\\boxed{\\textbf{(C)}\\ xz > yz}$. \n\n\nAs a check:\n\n\n$\\textbf{(D)}\\ \\frac {x}{z^2} > \\frac {y}{z^2}\\implies x>y$, we can divide $z^2$ safely and without worry because $z^2>0$.\n\n\n$\\textbf{(E)}\\ xz^2 > yz^2\\implies x>y$, similar reasoning as above but instead, multiply $z^2$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/21.json | AHSME |
1951_AHSME_Problems | 47 | 0 | Algebra | Multiple Choice | If $r$ and $s$ are the roots of the equation $ax^2+bx+c=0$, the value of $\frac{1}{r^{2}}+\frac{1}{s^{2}}$ is:
$\textbf{(A)}\ b^{2}-4ac\qquad\textbf{(B)}\ \frac{b^{2}-4ac}{2a}\qquad\textbf{(C)}\ \frac{b^{2}-4ac}{c^{2}}\qquad\textbf{(D)}\ \frac{b^{2}-2ac}{c^{2}}$
$\textbf{(E)}\ \text{none of these}$
| [
"Note that $\\frac{1}{r^2}+\\frac{1}{s^2} = \\frac{r^2+s^2}{r^2s^2} = \\frac{(r+s)^2-2(rs)}{(rs)^2}$. \n\n\nBy Vieta's, this is $\\frac{(-\\frac{b}{a})^2-2(\\frac{c}{a})}{(\\frac{c}{a})^2} = \\frac{\\frac{b^2}{a^2}-\\frac{2c}{a}}{\\frac{c^2}{a^2}} = \\frac{b^2-2ac}{c^2} \\implies \\boxed{(D)}$\n\n\n",
"$r$ and $s$ can be found in terms of $a$, $b$, and $c$ by using the quadratic formula; the roots are\n\n\n\\[\\frac{-b\\pm\\sqrt{b^2-4ac}}{2a}\\]\n\n\nBy Vieta's Formula, $r+s=-\\frac{b}{a}$ and $rs=\\frac{c}{a}$. Now let's algebraically manipulate what we want to find:\n\n\n\\[\\frac{1}{r^2}+\\frac{1}{s^2}=\\frac{r^2+s^2}{r^2s^2}=\\frac{(r+s)^2-2rs}{(rs)^2}\\]\n\n\nPlugging in the values for $r+s$ and $rs$ gives\n\n\n\\[\\frac{1}{r^2}+\\frac{1}{s^2}=\\frac{(-b/a)^2-2(c/a)}{(c/a)^2}=\\boxed{\\frac{b^2-2ca}{c^2}\\textbf{(D)}}\\]\n\n\n",
"\\begin{verbatim}\nhere $r$ and $s$ are the roots of the equation $ax^2+bx+c=0$,\n\\end{verbatim}\n\\begin{verbatim}\nnow we can convert this equation with roots $r^2$ and $s^2$.\n\\end{verbatim}\n\\begin{verbatim}\nlet, $y=x^2$ then above equation becomes\n\\end{verbatim}\n\\begin{verbatim}\n\\[ay+c=-b\\sqrt y\\] square on both sides we get \\[a^2y^2 +(2ac-b^2)y +c^2 =0\\]\n\\end{verbatim}\n\\begin{verbatim}\nAgain we can change this equation with roots $\\frac{1}{r^2}$ and $\\frac{1}{s^2}$ .\n\\end{verbatim}\nlet $z=\\frac{1}{y}$ then, $a^2\\frac{1}{z^2}+\\frac{2ac-b^2}{z}+c^2=0$ then\n\n\n\\begin{verbatim}\n \\[c^2z^2+z(2ac-b^2)+a^2=0\\]\n\\end{verbatim}\nthen the sum of roots of the above equation is $\\frac{1}{r^2}+\\frac{1}{s^2}=\\frac{b^2-2ac}{c^2}$\n\n\n\\begin{verbatim}\nhence, \\[\\frac{1}{r^2}+\\frac{1}{s^2}=\\boxed{\\frac{b^2-2ca}{c^2}\\textbf{(D)}}\\]\n\\end{verbatim}\n"
] | 3 | ./CreativeMath/AHSME/1951_AHSME_Problems/47.json | AHSME |
1951_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | Of the following statements, the one that is incorrect is:
$\textbf{(A)}\ \text{Doubling the base of a given rectangle doubles the area.}$
$\textbf{(B)}\ \text{Doubling the altitude of a triangle doubles the area.}$
$\textbf{(C)}\ \text{Doubling the radius of a given circle doubles the area.}$
$\textbf{(D)}\ \text{Doubling the divisor of a fraction and dividing its numerator by 2 changes the quotient.}$
$\textbf{(E)}\ \text{Doubling a given quantity may make it less than it originally was.}$
| [
"The well-known area formula for a circle is $A = \\pi r^2$, so doubling the radius will result it quadrupling the area (since $A' = \\pi (2r)^2 = 4 \\pi r^2 = 4A$). Statement $\\boxed{\\textbf{(C)}}$ is therefore incorrect, and is the correct answer choice.\n\n\n\n\nStatements $\\textbf{(A)}$ and $\\textbf{(B)}$ are evidently correct, since in triangles the area is directly proportional to both the base and the height ($A = \\frac{1}{2} bh$). Statement $\\textbf{(D)}$ is also correct: let $q = \\frac{a}{b}$. Then $q' = \\frac{a \\div 2}{2b} = \\frac{a}{4b} = \\frac{q}{4}$, which is a change from $q$. Finally, statement $\\textbf{(E)}$ is true: $2x < x$ if $x < 0$, as doubling a negative number makes it even more negative (and therefore less than it originally was).\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/10.json | AHSME |
1951_AHSME_Problems | 26 | 0 | Algebra | Multiple Choice | In the equation $\frac {x(x - 1) - (m + 1)}{(x - 1)(m - 1)} = \frac {x}{m}$ the roots are equal when
$\textbf{(A)}\ m = 1\qquad\textbf{(B)}\ m =\frac{1}{2}\qquad\textbf{(C)}\ m = 0\qquad\textbf{(D)}\ m =-1\qquad\textbf{(E)}\ m =-\frac{1}{2}$
| [
"Multiplying both sides by $(x-1)(m-1)m$ gives us\n\\[xm(x-1)-m(m+1)=x(x-1)(m-1)\\]\n\\[x^2m-xm-m^2-m=x^2m-xm-x^2+x\\]\n\\[-m^2-m=-x^2+x\\]\n\\[x^2-x-(m^2+m)=0\\]\nThe roots of this quadratic are equal if and only if its discriminant\n($b^2-4ac$) evaluates to 0.\nThis means\n\\[(-1)^2-4(-m^2-m)=0\\]\n\\[4m^2+4m+1=0\\]\n\\[(2m+1)^2=0\\]\n\\[m=-\\frac{1}{2} \\Rightarrow \\boxed{\\textbf{(E)}}\\]\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/26.json | AHSME |
1951_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | If two poles $20''$ and $80''$ high are $100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:
$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$
| [
"The two poles formula says this height is half the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is $\\frac1{\\frac1{20}+\\frac1{80}}$, or $\\frac1{\\frac1{16}}=\\boxed{16 \\textbf{ (C)}}$.\n\n\n",
"The two lines can be represented as $y=\\frac{-x}{5}+20$ and $y=\\frac{4x}{5}$.\nSolving the system,\n\n\n$\\frac{-x}{5}+20=\\frac{4x}{5}$\n\n\n$20=x.$\n\n\nSo the lines meet at an $x$-coordinate of 20.\n\n\nSolving for the height they meet,\n\n\n\\[y=\\frac{4\\cdot 20}{5}\\]\n\\[y=\\boxed{16 \\textbf{ (C)}}.\\]\n\n\n"
] | 2 | ./CreativeMath/AHSME/1951_AHSME_Problems/30.json | AHSME |
1951_AHSME_Problems | 31 | 0 | Algebra | Multiple Choice | A total of $28$ handshakes were exchanged at the conclusion of a party. Assuming that each participant was equally polite toward all the others, the number of people present was:
$\textbf{(A)}\ 14\qquad\textbf{(B)}\ 28\qquad\textbf{(C)}\ 56\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ 7$
| [
"The handshake equation is $h=\\frac{n(n-1)}{2}$, where $n$ is the number of people and $h$ is the number of handshakes. There were $28$ handshakes, so $28=\\frac{n(n-1)}{2}$\n$56=n(n-1)$\nThe factors of $56$ are:\n$1, 2, 4, 7, 8, 14, 28, 56$.\nAs we can see, only $7, 8$ fit the requirements $n(n-1)$ if $n$ was an integer.\nTherefore, the answer is $\\textbf{(D)}\\ 8$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/31.json | AHSME |
1951_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | Through a point inside a triangle, three lines are drawn from the vertices to the opposite sides forming six triangular sections. Then:
$\textbf{(A)}\ \text{the triangles are similar in opposite pairs}\qquad\textbf{(B)}\ \text{the triangles are congruent in opposite pairs}$
$\textbf{(C)}\ \text{the triangles are equal in area in opposite pairs}\qquad\textbf{(D)}\ \text{three similar quadrilaterals are formed}$
$\textbf{(E)}\ \text{none of the above relations are true}$
| [
"Say we draw 3 different types of triangles as the following\n\n\n1.An equilateral triangle with side lengths $x$\n\n\n2.An isosceles triangle with side lengths $x+3,x+3,x-1$\n\n\n3.A scalene triangle with sides $x+7,x-3,x+4$\n\n\n1. Gives us 6 congruent $30-60-90$ triangles, which means\n\n\n$\\textbf{(A)},\\textbf{(B)},\\textbf{(C)},\\textbf{(D)}$ are all true.\ntherefore 1. does not give any unique criteria.\n\n\n$\\textbf{(A)}$-Since the triangles are congruent with each other, they are similar by a ratio of $1$.\n\n\n$\\textbf{(B)}$-Since all the triangles are congruent they must always be congruent in any way.\n\n\n$\\textbf{(C)}$-The triangles are all congruent meaning that they always have equal area.\n\n\n$\\textbf{(D)}$-The top $2$ triangles, bottom-right $2$ triangles, and bottom-left $2$ triangles all make 3 identical congruent quadrilaterals.\n\n\n2. Gives us 3 pairs of adjacent congruent triangles, which means\n\n\n$\\textbf{(E)}$ is the only choice that works.\n\n\n$\\textbf{(E)}$-All of the relationships don't work so we must choose the last choice, no relations.\n\n\n3. Is not needed as we have found that all of the relationships given are not necessary all the time, so $\\textbf(E)$ is the only choice that works.\n\n\nSo the answer is $\\fbox{\\textbf{(E)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/27.json | AHSME |
1951_AHSME_Problems | 1 | 0 | Arithmetic | Multiple Choice | The percent that $M$ is greater than $N$ is:
$(\mathrm{A})\ \frac{100(M-N)}{M} \qquad (\mathrm{B})\ \frac{100(M-N)}{N} \qquad (\mathrm{C})\ \frac{M-N}{N} \qquad (\mathrm{D})\ \frac{M-N}{N} \qquad (\mathrm{E})\ \frac{100(M+N)}{N}$
| [
"$M-N$ is the amount by which $M$ is greater than $N$. We divide this by $N$ to get the percent by which $N$ is increased in the form of a decimal, and then multiply by $100$ to make it a percentage. Therefore, the answer is $\\boxed{\\mathrm{(B)}\\ \\dfrac{100(M-N)}{N}}$.\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/1.json | AHSME |
1951_AHSME_Problems | 50 | 0 | Algebra | Multiple Choice | Tom, Dick and Harry started out on a $100$-mile journey. Tom and Harry went by automobile at the rate of $25$ mph, while Dick walked at the rate of $5$ mph. After a certain distance, Harry got off and walked on at $5$ mph, while Tom went back for Dick and got him to the destination at the same time that Harry arrived. The number of hours required for the trip was:
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 8\qquad\textbf{(E)}\ \text{none of these answers}$
| [
"Let $d_1$ be the distance (in miles) that Harry traveled on car, and let $d_2$ be the distance (in miles) that Tom backtracked to get Dick. Let $T$ be the time (in hours) that it took the three to complete the journey. We now examine Harry's journey, Tom's journey, and Dick's journey. These yield, respectively, the equations\n\n\n\\[\\frac{d_1}{25}+\\frac{100-d_1}{5}=T,\\]\n\n\n\\[\\frac{d_1}{25}+\\frac{d_2}{25}+\\frac{100-(d_1-d_2)}{25}=T,\\]\n\n\n\\[\\frac{d_1-d_2}{5}+\\frac{100-(d_1-d_2)}{25}=T.\\]\n\n\nWe combine these three equations:\n\n\n\\[\\frac{d_1}{25}+\\frac{100-d_1}{5}=\\frac{d_1}{25}+\\frac{d_2}{25}+\\frac{100-(d_1-d_2)}{25}=\\frac{d_1-d_2}{5}+\\frac{100-(d_1-d_2)}{25}\\]\n\n\nAfter multiplying everything by 25 and simplifying, we get\n\n\n\\[500-4d_1=100+2d_2=100+4d_1-4d_2\\]\n\n\nWe have that $100+2d_2=100+4d_1-4d_2$, so $4d_1=6d_2\\Rightarrow d_1=\\frac{3}{2}d_2$. This then shows that $500-4d_1=100+\\frac{4}{3}d_1\\Rightarrow 400=\\frac{16}{3}d_1\\Rightarrow d_1=75$, which in turn gives that $d_2=50$. Now we only need to solve for $T$:\n\n\n\\[T=\\frac{d_1}{25}+\\frac{100-d_1}{5}=\\frac{75}{25}+\\frac{100-75}{5}=3+5=8\\]\n\n\nThe journey took 8 hours, so the correct answer is $\\boxed{\\textbf{(D)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/50.json | AHSME |
1951_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | The limit of the sum of an infinite number of terms in a geometric progression is $\frac {a}{1 - r}$ where $a$ denotes the first term and $- 1 < r < 1$ denotes the common ratio. The limit of the sum of their squares is:
$\textbf{(A)}\ \frac {a^2}{(1 - r)^2} \qquad\textbf{(B)}\ \frac {a^2}{1 + r^2} \qquad\textbf{(C)}\ \frac {a^2}{1 - r^2} \qquad\textbf{(D)}\ \frac {4a^2}{1 + r^2} \qquad\textbf{(E)}\ \text{none of these}$
| [
"Let the original geometric series be $a,ar,ar^2,ar^3,ar^4\\cdots$. Therefore, their squares are $a^2,a^2r^2,a^2r^4,a^2r^6,\\cdots$, which is a geometric sequence with first term $a^2$ and common ratio $r^2$. Thus, the sum is $\\boxed{\\textbf{(C)}\\ \\frac {a^2}{1 - r^2}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/11.json | AHSME |
1951_AHSME_Problems | 46 | 0 | Geometry | Multiple Choice | $AB$ is a fixed diameter of a circle whose center is $O$. From $C$, any point on the circle, a chord $CD$ is drawn perpendicular to $AB$. Then, as $C$ moves over a semicircle, the bisector of angle $OCD$ cuts the circle in a point that always:
$\textbf{(A)}\ \text{bisects the arc }AB\qquad\textbf{(B)}\ \text{trisects the arc }AB\qquad\textbf{(C)}\ \text{varies}$
$\textbf{(D)}\ \text{is as far from }AB\text{ as from }D\qquad\textbf{(E)}\ \text{is equidistant from }B\text{ and }C$
| [
"[asy] pair O=(0,0), A=(-1,0), B=(1,0), C=(-0.5,0.5sqrt(3)), D=(-0.5,-0.5sqrt(3)), E=(0.5,-0.5sqrt(3)), P=(0,-1); draw(circle(O,1)); label(\"$A$\",A,W); label(\"$B$\",B,E); label(\"$O$\",O,NE); label(\"$P$\",P,S); label(\"$C$\",C,NW); label(\"$D$\",D,SW); label (\"$E$\",E,SE); dot(A); dot(B); dot(C); dot(D); dot(E); dot(O); dot(P); draw(A--B); draw(C--D--E--cycle); draw(C--P,dashed); [/asy]\nDraw in diameter $CE$. Note that, as it is inscribed in a semicircle, $\\bigtriangleup CDE$ always has a right angle at $D$. Since it is given that $AB \\perp CD$, $AB \\parallel DE$ by congruent corresponding angles. $CP$ bisects $DE$, as well as the arc subtended by $DE$, $\\widehat{DPE}$. Because $AB \\parallel DE$, $CP$ always $\\boxed{\\textbf{(A)}\\ \\text{bisects the arc }AB}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/46.json | AHSME |
1951_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | A rectangular field is half as wide as it is long and is completely enclosed by $x$ yards of fencing. The area in terms of $x$ is:
$(\mathrm{A})\ \frac{x^2}2 \qquad (\mathrm{B})\ 2x^2 \qquad (\mathrm{C})\ \frac{2x^2}9 \qquad (\mathrm{D})\ \frac{x^2}{18} \qquad (\mathrm{E})\ \frac{x^2}{72}$
| [
"Let $w$ be the width. Then $l = 2w$, and the perimeter is $x = 2(2w)+2w = 6w \\implies w = \\frac{x}6$. The area is $wl = w(2w) = 2w^2 = 2\\left(\\frac{x^2}{36}\\right) = \\frac{x^2}{18}$, so the answer is $\\mathrm{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/2.json | AHSME |
1951_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | The pressure $(P)$ of wind on a sail varies jointly as the area $(A)$ of the sail and the square of the velocity $(V)$ of the wind. The pressure on a square foot is $1$ pound when the velocity is $16$ miles per hour. The velocity of the wind when the pressure on a square yard is $36$ pounds is:
$\textbf{(A)}\ 10\frac{2}{3}\text{ mph}\qquad\textbf{(B)}\ 96\text{ mph}\qquad\textbf{(C)}\ 32\text{ mph}\qquad\textbf{(D)}\ 1\frac{2}{3}\text{ mph}\qquad\textbf{(E)}\ 16\text{ mph}$
| [
"Because $P$ varies jointly as $A$ and $V^2$, that means that there is a number $k$ such that $P=kAV^2$. You are given that $P=1$ when $A=1$ and $V=16$. That means that $1=k(1)(16^2) \\rightarrow k=\\frac{1}{256}$. Then, substituting into the original equation with $P=36$ and $A=9$ (because a square yard is $9$ times a square foot), you get $4=\\frac{1}{256}(V^2)$. Solving for $V$, we get $V^2=1024$, so $V=32$. Hence, the answer is $\\boxed{C}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/28.json | AHSME |
1951_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | At $2: 15$ o'clock, the hour and minute hands of a clock form an angle of:
$\textbf{(A)}\ 30^{\circ} \qquad\textbf{(B)}\ 5^{\circ} \qquad\textbf{(C)}\ 22\frac {1}{2}^{\circ} \qquad\textbf{(D)}\ 7\frac {1}{2} ^{\circ} \qquad\textbf{(E)}\ 28^{\circ}$
| [
"Using the formula $\\frac{|60h-11m|}{2}$, where $h$ is the hour time and $m$ is the minute time, we get that the angle is $\\frac{|60(2)-11(15)|}{2}=\\frac{|120-165|}{2}=\\frac{45}{2}=\\boxed{\\textbf{(C)}\\ 22\\frac {1}{2}^{\\circ} }$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/12.json | AHSME |
1951_AHSME_Problems | 45 | 0 | Algebra | Multiple Choice | If you are given $\log 8\approx .9031$ and $\log 9\approx .9542$, then the only logarithm that cannot be found without the use of tables is:
$\textbf{(A)}\ \log 17\qquad\textbf{(B)}\ \log\frac{5}{4}\qquad\textbf{(C)}\ \log 15\qquad\textbf{(D)}\ \log 600\qquad\textbf{(E)}\ \log .4$
| [
"While $\\log 17 = \\log(8 + 9)$, we cannot easily deal with the logarithm of a sum. Furthermore, $17$ is prime, so none of the logarithm rules involving products or differences works. It therefore cannot be found without the use of a table (note: in 1951, calculators were very rare). The correct answer is therefore $\\boxed{\\textbf{(A)}\\ \\log 17}$.\n\n\n\n\nAs for the rest of the cases: \\[\\log\\frac{5}{4} = \\log\\frac{10}{8} = \\log 10 - \\log 8 = 1 - \\log 8\\] can be found; \\[\\log 15 = \\log 3 + \\log 5 = \\frac{1}{2}\\log 3^2 + \\log\\frac{10}{2} = \\frac{1}{2} \\log 9 + \\log 10 - \\log 2 = \\frac{1}{2} \\log 9 + 1 - \\frac{1}{3} \\log 8\\] can be found; \\[\\log 600 = \\log 100 + \\log 6 = 2 + \\log 2 + \\log 3 = 2 + \\frac{1}{3} \\log 8 + \\frac{1}{2} \\log 9\\] can be found; and \\[\\log .4 = \\log\\frac{4}{10} = \\log 4 - \\log 10 = 2 \\log 2 - 1 = \\frac{2}{3} \\log 8 - 1\\] can be found.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/45.json | AHSME |
1951_AHSME_Problems | 32 | 0 | Geometry | Multiple Choice | If $\triangle ABC$ is inscribed in a semicircle whose diameter is $AB$, then $AC+BC$ must be
$\textbf{(A)}\ \text{equal to }AB\qquad\textbf{(B)}\ \text{equal to }AB\sqrt{2}\qquad\textbf{(C)}\ \geq AB\sqrt{2}\qquad\textbf{(D)}\ \leq AB\sqrt{2}$ $\textbf{(E)}\ AB^{2}$
| [
"Because $AB$ is the diameter of the semi-circle, it follows that $\\angle C = 90$. Now we can try to eliminate all the solutions except for one by giving counterexamples.\n\n\n$\\textbf{(A):}$ \nSet point $C$ anywhere on the perimeter of the semicircle except on $AB$. By triangle inequality, $AC+BC>AB$, so $\\textbf{(A)}$ is wrong.\n$\\textbf{(B):}$\nSet point $C$ on the perimeter of the semicircle infinitesimally close to $AB$, and so $AC+BC$ almost equals $AB$, therefore $\\textbf{(B)}$ is wrong.\n$\\textbf{(C):}$\nBecause we proved that $AC+BC$ can be very close to $AB$ in case $\\textbf{(B)}$, it follows that $\\textbf{(C)}$ is wrong.\n$\\textbf{(E):}$\nBecause we proved that $AC+BC$ can be very close to $AB$ in case $\\textbf{(B)}$, it follows that $\\textbf{(E)}$ is wrong.\nTherefore, the only possible case is $\\boxed{\\textbf{(D)}\\ \\leq AB\\sqrt{2}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/32.json | AHSME |
1951_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | $\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}$ when simplified is:
$\textbf{(A)}\ 2^{n+1}-\frac{1}{8}\qquad\textbf{(B)}\ -2^{n+1}\qquad\textbf{(C)}\ 1-2^{n}\qquad\textbf{(D)}\ \frac{7}{8}\qquad\textbf{(E)}\ \frac{7}{4}$
| [
"We have $2(2^n)=2^{n+1}$, and $2(2^{n+3})=2^{n+4}$. Thus, $\\frac{2^{n+4}-2(2^{n})}{2(2^{n+3})}=\\dfrac{2^{n+4}-2^{n+1}}{2^{n+4}}$. Factoring out a $2^{n+1}$ in the numerator, we get $\\dfrac{2^{n+1}(2^3-1)}{2^{n+4}}=\\dfrac{8-1}{2^3}=\\boxed{\\textbf{(D)}\\ \\frac{7}{8}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/24.json | AHSME |
1951_AHSME_Problems | 49 | 0 | Geometry | Multiple Choice | The medians of a right triangle which are drawn from the vertices of the acute angles are $5$ and $\sqrt{40}$. The value of the hypotenuse is:
$\textbf{(A)}\ 10\qquad\textbf{(B)}\ 2\sqrt{40}\qquad\textbf{(C)}\ \sqrt{13}\qquad\textbf{(D)}\ 2\sqrt{13}\qquad\textbf{(E)}\ \text{none of these}$
| [
"We will proceed by coordinate bashing.\n\n\nCall the first leg $2a$, and the second leg $2b$ (We are using the double of a variable to avoid any fractions)\n\n\nNotice that we want to find $\\sqrt{(2a)^2+(2b)^2}$\n\n\nTwo equations can be written for the two medians: $a^2 + 4b^2 = 40$ and $4a^2+b^2= 25$. \n\n\nAdd them together and we get $5a^2+5b^2=65$,\n\n\nDividing by 5 gives $x^2+y^2=13$\n\n\nMultiplying this by 4 gives $4x^2+4y^2=52\\implies (2x)^2+(2y)^2=52$, just what we need to find the hypotenuse. Recall that he hypotenuse is $\\sqrt{(2a)^2+(2b)^2}$. The value inside the radical is equal to $52$, so the hypotenuse is equal to $\\sqrt{52}=\\boxed{\\textbf{(D)}\\ 2\\sqrt{13}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/49.json | AHSME |
1951_AHSME_Problems | 48 | 0 | Geometry | Multiple Choice | The area of a square inscribed in a semicircle is to the area of the square inscribed in the entire circle as:
$\textbf{(A)}\ 1: 2\qquad\textbf{(B)}\ 2: 3\qquad\textbf{(C)}\ 2: 5\qquad\textbf{(D)}\ 3: 4\qquad\textbf{(E)}\ 3: 5$
| [
"Let the radius of the circle be $r$. Let $s_1$ be the side length of the square inscribed in the semicircle and $s_2$ be the side length of the square inscribed in the entire circle. For the square in the semicircle, we have $s_1^2 + (\\frac{s_1}{2})^2 = r^2 \\Rightarrow s_1^2 = \\frac{4}{5}r^2$. For the square in the circle, we have $s_2 \\sqrt{2} = 2r \\Rightarrow s_2^2 = 2r^2$.\n\n\nTherefore, $s_1^2 : s_2^2 = \\frac{4}{5}r^2 : 2r^2 = \\boxed{\\textbf{(C)}\\ 2: 5}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/48.json | AHSME |
1951_AHSME_Problems | 25 | 0 | Geometry | Multiple Choice | The apothem of a square having its area numerically equal to its perimeter is compared with the apothem of an equilateral triangle having its area numerically equal to its perimeter. The first apothem will be:
$\textbf{(A)}\ \text{equal to the second}\qquad\textbf{(B)}\ \frac{4}{3}\text{ times the second}\qquad\textbf{(C)}\ \frac{2}{\sqrt{3}}\text{ times the second}\\ \textbf{(D)}\ \frac{\sqrt{2}}{\sqrt{3}}\text{ times the second}\qquad\textbf{(E)}\ \text{indeterminately related to the second}$
| [
"First we try to find the size of the square. Let $s$ be the side length of the square. It states that $s^2=4s$, therefore $s=0,4$. We cross out the trivial case $s=0$, so the side length of the square is $4$. The apothem of the square is simply half its side length, or $2$.\n\n\nLet the side length of the equilateral triangle be $t$. The problem states that $3t=\\frac{t^2\\sqrt{3}}{4}$, so $12t=t^2\\sqrt{3}$, therefore $t=0, \\frac{12}{\\sqrt{3}}$. Again we cross out the trivial case $t=0$, so we have the side length of the triangle as $\\frac{12}{\\sqrt{3}}$. The apothem of the triangle is $\\frac{1}{3}$ of the height. The height of the triangle is $\\frac{\\sqrt{3}}{2}\\cdot \\frac{12}{\\sqrt{3}} = 6$, so the apothem is $2$ Therefore, the answer is $\\textbf{(A)}\\ \\text{equal to the second}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/25.json | AHSME |
1951_AHSME_Problems | 33 | 0 | Algebra | Multiple Choice | The roots of the equation $x^{2}-2x = 0$ can be obtained graphically by finding the abscissas of the points of intersection of each of the following pairs of equations except the pair:
$\textbf{(A)}\ y = x^{2}, y = 2x\qquad\textbf{(B)}\ y = x^{2}-2x, y = 0\qquad\textbf{(C)}\ y = x, y = x-2\qquad\textbf{(D)}\ y = x^{2}-2x+1, y = 1$
$\textbf{(E)}\ y = x^{2}-1, y = 2x-1$
\textit{[Note: Abscissas means x-coordinate.]}
| [
"If you find the intersections of the curves listed in the answers $\\textbf{(A)}$, $\\textbf{(B)}$, $\\textbf{(D)}$, and $\\textbf{(E)}$, you will find that their abscissas are $0$ and $2$. Also you can note that the curves in $\\textbf{(C)}$ don't actually intersect. Therefore the answer is $\\boxed{\\textbf{(C)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/33.json | AHSME |
1951_AHSME_Problems | 44 | 0 | Algebra | Multiple Choice | If $\frac{xy}{x+y}= a,\frac{xz}{x+z}= b,\frac{yz}{y+z}= c$, where $a, b, c$ are other than zero, then $x$ equals:
$\textbf{(A)}\ \frac{abc}{ab+ac+bc}\qquad\textbf{(B)}\ \frac{2abc}{ab+bc+ac}\qquad\textbf{(C)}\ \frac{2abc}{ab+ac-bc}$
$\textbf{(D)}\ \frac{2abc}{ab+bc-ac}\qquad\textbf{(E)}\ \frac{2abc}{ac+bc-ab}$
| [
"Note that $\\frac{1}{a}=\\frac{1}{x}+\\frac{1}{y}$, $\\frac{1}{b}=\\frac{1}{x}+\\frac{1}{z}$, and $\\frac{1}{c}=\\frac{1}{y}+\\frac{1}{z}$. Therefore\n\n\n\\[\\frac{1}{x}+\\frac{1}{y}+\\frac{1}{z}=\\frac{\\frac{1}{a}+\\frac{1}{b}+\\frac{1}{c}}{2}\\]\n\n\nTherefore\n\n\n\\[\\frac{1}{x}=\\frac{1}{2a}+\\frac{1}{2b}+\\frac{1}{2c}-\\frac{1}{c}=\\frac{1}{2a}+\\frac{1}{2b}-\\frac{1}{2c}\\]\n\n\nA little algebraic manipulation yields that\n\n\n\\[x=\\boxed{\\textbf{(E)}\\ \\frac{2abc}{ac+bc-ab}}\\]\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/44.json | AHSME |
1951_AHSME_Problems | 13 | 0 | Arithmetic | Multiple Choice | $A$ can do a piece of work in $9$ days. $B$ is $50\%$ more efficient than $A$. The number of days it takes $B$ to do the same piece of work is:
$\textbf{(A)}\ 13\frac {1}{2} \qquad\textbf{(B)}\ 4\frac {1}{2} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ \text{none of these answers}$
| [
"Because $B$ is $50\\%$ more efficient, he can do $1.5$ pieces of work in $9$ days. This is equal to 1 piece of work in $\\textbf{(C)}\\ 6$ days\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/13.json | AHSME |
1951_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | Of the following sets of data the only one that does not determine the shape of a triangle is:
$\textbf{(A)}\ \text{the ratio of two sides and the inc{}luded angle}\\ \qquad\textbf{(B)}\ \text{the ratios of the three altitudes}\\ \qquad\textbf{(C)}\ \text{the ratios of the three medians}\\ \qquad\textbf{(D)}\ \text{the ratio of the altitude to the corresponding base}\\ \qquad\textbf{(E)}\ \text{two angles}$
| [
"The answer is $\\boxed{\\textbf{(D)}}$. The ratio of the altitude to the base is insufficient to determine the shape of a triangle; you also need to know the ratio of the two segments into which the altitude divides the base.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/29.json | AHSME |
1951_AHSME_Problems | 3 | 0 | Geometry | Multiple Choice | If the length of a diagonal of a square is $a + b$, then the area of the square is:
$\mathrm{(A) \ (a+b)^2 } \qquad \mathrm{(B) \ \frac{1}{2}(a+b)^2 } \qquad \mathrm{(C) \ a^2+b^2 } \qquad \mathrm{(D) \ \frac {1}{2}(a^2+b^2) } \qquad \mathrm{(E) \ \text{none of these} }$
| [
"Let a side be $s$; then by the Pythagorean Theorem, $s^2 + s^2 = 2s^2 = (a+b)^2$. The area of a square is $s^2 = \\frac{(a+b)^2}{2} \\Rightarrow \\mathrm{(B)}$.\n\n\nAlternatively, using the area formula for a kite, the area is $\\frac{1}{2}d_1d_2 = \\frac{1}{2}(a+b)^2$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/3.json | AHSME |
1951_AHSME_Problems | 34 | 0 | Algebra | Multiple Choice | The value of $10^{\log_{10}7}$ is:
$\textbf{(A)}\ 7\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ \log_{10}7\qquad\textbf{(E)}\ \log_{7}10$
| [
"$\\log_{10}7=x \\Rightarrow 10^x=7$\nSubstitute $\\log_{10}7=x$ in $10^{\\log_{10}7} \\Rightarrow 10^x=?$\nIt was already stated that $10^x=7$, so our answer is $\\textbf{(A)}\\ 7$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/34.json | AHSME |
1951_AHSME_Problems | 8 | 0 | Arithmetic | Multiple Choice | The price of an article is cut $10 \%.$ To restore it to its former value, the new price must be increased by:
$\textbf{(A) \ } 10 \% \qquad\textbf{(B) \ } 9 \% \qquad \textbf{(C) \ } 11\frac{1}{9} \% \qquad\textbf{(D) \ } 11 \% \qquad\textbf{(E) \ } \text{none of these answers}$
| [
"Without loss of generality, let the price of the article be 100. Thus, the new price is $.9\\cdot 100= 90$. Then, to restore it to the original price, we solve $90x=100$. We find $x=1\\dfrac{1}{9}$, thus, the percent increase is $1\\dfrac{1}{9}-1=\\dfrac{1}{9}=\\boxed{\\textbf{(C) \\ } 11\\frac{1}{9}\\%}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/8.json | AHSME |
1951_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | The values of $a$ in the equation: $\log_{10}(a^2 - 15a) = 2$ are:
$\textbf{(A)}\ \frac {15\pm\sqrt {233}}{2} \qquad\textbf{(B)}\ 20, - 5 \qquad\textbf{(C)}\ \frac {15 \pm \sqrt {305}}{2}$
$\textbf{(D)}\ \pm20 \qquad\textbf{(E)}\ \text{none of these}$
| [
"Putting into exponential form, we get that $10^2=a^2-15a\\Rightarrow a^2-15a-100=0$\n\n\nNow we use the quadratic formula to solve for $a$, and we get $a=\\frac{15\\pm\\sqrt{625}}{2}\\implies a=\\boxed{\\textbf{(B)}\\ 20, - 5}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/22.json | AHSME |
1951_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | The expression $21x^2 +ax +21$ is to be factored into two linear prime binomial factors with integer coefficients. This can be done if $a$ is:
$\textbf{(A)}\ \text{any odd number} \qquad\textbf{(B)}\ \text{some odd number} \qquad\textbf{(C)}\ \text{any even number}$
$\textbf{(D)}\ \text{some even number} \qquad\textbf{(E)}\ \text{zero}$
| [
"We can factor $21x^2 + ax + 21$ as $(7x+3)(3x+7)$, which expands to $21x^2+42x+21$. So the answer is $\\textbf{(D)}\\ \\text{some even number}$\n\n\n",
"Factoring $21x^2+ax+21$ by grouping, we need to find some $b,c$ such that $b\\cdot c = 21^2$, and that $b+c=a$.\nSince $21^2\\equiv 1\\;(mod\\;2)$, $b\\land c \\equiv 1\\;(mod\\;2)$, and $b+c \\equiv 0\\;(mod\\;2)$. So $a$ must be even. $a$ cannot be $any$ even number, since $21^2$ only has 4 odd factors, so the answer is $\\textbf{(D)}\\ \\text{some even number}$\n\n\n"
] | 2 | ./CreativeMath/AHSME/1951_AHSME_Problems/18.json | AHSME |
1951_AHSME_Problems | 38 | 0 | Algebra | Multiple Choice | A rise of $600$ feet is required to get a railroad line over a mountain. The grade can be kept down by lengthening the track and curving it around the mountain peak. The additional length of track required to reduce the grade from $3\%$ to $2\%$ is approximately:
$\textbf{(A)}\ 10000\text{ ft.}\qquad\textbf{(B)}\ 20000\text{ ft.}\qquad\textbf{(C)}\ 30000\text{ ft.}\qquad\textbf{(D)}\ 12000\text{ ft.}\qquad\textbf{(E)}\ \text{none of these}$
| [
"A grade is the rise divided by the horizontal length for a given segment of track. This means we can get the horizontal length of the track by dividing the rise by the grade.\n\n\nAt a $3\\%$ grade, the horizontal track length is $20000$ feet. At a $2\\%$ grade, the horizontal track length is $30000$ feet. The difference is $10000$ feet of horizontal track. Compared to $10000$, $600$ feet is insignificant and can be safely covered by the problem's use of the word \"approximately.\" Therefore, the answer is $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/38.json | AHSME |
1951_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | A barn with a roof is rectangular in shape, $10$ yd. wide, $13$ yd. long and $5$ yd. high. It is to be painted inside and outside, and on the ceiling, but not on the roof or floor. The total number of sq. yd. to be painted is:
$\mathrm{(A) \ } 360 \qquad \mathrm{(B) \ } 460 \qquad \mathrm{(C) \ } 490 \qquad \mathrm{(D) \ } 590 \qquad \mathrm{(E) \ } 720$
| [
"The walls are $13*5=65$ and $10*5=50$ in area, and the ceiling has an area of $10*13=130$.\n\n\n$((65+50)2)2+130=590 \\Rightarrow \\boxed{\\mathrm{(D)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/4.json | AHSME |
1951_AHSME_Problems | 14 | 0 | Geometry | Multiple Choice | In connection with proof in geometry, indicate which one of the following statements is \textit{incorrect}:
$\textbf{(A)}\ \text{Some statements are accepted without being proved.}$
$\textbf{(B)}\ \text{In some instances there is more than one correct order in proving certain propositions.}$
$\textbf{(C)}\ \text{Every term used in a proof must have been defined previously.}$
$\textbf{(D)}\ \text{It is not possible to arrive by correct reasoning at a true conclusion if, in the given, there is an untrue proposition.}$
$\textbf{(E)}\ \text{Indirect proof can be used whenever there are two or more contrary propositions.}$
| [
"After reading the options, it is very apparent that $\\textbf{(E)}\\ \\text{Indirect proof can be used whenever there are two or more contrary propositions.}$ is the correct answer; rigorous proof is needed no matter what.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/14.json | AHSME |
1951_AHSME_Problems | 43 | 0 | Algebra | Multiple Choice | Of the following statements, the only one that is incorrect is:
$\textbf{(A)}\ \text{An inequality will remain true after each side is increased,}$ $\text{ decreased, multiplied or divided zero excluded by the same positive quantity.}$
$\textbf{(B)}\ \text{The arithmetic mean of two unequal positive quantities is greater than their geometric mean.}$
$\textbf{(C)}\ \text{If the sum of two positive quantities is given, ther product is largest when they are equal.}$
$\textbf{(D)}\ \text{If }a\text{ and }b\text{ are positive and unequal, }\frac{1}{2}(a^{2}+b^{2})\text{ is greater than }[\frac{1}{2}(a+b)]^{2}.$
$\textbf{(E)}\ \text{If the product of two positive quantities is given, their sum is greatest when they are equal.}$
| [
"The answer is $\\boxed{\\textbf{(E)}}$.\nQuite the opposite of statement (E) is true--the sum $a+b$ is \\textit{minimized} when $a=b$, but it approaches $\\infty$ when one of $a,b$ gets arbitrarily small.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/43.json | AHSME |
1951_AHSME_Problems | 42 | 0 | Algebra | Multiple Choice | If $x =\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\cdots}}}}$, then:
$\textbf{(A)}\ x = 1\qquad\textbf{(B)}\ 0 < x < 1\qquad\textbf{(C)}\ 1 < x < 2\qquad\textbf{(D)}\ x\text{ is infinite}$
$\textbf{(E)}\ x > 2\text{ but finite}$
| [
"We note that $x^2=1+\\sqrt{1+\\sqrt{1+\\sqrt{1+\\cdots}}}=1+x$. By the quadratic formula, $x=\\dfrac{1\\pm \\sqrt{5}}{2}$. Because there are only positive square roots in $x$, $x$ must be positive, thus, it is $\\dfrac{1+\\sqrt{5}}{2}\\approx 1.618$. Thus, $\\boxed{\\textbf{(C)}\\ 1 < x < 2}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/42.json | AHSME |
1951_AHSME_Problems | 15 | 0 | Number Theory | Multiple Choice | The largest number by which the expression $n^3 - n$ is divisible for all possible integral values of $n$, is:
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 5 \qquad\textbf{(E)}\ 6$
| [
"Factoring the polynomial gives $(n+1)(n)(n-1)$ According to the factorization, one of those factors must be a multiple of two because there are more than 2 consecutive integers. In addition, because there are three consecutive integers, one of the integers must be a multiple of 3. Therefore $6$ must divide the given expression.\nPlugging in $n=2$ yields $6$. So the largest possibility is $6$.\n\n\nClearly the answer is $\\boxed{\\textbf{(E)} \\ 6}$\n\n\n",
"In general, $r!$ | $n(n+1)(n+2)...(n+r-1)$ were $r$ and $n$ are integers. So here $3!$ | $n^3$ - $n$ always for any integer $n$.Hence,the correct answer is $6$. \n$\\boxed{\\textbf{(E)} \\ 6}$\n\n\n~geometry wizard.\n\n\n"
] | 2 | ./CreativeMath/AHSME/1951_AHSME_Problems/15.json | AHSME |
1951_AHSME_Problems | 5 | 0 | Arithmetic | Multiple Choice | Mr. $A$ owns a home worth $$10,000$. He sells it to Mr. $B$ at a $10\%$ profit based on the worth of the house. Mr. $B$ sells the house back to Mr. $A$ at a $10\%$ loss. Then:
$\mathrm{(A) \ A\ comes\ out\ even } \qquad$ $\mathrm{(B) \ A\ makes\ 1100\ on\ the\ deal}$ $\qquad \mathrm{(C) \ A\ makes\ 1000\ on\ the\ deal } \qquad$ $\mathrm{(D) \ A\ loses\ 900\ on\ the\ deal }$ $\qquad \mathrm{(E) \ A\ loses\ 1000\ on\ the\ deal }$
| [
"Mr. $A$ sells his home for $(1 + 10$%$)$ $\\cdot$ $10,000$ dollars $=$ $1.1$ $\\cdot$ $10,000$ dollars $=$ $11,000$ dollars to Mr. $B$. Then, Mr. $B$ sells it at a price of $(1-10$%$)$ $\\cdot$ $11,000$ dollars $=$ $0.9$ $\\cdot$ $11,000$ dollars $=$ $9,900$ dollars, thus $11,000 - 9,900$ $=$ $\\boxed{\\textrm{(B)}\\ \\text{A makes 1100 on the deal}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/5.json | AHSME |
1951_AHSME_Problems | 39 | 0 | Algebra | Multiple Choice | A stone is dropped into a well and the report of the stone striking the bottom is heard $7.7$ seconds after it is dropped. Assume that the stone falls $16t^2$ feet in t seconds and that the velocity of sound is $1120$ feet per second. The depth of the well is:
$\textbf{(A)}\ 784\text{ ft.}\qquad\textbf{(B)}\ 342\text{ ft.}\qquad\textbf{(C)}\ 1568\text{ ft.}\qquad\textbf{(D)}\ 156.8\text{ ft.}\qquad\textbf{(E)}\ \text{none of these}$
| [
"Let $d$ be the depth of the well in feet, let $t_1$ be the number of seconds the rock took to fall to the bottom of the well, and let $t_2$ be the number of seconds the sound took to travel back up the well. We know $t_1+t_2=7.7$. Now we can solve $t_1$ for $d$:\n\\[d=16t_1^2\\]\n\\[\\frac{d}{16}=t_1^2\\]\n\\[t_1=\\frac{\\sqrt{d}}4\\]\nAnd similarly $t_2$:\n\\[d=1120t_2\\]\n\\[t_2=\\frac{d}{1120}\\]\nSo $\\frac{d}{1120}+\\frac{\\sqrt{d}}4-7.7=0$. If we let $u=\\sqrt{d}$, this becomes a quadratic. \n\\[\\frac{u^2}{1120}+\\frac{u}4-7.7=0\\]\n\\[u=\\frac{-\\frac14\\pm\\sqrt{\\left(\\frac14\\right)^2-4(\\frac1{1120})(-7.7)}}{\\frac2{1120}}\\]\n\\[u=560\\cdot\\left(-\\frac14\\pm\\sqrt{\\frac1{16}+\\frac{7.7}{280}}\\right)\\]\n\\[u=560\\cdot\\left(-\\frac14\\pm\\sqrt{\\frac{36}{400}}\\right)\\]\n\\[u=560\\cdot\\left(-\\frac14\\pm\\frac3{10}\\right)\\]\nWe know $u$ is the positive square root of $d$, so we can replace the $\\pm$ with a $+$.\n\\[u=560\\cdot\\left(\\frac1{20}\\right)\\]\n\\[u=\\frac{560}{20}\\]\n\\[u=28\\]\nThen $d=28^2=784$, and the answer is $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/39.json | AHSME |
1951_AHSME_Problems | 19 | 0 | Number Theory | Multiple Choice | A six place number is formed by repeating a three place number; for example, $256256$ or $678678$, etc. Any number of this form is always exactly divisible by:
$\textbf{(A)}\ 7 \text{ only} \qquad\textbf{(B)}\ 11 \text{ only} \qquad\textbf{(C)}\ 13 \text{ only} \qquad\textbf{(D)}\ 101 \qquad\textbf{(E)}\ 1001$
| [
"We can express any of these types of numbers in the form $\\overline{abc}\\times 1001$, where $\\overline{abc}$ is a 3-digit number. Therefore, the answer is $\\textbf{(E)}\\ 1001$.\n\n\n",
"$\\overline{abcabc}\\ = \\overline{abc}\\* 1001$ .(E)\n\n\n\n\n~GEOMETRY-WIZARD.\n\n\n"
] | 2 | ./CreativeMath/AHSME/1951_AHSME_Problems/19.json | AHSME |
1951_AHSME_Problems | 23 | 0 | Algebra | Multiple Choice | The radius of a cylindrical box is $8$ inches and the height is $3$ inches. The number of inches that may be added to either the radius or the height to give the same nonzero increase in volume is:
$\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 5\frac {1}{3} \qquad\textbf{(C)}\ \text{any number} \qquad\textbf{(D)}\ \text{non-existent} \qquad\textbf{(E)}\ \text{none of these}$
| [
"Let $x$ be the number of inches increased. We can set up an equation for $x$: \\[8^2 \\times (3+x)=(8+x)^2\\times 3\\]\n\n\nExpanding gives $3x^2+48x+192=64x+192$.\n\n\nCombining like terms gives the quadratic $3x^2-16x=0$\n\n\nFactoring out an $x$ gives $x(3x-16)=0$.\n\n\nSo either $x=0$, or $3x-16=0$.\n\n\nThe first equation is not possible, because the problem states that the value has to be non-zero. The second equation gives the answer, $x = \\boxed{5\\frac{1}{3}}.$ Thus the answer is $\\textbf{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/23.json | AHSME |
1951_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | An equilateral triangle is drawn with a side of length $a$. A new equilateral triangle is formed by joining the midpoints of the sides of the first one. Then a third equilateral triangle is formed by joining the midpoints of the sides of the second; and so on forever. The limit of the sum of the perimeters of all the triangles thus drawn is:
$\textbf{(A)}\ \text{Infinite} \qquad\textbf{(B)}\ 5\frac {1}{4}a \qquad\textbf{(C)}\ 2a \qquad\textbf{(D)}\ 6a \qquad\textbf{(E)}\ 4\frac {1}{2}a$
| [
"The perimeter of the first triangle is $3a$. The perimeter of the 2nd triangle is half of that, after drawing a picture. The 3rd triangle's perimeter is half the second's, and so on. Therefore, we are computing $3a+\\frac{3a}{2}+\\frac{3a}{4}+\\cdots$. \n\n\nThe starting term is $3a$, and the common ratio is $1/2$. Therefore, the sum is $\\frac{3a}{1-\\frac{1}{2}}=\\boxed{\\textbf{(D)}\\ 6a}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/9.json | AHSME |
1951_AHSME_Problems | 35 | 0 | Algebra | Multiple Choice | If $a^{x}= c^{q}= b$ and $c^{y}= a^{z}= d$, then
$\textbf{(A)}\ xy = qz\qquad\textbf{(B)}\ \frac{x}{y}=\frac{q}{z}\qquad\textbf{(C)}\ x+y = q+z\qquad\textbf{(D)}\ x-y = q-z$
$\textbf{(E)}\ x^{y}= q^{z}$
| [
"Try solving both equations for $a$. Taking the $x$-th root of both sides in the first equation and the $z$-the root of both sides in the second gives $a=c^{\\frac{q}x}$ and $a=c^{\\frac{y}z}$.\nSo $\\frac{q}x=\\frac{y}z$. Multiplying both sides by $xz$, $qz=xy$. $\\boxed{\\textbf{(A)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1951_AHSME_Problems/35.json | AHSME |
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