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 |
|---|---|---|---|---|---|---|---|---|---|
1965_AHSME_Problems | 9 | 0 | Algebra | Multiple Choice | The vertex of the parabola $y = x^2 - 8x + c$ will be a point on the $x$-axis if the value of $c$ is:
$\textbf{(A)}\ - 16 \qquad \textbf{(B) }\ - 4 \qquad \textbf{(C) }\ 4 \qquad \textbf{(D) }\ 8 \qquad \textbf{(E) }\ 16$
| [
"Notice that if the vertex of a parabola is on the x-axis, then the x-coordinate of the vertex must be a solution to the quadratic. Since the quadratic is strictly increasing on either side of the vertex, the solution must have double multiplicity, or the quadratic is a perfect square trinomial. This means that for the vertex of $y = x^2 - 8x + c$ to be on the x-axis, the trinomial must be a perfect square, and have discriminant of zero. So,\n\n\n\\begin{align*} 0 &= b^2-4ac\\\\ 0 &= (-8)^2-4c\\\\ c &= 64\\\\ c &= 16\\\\ \\end{align*}\n\n\nTherefore $c=16$, and our answer is $\\boxed{\\textbf{(E)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1965_AHSME_Problems/9.json | AHSME |
1970_AHSME_Problems | 20 | 0 | Geometry | Multiple Choice | Lines $HK$ and $BC$ lie in a plane. $M$ is the midpoint of the line segment $BC$, and $BH$ and $CK$ are perpendicular to $HK$.
Then we
$\text{(A) always have } MH=MK\quad\\ \text{(B) always have } MH>BK\quad\\ \text{(C) sometimes have } MH=MK \text{ but not always}\quad\\ \text{(D) always have } MH>MB\quad\\ \text{(E) always have } BH<BC$
| [
"$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/20.json | AHSME |
1970_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | If $F(n)$ is a function such that $F(1)=F(2)=F(3)=1$, and such that $F(n+1)= \frac{F(n)\cdot F(n-1)+1}{F(n-2)}$ for $n\ge 3,$
then $F(6)=$
$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 7\quad \text{(D) } 11\quad \text{(E) } 26$
\section{Solution}
Plugging in $n=3$ gives $F(4) = \frac{F(3) \cdot F(2) + 1}{F(1)} = \frac{1 \cdot 1 + 1}{1} = 2$.
Plugging in $n=4$ gives $F(5) = \frac{F(4) \cdot F(3) + 1}{F(2)} = \frac{2 \cdot 1 + 1}{1} = 3$.
Plugging in $n=5$ gives $F(6) = \frac{F(5) \cdot F(4) + 1}{F(3)} = \frac{3 \cdot 2 + 1}{1} = 7$.
Thus, the answer is $\fbox{C}$.
| [
"Plugging in $n=3$ gives $F(4) = \\frac{F(3) \\cdot F(2) + 1}{F(1)} = \\frac{1 \\cdot 1 + 1}{1} = 2$.\n\n\nPlugging in $n=4$ gives $F(5) = \\frac{F(4) \\cdot F(3) + 1}{F(2)} = \\frac{2 \\cdot 1 + 1}{1} = 3$.\n\n\nPlugging in $n=5$ gives $F(6) = \\frac{F(5) \\cdot F(4) + 1}{F(3)} = \\frac{3 \\cdot 2 + 1}{1} = 7$.\n\n\nThus, the answer is $\\fbox{C}$.\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/16.json | AHSME |
1970_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | The smallest value of $x^2+8x$ for real values of $x$ is
$\text{(A) } -16.25\quad \text{(B) } -16\quad \text{(C) } -15\quad \text{(D) } -8\quad \text{(E) None of these}$
| [
"Let's imagine this as a quadratic equation. To find the minimum or maximum value, we always need to find the vertex of the quadratic equation. The vertex of the quadratic is $\\frac{-b}{2a}$ in $ax^2+bx+c=0$. Then to find the output, or the y value of the quadratic, we plug the vertex \"x\" value back into the equation. In this quadratic, a=1, b=8, and c=0. So the \"x\" value is $\\frac{-b}{2a} \\Rightarrow \\frac{-8}{2} = -4$. Plugging it back into $x^2 + 8x$, we get $16-32 = -16 \\Rightarrow$ $\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/6.json | AHSME |
1970_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | Inside square $ABCD$ with side $s$, quarter-circle arcs with radii $s$ and centers at $A$ and $B$ are drawn. These arcs intersect at a point $X$ inside the square. How far is $X$ from the side of $CD$?
$\text{(A) } \tfrac{1}{2} s(\sqrt{3}+4)\quad \text{(B) } \tfrac{1}{2} s\sqrt{3}\quad \text{(C) } \tfrac{1}{2} s(1+\sqrt{3})\quad \text{(D) } \tfrac{1}{2} s(\sqrt{3}-1)\quad \text{(E) } \tfrac{1}{2} s(2-\sqrt{3})$
| [
"All answers are proportional to $s$, so for ease, let $s=1$.\n\n\nLet $ABCD$ be oriented so that $A(0, 0), B(1, 0), C(1, 1), D(0, 1)$.\n\n\nThe circle centered at $A$ with radius $1$ is $x^2 + y^2 = 1$. The circle centered at $B$ with radius $1$ is $(x - 1)^2 + y^2 = 1$. Solving each equation for $1 - y^2$ to find the intersection leads to $(x - 1)^2 = x^2$, which leads to $x = \\frac{1}{2}$.\n\n\nPlugging that back in to $x^2 + y^2 = 1$ leads to $y^2 = 1 - \\frac{1}{4}$, or $y = \\pm \\frac{\\sqrt{3}}{2}$. Since we want the intersection within the square where $0 < x, y < 1$, we take the positive solution, and the intersection is at $X(\\frac{1}{2}, \\frac{\\sqrt{3}}{2})$.\n\n\nThe distance from $X$ to side $CD$, which lies along the line $y=1$, is $1 - \\frac{\\sqrt{3}}{2}$, or $\\frac{2 -\\sqrt{3}}{2}$. This is answer $\\fbox{E}$ when $s=1$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/7.json | AHSME |
1970_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | If $r>0$, then for all $p$ and $q$ such that $pq\ne 0$ and $pr>qr$, we have
$\text{(A) } -p>-q\quad \text{(B) } -p>q\quad \text{(C) } 1>-q/p\quad \text{(D) } 1<q/p\quad \text{(E) None of these}$
| [
"If $r>0$ and $pr>qr$, we can divide by the positive number $r$ and not change the inequality direction to get $p>q$. Multiplying by $-1$ (and flipping the inequality sign because we're multiplying by a negative number) leads to $-p < -q$, which directly contradicts $A$. Thus, $A$ is always false.\n\n\nIf $p < 0$ (which is possible but not guaranteed), we can divide the true statement $p>q$ by $p$ to get $1 > \\frac{q}{p}$. This contradicts $D$. Thus, $D$ is sometimes false, which is bad enough to be eliminated.\n\n\nIf $(p, q, r) = (3, 2, 1)$, then the condition that $pr > qr$ is satisfied. However, $-p = -3$ and $q = 2$, so $-p > q$ is false for at least this case, eliminating $B$.\n\n\nIf $(p, q, r) = (2, -3, 1)$, then $pr > qr$ is also satisfied. However, $-\\frac{q}{p} = 1.5$, so $1 > -\\frac{q}{p}$ is false, eliminating $C$.\n\n\nAll four options do not follow from the premises, leading to $\\fbox{E}$ as the correct answer.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/17.json | AHSME |
1970_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | On an auto trip, the distance read from the instrument panel was $450$ miles. With snow tires on for the return trip over the same route, the reading was $440$ miles. Find, to the nearest hundredth of an inch, the increase in radius of the wheels if the original radius was 15 inches.
$\text{(A) } .33\quad \text{(B) } .34\quad \text{(C) } .35\quad \text{(D) } .38\quad \text{(E) } .66$
| [
"$\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/21.json | AHSME |
1970_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | Let $F=.48181\cdots$ be an infinite repeating decimal with the digits $8$ and $1$ repeating. When $F$ is written as a fraction in lowest terms, the denominator exceeds the numerator by
$\text{(A) } 13\quad \text{(B) } 14\quad \text{(C) } 29\quad \text{(D) } 57\quad \text{(E) } 126$
| [
"Multiplying by $100$ gives $100F = 48.181818...$. Subtracting the first equation from the second gives $99F = 47.7$, and all the other repeating parts cancel out. This gives $F = \\frac{47.7}{99} = \\frac{477}{990} = \\frac{159}{330} = \\frac{53}{110}$. Subtracting the numerator from the denominator gives $\\fbox{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/10.json | AHSME |
1970_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | The number of distinct points in the $xy$-plane common to the graphs of $(x+y-5)(2x-3y+5)=0$ and $(x-y+1)(3x+2y-12)=0$ is
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 4\quad \text{(F) } \infty$
| [
"The graph $(x + y - 5)(2x - 3y + 5) = 0$ is the combined graphs of $x + y - 5=0$ and $2x - 3y + 5 = 0$. Likewise, the graph $(x -y + 1)(3x + 2y - 12) = 0$ is the combined graphs of $x-y+1=0$ and $3x+2y-12=0$. All these lines intersect at one point, $(2,3)$.\nTherefore, the answer is $\\fbox{(B) 1}$.\n\n\n",
"We need to satisfy both $(x+y-5)(2x-3y+5)=0$ and $(x-y+1)(3x+2y-12)=0$. In order to do this, let us look at the first equation. Either $x+y=5,$ or $2x-3y=-5.$ For the second equation, either $x-y=-1,$ or $3x+2y=12.$ Thus, we need to solve 4 systems of equations: $x+y=5$ and $x-y=-1$, $x+y=5$ and $3x+2y=12$, $2x-3y=-5$ and $x-y=-1$, and finally $2x-3y=-5$ and $3x+2y=12.$ Solving all of these systems of equations is pretty trivial, and all of them come out to be $(2,3).$ Thus, they only intersect at $1$ point, and our answer is $\\fbox{(B) 1}$.\n\n\n~SirAppel\n\n\n"
] | 2 | ./CreativeMath/AHSME/1970_AHSME_Problems/26.json | AHSME |
1970_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | [asy] draw((0,0)--(2,2)--(5/2,1/2)--(2,0)--cycle,dot); MP("A",(0,0),W);MP("B",(2,2),N);MP("C",(5/2,1/2),SE);MP("D",(2,0),S); MP("a",(1,0),N);MP("b",(17/8,1/8),N); [/asy]
In the accompanying figure, segments $AB$ and $CD$ are parallel, the measure of angle $D$ is twice that of angle $B$, and the measures of segments $AD$ and $CD$ are $a$ and $b$ respectively. Then the measure of $AB$ is equal to
$\text{(A) } \tfrac{1}{2}a+2b\quad \text{(B) } \tfrac{3}{2}b+\tfrac{3}{4}a\quad \text{(C) } 2a-b\quad \text{(D) } 4b-\tfrac{1}{2}a\quad \text{(E) } a+b$
| [
"With reference to the diagram above, let $E$ be the point on $AB$ such that $DE||BC$. Let $\\angle ABC=\\alpha$. We then have $\\alpha =\\angle AED = \\angle EDC$ since $AB||CD$, so $\\angle ADE=\\angle ADC-\\angle BDC=2\\alpha-\\alpha = \\alpha$, which means $\\triangle AED$ is isosceles. \n\n\nTherefore, $AB=AE+EB=a+b$, hence our answer is $\\fbox{E}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/30.json | AHSME |
1970_AHSME_Problems | 31 | 0 | Probability | Multiple Choice | If a number is selected at random from the set of all five-digit numbers in which the sum of the digits is equal to 43, what is the probability that this number will be divisible by 11?
$\text{(A) } \frac{2}{5}\quad \text{(B) } \frac{1}{5}\quad \text{(C) } \frac{1}{6}\quad \text{(D) } \frac{1}{11}\quad \text{(E) } \frac{1}{15}$
| [
"For the sums of the digits of be equal to 43, this means that the number's digits must fall under one of two categories. It either has 4 9's and a 7, or 3 9's and 2 8's. You can use casework from there to find that there are 15 5-digit numbers in which the sum of the digits is equal to 43. To find the number of ways that the number is divisible by 11, use the divisibility rule for 11, letting the number be represented by $abcde$. This gives 3 possible numbers that are divisible by 11, so the answer is $\\frac{3}{15}=\\frac{1}{5}=\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/31.json | AHSME |
1970_AHSME_Problems | 27 | 0 | Geometry | Multiple Choice | In a triangle, the area is numerically equal to the perimeter. What is the radius of the inscribed circle?
$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 4\quad \text{(D) } 5\quad \text{(E) } 6$
| [
"One of the most common formulas involving the inradius of a triangle is $A = rs$, where $A$ is the area of the triangle, $r$ is the inradius, and $s$ is the semiperimeter.\n\n\nThe problem states that $A = p = 2s$. This means $2s = rs$, or $r = 2$, which is option $\\fbox{A}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/27.json | AHSME |
1970_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | The fourth power of $\sqrt{1+\sqrt{1+\sqrt{1}}}$ is
$\text{(A) } \sqrt{2}+\sqrt{3}\quad \text{(B) } \tfrac{1}{2}(7+3\sqrt{5})\quad \text{(C) } 1+2\sqrt{3}\quad \text{(D) } 3\quad \text{(E) } 3+2\sqrt{2}$
| [
"We can simplify the expression to $\\sqrt{1+\\sqrt{2}}$ Then we square it so it is now $1+\\sqrt{2}$. We still have to square it one more time, getting $\\boxed{(E)\\ 3+2\\sqrt{2}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/1.json | AHSME |
1970_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | If two factors of $2x^3-hx+k$ are $x+2$ and $x-1$, the value of $|2h-3k|$ is
$\text{(A) } 4\quad \text{(B) } 3\quad \text{(C) } 2\quad \text{(D) } 1\quad \text{(E) } 0$
| [
"From the Remainder Theorem, we have $2(-2)^3 - h(-2) + k = 0$ and $2(1)^3 - h(1) + k = 0$. Simplifying both of those equations gives $-16 + 2h + k = 0$ and $2 - h + k = 0$. Since $k = 16 - 2h$ and $k = h - 2$, we set those equal to get:\n\n\n$16 - 2h = h - 2$\n\n\n$3h = 18$\n\n\n$h = 6$\n\n\nThis gives $k = 4$ when substituting back into either of the two equations, and $|2h - 3k| = 0$, which is answer $\\fbox{E}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/11.json | AHSME |
1970_AHSME_Problems | 2 | 0 | Geometry | Multiple Choice | A square and a circle have equal perimeters. The ratio of the area of the circle to the are of the square is
$\text{(A) } \frac{4}{\pi}\quad \text{(B) } \frac{\pi}{\sqrt{2}}\quad \text{(C) } \frac{4}{1}\quad \text{(D) } \frac{\sqrt{2}}{\pi}\quad \text{(E) } \frac{\pi}{4}$
| [
"Let's say the circle has a circumference ( or perimeter of $4\\pi$). Since the perimeter of the square is the same as the perimeter of the circle, the side length of the square is $\\pi$. That means that the area of the square is $\\pi^2$. The area of the circle is $4\\pi$. So the area of the circle over the area of the square is $\\frac{4\\pi}{\\pi^2} \\Rightarrow \\frac{4}{\\pi} \\Rightarrow$ $\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/2.json | AHSME |
1970_AHSME_Problems | 28 | 0 | Geometry | Multiple Choice | In triangle $ABC$, the median from vertex $A$ is perpendicular to the median from vertex $B$. If the lengths of sides $AC$ and $BC$ are $6$ and $7$ respectively, then the length of side $AB$ is
$\text{(A) } \sqrt{17}\quad \text{(B) } 4\quad \text{(C) } 4\tfrac{1}{2}\quad \text{(D) } 2\sqrt{5}\quad \text{(E) } 4\tfrac{1}{4}$
| [
"$\\fbox{A}$\nlet the midpoint be M,N ( i.e. AM,BN are the medians); connecting MN we know that AB = 2x and MN = x hence apply stewart's theorem in triangle ABC with median MN first and then apply stewart's in triangle BNC with median MN\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/28.json | AHSME |
1970_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | A circle with radius $r$ is tangent to sides $AB,AD$ and $CD$ of rectangle $ABCD$ and passes through the midpoint of diagonal $AC$. The area of the rectangle, in terms of $r$, is
$\text{(A) } 4r^2\quad \text{(B) } 6r^2\quad \text{(C) } 8r^2\quad \text{(D) } 12r^2\quad \text{(E) } 20r^2$
| [
"$\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/12.json | AHSME |
1970_AHSME_Problems | 32 | 0 | Algebra | Multiple Choice | $A$ and $B$ travel around a circular track at uniform speeds in opposite directions, starting from diametrically opposite points. If they start at the same time, meet first after $B$ has travelled $100$ yards, and meet a second time $60$ yards before $A$ completes one lap, then the circumference of the track in yards is
$\text{(A) } 400\quad \text{(B) } 440\quad \text{(C) } 480\quad \text{(D) } 560\quad \text{(E) } 880$
| [
"$\\fbox{C}$\n\n\nLet $x$ be half the circumference of the track. They first meet after $B$ has run $100$ yards, meaning that in the time $B$ has run $100$ yards, $A$ has run $x-100$ yards. The second time they meet is when $A$ is 60 yards before he completes the lap. This means that in the time that $A$ has run $2x-60$ yards, $B$ has run $x+60$ yards. \nBecause they run at uniform speeds, we can write the equation\n\\[\\frac{100}{x-100}=\\frac{x+60}{2x-60} .\\]\nCross multiplying,\n\\[200x-6000=x^2-40x-6000\\]\nAdding $6000$ to both sides and simplifying, we have\n\\[200x=x^2-40x\\]\n\\[240x=x^2\\]\n\\[x=240.\\]\nBecause $x$ is only half of the circumference of the track, the answer we are looking for is $2 \\cdot 240 = 480, \\text{or } \\fbox{C}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/32.json | AHSME |
1970_AHSME_Problems | 24 | 0 | Geometry | Multiple Choice | An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is $2$, then the area of the hexagon is
$\text{(A) } 2\quad \text{(B) } 3\quad \text{(C) } 4\quad \text{(D) } 6\quad \text{(E) } 12$
| [
"Let $ABCDEF$ be our regular hexagon, with centre $O$ - and join $AO, BO, CO, DO, EO,$ and $FO$. Note that we form six equilateral triangles with sidelength $\\frac{s}{2}$, where $s$ is the sidelength of the triangle (since the perimeter of the two polygons are equal). If the area of the original equilateral triangle is $2$, then the area of each of the six smaller triangles is $1/4\\times 2 = \\frac{1}{2}$ (by similarity area ratios). \n\n\nThus, the area of the hexagon is $6\\times \\frac{1}{2}=3$, hence our answer is $\\fbox{B}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/24.json | AHSME |
1970_AHSME_Problems | 25 | 0 | Number Theory | Multiple Choice | For every real number $x$, let $[x]$ be the greatest integer which is less than or equal to $x$. If the postal rate for first class mail is six cents for every ounce or portion thereof, then the cost in cents of first-class postage on a letter weighing $W$ ounces is always
$\text{(A) } 6W\quad \text{(B) } 6[W]\quad \text{(C) } 6([W]-1)\quad \text{(D) } 6([W]+1)\quad \text{(E) } -6[-W]$
| [
"This question is trying to convert the floor function, which is more commonly notated as $\\lfloor x \\rfloor$, into the ceiling function, which is $\\lceil x \\rceil$. The identity is $\\lceil x \\rceil = -\\lfloor -x \\rfloor$, which can be verified graphically, or proven using the definition of floor and ceiling functions.\n\n\nHowever, for this problem, some test values will eliminate answers. If $W = 2.5$ ounces, the cost will be $18$ cents. Plugging in $W = 2.5$ into the five options gives answers of $15, 12, 6, 18, 18$. This leaves options $D$ and $E$ as viable. If $W = 2$ ounces, the cost is $12$ cents. Option $D$ remains $18$ cents, while option $E$ gives $12$ cents, the correct answer. Thus, the answer is $\\fbox{E}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/25.json | AHSME |
1970_AHSME_Problems | 33 | 0 | Arithmetic | Multiple Choice | Find the sum of digits of all the numbers in the sequence $1,2,3,4,\cdots ,10000$.
$\text{(A) } 180001\quad \text{(B) } 154756\quad \text{(C) } 45001\quad \text{(D) } 154755\quad \text{(E) } 270001$
| [
"We can find the sum using the following method. We break it down into cases. The first case is the numbers $1$ to $9$. The second case is the numbers $10$ to $99$. The third case is the numbers $100$ to $999$. The fourth case is the numbers $1,000$ to $9,999$. And lastly, the sum of the digits in $10,000$. The first case is just the sum of the numbers $1$ to $9$ which is, using $\\frac{n(n+1)}{2}$, $45$. In the second case, every number $1$ to $9$ is used $19$ times. $10$ times in the tens place, and $9$ times in the ones place. So the sum is just $19(45)$. Similarly, in the third case, every number $1$ to $9$ is used $100$ times in the hundreds place, $90$ times in the tens place, and $90$ times in the ones place, for a total sum of $280(45)$. By the same method, every number $1$ to $9$ is used $1,000$ times in the thousands place, $900$ times in the hundreds place, $900$ times in the tens place, and $900$ times in the ones place, for a total of $3700(45)$. Thus, our final sum is $45+19(45)+280(45)+3700(45)+1=4000(45)+1=\\boxed{\\text{A)}180001}.$",
"Consider the numbers from $0000-9999$. We have $40000$ digits and each has equal an probability of being $0,1,2....9$.\nThus, our desired sum is $4000\\left( \\frac{9 \\cdot 10}{2} \\right)+1=4000(45)+1=\\boxed{\\text{A)}180001}.$\n\\textit{Credit: Math1331Math}",
"As in Solution 2, we consider the four digit numbers from $0000-9999.$ We see that we have $10000\\times4=40000$ digits with each digit appearing equally.\n\n\nThus, the digit sum will be the average of the digits multiplied by $40000.$ The digit average comes out to be $\\frac{0+9}{2},$ since all digits are consecutive.\n\n\nSo, our answer will be $\\frac{9}{2} \\times 40000 = 180000.$ However, since we purposely did not include $10000,$ we add one to get our final answer as $\\boxed{\\textbf{(A)}180001}.$\n\n\nSolution by davidaops.\n\n\n\\textit{Credit to Math1331Math}\n\n\n"
] | 3 | ./CreativeMath/AHSME/1970_AHSME_Problems/33.json | AHSME |
1970_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | Given the binary operation $\star$ defined by $a\star b=a^b$ for all positive numbers $a$ and $b$. Then for all positive $a,b,c,n$, we have
$\text{(A) } a\star b=b\star a\quad\qquad\qquad\ \text{(B) } a\star (b\star c)=(a\star b) \star c\quad\\ \text{(C) } (a\star b^n)=(a \star n) \star b\quad \text{(D) } (a\star b)^n =a\star (bn)\quad\\ \text{(E) None of these}$
| [
"Let $a = 2, b = 3, c=n = 4$. If all of them are false, the answer must be $E$. If one does not fail, we will try to prove it.\n\n\nFor option $A$, we have $2^3 = 3^2$, which is clearly false.\n\n\nFor option $B$, we have $2^{81} = 8^{4}$, which is false.\n\n\nFor option $C$, we have $2^{81} = 16^3$, which is false.\n\n\nFor option $D$, we have $8^4 = 2^{12}$, which is true.\n\n\nThe LHS is $(a^b)^n$. By the elementary definition of exponentiation, this is $a^b$ multiplied by itself $n$ times. Since each $a^b$ is actually $a$ multiplied $b$ times, the expression $(a^b)^n$ is $a$ multiplied by itself $bn$ times. \n\n\nThe RHS is $a^{bn}$. This is $a$ multiplied by itself $bn$ times.\n\n\nThus, the LHS is always equal to the RHS, so $\\fbox{D}$ is the only correct statement.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/13.json | AHSME |
1970_AHSME_Problems | 29 | 0 | Geometry | Multiple Choice | It is now between 10:00 and 11:00 o'clock, and six minutes from now, the minute hand of a watch will be exactly opposite the place where the hour hand was three minutes ago. What is the exact time now?
$\text{(A) } 10:05\tfrac{5}{11}\quad \text{(B) } 10:07\tfrac{1}{2}\quad \text{(C) } 10:10\quad \text{(D) } 10:15\quad \text{(E) } 10:17\tfrac{1}{2}$
| [
"$\\fbox{D}$\n50 + (m-3)/12 = 30 + (m+6)\n= 10hours + m = 10: 15\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/29.json | AHSME |
1970_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | If $x=1+2^p$ and $y=1+2^{-p}$, then $y$ in terms of $x$ is
$\text{(A) } \frac{x+1}{x-1}\quad \text{(B) } \frac{x+2}{x-1}\quad \text{(C) } \frac{x}{x-1}\quad \text{(D) } 2-x\quad \text{(E) } \frac{x-1}{x}$
| [
"Since we want the $y$ expression in terms of $x$, let's convert the $y$ expression. We can convert it to $1+ \\frac{1}{2^p} \\Rightarrow \\frac{2^p+1}{2^p} \\Rightarrow \\frac{x}{x-1} \\Rightarrow$ $\\fbox{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/3.json | AHSME |
1970_AHSME_Problems | 34 | 0 | Number Theory | Multiple Choice | The greatest integer that will divide $13511$, $13903$ and $14589$ and leave the same remainder is
$\text{(A) } 28\quad \text{(B) } 49\quad \text{(C) } 98\quad\\ \text{(D) an odd multiple of } 7 \text{ greater than } 49\quad\\ \text{(E) an even multiple of } 7 \text{ greater than } 98$
\section{Solution}
We know that 13903 minus 13511 is equivalent to 392. Additionally, 14589 minus 13903 is equivalent to 686. Since we are searching for the greatest integer that divides these three integers and leaves the same remainder, the answer resides in the greatest common factor of 686 and 392. Therefore, the answer is 98, or $\fbox{C}$
"Credit: Skupp3"
| [
"We know that 13903 minus 13511 is equivalent to 392. Additionally, 14589 minus 13903 is equivalent to 686. Since we are searching for the greatest integer that divides these three integers and leaves the same remainder, the answer resides in the greatest common factor of 686 and 392. Therefore, the answer is 98, or $\\fbox{C}$\n\n\n\"Credit: Skupp3\"\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/34.json | AHSME |
1970_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | If $a=\log_8 225$ and $b=\log_2 15$, then
$\text{(A) } a=b/2\quad \text{(B) } a=2b/3\quad \text{(C) } a=b\quad \text{(D) } b=a/2\quad \text{(E) } a=3b/2$
| [
"The solutions imply that finding the ratio $\\frac{a}{b}$ will solve the problem. We compute $\\frac{a}{b}$, use change-of-base to a neutral base, rearrange the terms, and then use the reverse-change-of-base:\n\n\n$\\frac{\\log_8 225}{\\log_2 15}$\n\n\n$\\frac{\\frac{\\ln 225}{\\ln 8}}{\\frac{\\ln 15}{\\ln 2}}$\n\n\n$\\frac{\\ln 225 \\ln 2}{\\ln 15 \\ln 8}$\n\n\n$\\frac{\\ln 225}{\\ln 15} \\cdot \\frac{\\ln 2}{\\ln 8}$\n\n\n$\\ln_{15} 225 \\cdot \\ln_8 2$\n\n\nSince $15^2 = 225$, the first logarithm is $2$. Since $8^{\\frac{1}{3}} = 2$, the second logarithm is $\\frac{1}{3}$.\n\n\nThus, we have $\\frac{a}{b} = \\frac{2}{3}$, or $a = \\frac{2}{3}b$, which is option $\\fbox{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/8.json | AHSME |
1970_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | If the sum of the first $3n$ positive integers is $150$ more than the sum of the first $n$ positive integers, then the sum of the first $4n$ positive integers is
$\text{(A) } 300\quad \text{(B) } 350\quad \text{(C) } 400\quad \text{(D) } 450\quad \text{(E) } 600$
| [
"We can setup our first equation as\n\n\n$\\frac{3n(3n+1)}{2} = \\frac{n(n+1)}{2} + 150$\n\n\nSimplifying we get\n\n\n$9n^2 + 3n = n^2 + n + 300 \\Rightarrow 8n^2 + 2n - 300 = 0 \\Rightarrow 4n^2 + n - 150 = 0$\n\n\nSo our roots using the quadratic formula are\n\n\n$\\dfrac{-b\\pm\\sqrt{b^2 - 4ac}}{2a} \\Rightarrow \\dfrac{-1\\pm\\sqrt{1^2 - 4\\cdot(-150)\\cdot4}}{2\\cdot4} \\Rightarrow \\dfrac{-1\\pm\\sqrt{1+2400}}{8} \\Rightarrow 6, -25/4$\n\n\nSince the question said positive integers, $n = 6$, so $4n = 24$\n\n\n$\\frac{24\\cdot 25}{2} = 300$\n\n\n$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/22.json | AHSME |
1970_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | $\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}$ is equal to
$\text{(A) } 2\quad \text{(B) } 2\sqrt{3}\quad \text{(C) } 4\sqrt{2}\quad \text{(D) } \sqrt{6}\quad \text{(E) } 2\sqrt{2}$
| [
"Square the expression:\n\n\n$(\\sqrt{3+2\\sqrt{2}}-\\sqrt{3-2\\sqrt{2}})^2=3+\\sqrt{2}-2\\sqrt{(3+\\sqrt{2})(3-\\sqrt{2})}+3-2\\sqrt{2}=6-2\\sqrt{9-8}=6-2\\sqrt{1}=4$\n\n\n$\\Rightarrow\\sqrt{3+2\\sqrt{2}}-\\sqrt{3-2\\sqrt{2}}=\\sqrt{4}=2\\Rightarrow\\boxed{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/18.json | AHSME |
1970_AHSME_Problems | 4 | 0 | Number Theory | Multiple Choice | Let $S$ be the set of all numbers which are the sum of the squares of three consecutive integers. Then we can say that
$\text{(A) No member of S is divisible by } 2\quad\\ \text{(B) No member of S is divisible by } 3 \text{ but some member is divisible by } 11\quad\\ \text{(C) No member of S is divisible by } 3 \text{ or } 5\quad\\ \text{(D) No member of S is divisible by } 3 \text{ or } 7 \quad\\ \text{(E) None of these}$
| [
"Consider $3$ consecutive integers $a, b,$ and $c$. Exactly one of these integers must be divisible by 3; WLOG, suppose $a$ is divisible by 3. Then $a \\equiv 0 \\pmod {3}, b \\equiv 1 \\pmod{3},$ and $c \\equiv 2 \\pmod{3}$. Squaring, we have that $a^{2} \\equiv 0 \\pmod{3}, b^{2} \\equiv 1 \\pmod{3},$ and $c^{2} \\equiv 1 \\pmod{3}$, so $a^{2} + b^{2} + c^{2} \\equiv 2 \\pmod{3}$. Therefore, no member of $S$ is divisible by 3.\n\n\nNow consider $3$ more consecutive integers $a, b,$ and $c$, which we will consider mod 11. We will assign $k$ such that $a \\equiv k \\pmod{11}, b \\equiv k + 1 \\pmod{11},$ and $c \\equiv k + 2 \\pmod{11}$. Some experimentation shows that when $k = 4, a \\equiv 4 \\pmod{11}$ so $a^{2} \\equiv 5 \\pmod{11}$. Similarly, $b \\equiv 5 \\pmod{11}$ so $b^{2} \\equiv 3 \\pmod{11}$, and $c \\equiv 6 \\pmod{11}$ so $c^{2} \\equiv 3 \\pmod{11}$. Therefore, $a^{2} + b^{2} + c^{2} \\equiv 0 \\pmod{11}$, so there is at least one member of $S$ which is divisible by 11. Thus, $\\fbox{B}$ is correct.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/4.json | AHSME |
1970_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | Consider $x^2+px+q=0$, where $p$ and $q$ are positive numbers. If the roots of this equation differ by 1, then $p$ equals
$\text{(A) } \sqrt{4q+1}\quad \text{(B) } q-1\quad \text{(C) } -\sqrt{4q+1}\quad \text{(D) } q+1\quad \text{(E) } \sqrt{4q-1}$
| [
"From the quadratic equation, the two roots of the equation are $\\frac{-p\\pm\\sqrt{p^2-4q}}{2}$. The positive difference between these roots is $\\sqrt{p^2 - 4q}$. Setting $\\sqrt{p^2-4q}=1$ and isolating $p$ gives $\\sqrt{4q+1}$, or choice $\\boxed{\\text{(A)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/14.json | AHSME |
1970_AHSME_Problems | 15 | 0 | Geometry | Multiple Choice | Lines in the $xy$-plane are drawn through the point $(3,4)$ and the trisection points of the line segment joining the points $(-4,5)$ and $(5,-1)$. One of these lines has the equation
$\text{(A) } 3x-2y-1=0\quad \text{(B) } 4x-5y+8=0\quad \text{(C) } 5x+2y-23=0\quad\\ \text{(D) } x+7y-31=0\quad \text{(E) } x-4y+13=0$
| [
"The trisection points of $(-4, 5)$ and $(5, -1)$ can be found by trisecting the x-coordinates and the y-coordinates separately. The difference of the x-coordinates is $9$, so the trisection points happen at $-4 + \\frac{9}{3}$ and $-4 + \\frac{9}{3} + \\frac{9}{3}$, which are $-1$ and $2$. Similarly, the y-coordinates have a difference of $6$, so the trisections happen at $3$ and $1$. So, the two points are $(-1, 3)$ and $(2, 1)$.\n\n\nWe now check which line has both $(3, 4)$ and one of the two trisection points on it. Plugging in $(x, y) = (3, 4)$ into all five of the equations works. The point $(2, 1)$ doesn't work in any of the five lines. However, $(-1, 3)$ works in line $E$. Thus, the answer is $\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/15.json | AHSME |
1970_AHSME_Problems | 5 | 0 | Algebra | Multiple Choice | If $f(x)=\frac{x^4+x^2}{x+1}$, then $f(i)$, where $i=\sqrt{-1}$, is equal to
$\text{(A) } 1+i\quad \text{(B) } 1\quad \text{(C) } -1\quad \text{(D) } 0\quad \text{(E) } -1-i$
| [
"$i^4 = 1$ and $i^2=-1$, so the numerator is $0$. As long as the denominator is not $0$, which it isn't, the answer is $0 \\Rightarrow$ $\\fbox{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/5.json | AHSME |
1970_AHSME_Problems | 19 | 0 | Algebra | Multiple Choice | The sum of an infinite geometric series with common ratio $r$ such that $|r|<1$ is $15$, and the sum of the squares of the terms of this series is $45$. The first term of the series is
$\textbf{(A) } 12\quad \textbf{(B) } 10\quad \textbf{(C) } 5\quad \textbf{(D) } 3\quad \textbf{(E) 2}$
| [
"We know that the formula for the sum of an infinite geometric series is $S = \\frac{a}{1-r}$.\n\n\nSo we can apply this to the conditions given by the problem. \n\n\nWe have two equations:\n\n\n\\begin{align*} 15 &= \\frac{a}{1-r} \\\\ 45 &= \\frac{a^{2}}{1-r^{2}} \\end{align*}\n\n\nWe get\n\n\n\\begin{align*} a &= 15 - 15r \\\\ a^{2} &= 45 - 45r^{2} \\\\ \\\\ (15 - 15r)^{2} &= 45 - 45r^{2} \\\\ 270r^{2} - 450r + 180 &= 0 \\\\ 3r^{2} - 5r + 2 &= 0 \\\\ (3r - 2)(r - 1) &= 0 \\end{align*}\n\n\nSince $|r| < 1$, $r = \\frac{2}{3}$, so plug this into the equation above and we get $a = 15 - 15r = 15 - 10 = \\boxed{\\textbf{(C)} \\quad 5}$\n\n\nSolution by $\\underline{\\textbf{Invoker}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/19.json | AHSME |
1970_AHSME_Problems | 23 | 0 | Number Theory | Multiple Choice | The number $10!$ ($10$ is written in base $10$), when written in the base $12$ system, ends with exactly $k$ zeros. The value of $k$ is
$\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$
| [
"A number in base $b$ that ends in exactly $k$ zeros will be divisible by $b^k$, but not by $b^{k+1}$. Thus, we want to find the highest $k$ for which $12^k | 10!$.\n\n\nThere are $4$ factors of $3$: $3, 6, 9$, and an extra factor from $9$.\n\n\nThere are $8$ factors of $2$: $2, 4, 6, 8, 10$, an extra factor from $4, 8$, and a third extra factor from $8$.\n\n\nSo, $2^8 \\cdot 3^4 = 4^4 \\cdot 3^4 = 12^4$ will divide $10!$. Thus, the answer is $\\fbox{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/23.json | AHSME |
1970_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | Points $P$ and $Q$ are on line segment $AB$, and both points are on the same side of the midpoint of $AB$. Point $P$ divides $AB$ in the ratio $2:3$, and $Q$ divides $AB$ in the ratio $3:4$. If $PQ$=2, then the length of segment $AB$ is
$\text{(A) } 12\quad \text{(B) } 28\quad \text{(C) } 70\quad \text{(D) } 75\quad \text{(E) } 105$
| [
"In order, the points from left to right are $A, P, Q, B$. Let the lengths between successive points be $x, 2, y$, respectively. \n\n\nSince $\\frac{AP}{PB} = \\frac{2}{3}$, we have $\\frac{x}{2 + y} = \\frac{2}{3}$.\n\n\nSince $\\frac{AQ}{QB} = \\frac{3}{4}$, we have $\\frac{x + 2}{y} = \\frac{3}{4}$.\n\n\nThe first equation gives $x = \\frac{2}{3}(2 + y)$. Plugging that in to the second equation gives:\n\n\n$\\frac{\\frac{2}{3}(2 + y) + 2}{y} = \\frac{3}{4}$.\n\n\n$\\frac{2}{3}(2 + y) + 2 = \\frac{3}{4}y$\n\n\n$\\frac{4}{3} + \\frac{2}{3}y + 2 = \\frac{3}{4}y$\n\n\n$\\frac{10}{3} = \\frac{1}{12}y$\n\n\n$y = 40$\n\n\nSince $x = \\frac{2}{3}(2 + y)$, we have $x = 28$. The length of the entire segment is $x + 2 + y$, which is $70$, or option $\\fbox{C}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/9.json | AHSME |
1970_AHSME_Problems | 35 | 0 | Algebra | Multiple Choice | A retiring employee receives an annual pension proportional to the square root of the number of years of his service. Had he served $a$ years more, his pension would have been $p$ dollars greater, whereas had he served $b$ years more $(b\ne a)$, his pension would have been $q$ dollars greater than the original annual pension. Find his annual pension in terms of $a,b,p$ and $q$.
$\text{(A) } \frac{p^2-q^2}{2(a-b)}\quad \text{(B) } \frac{(p-q)^2}{2\sqrt{ab}}\quad \text{(C) } \frac{ap^2-bq^2}{2(ap-bq)}\quad \text{(D) } \frac{aq^2-bp^2}{2(bp-aq)}\quad \text{(E) } \sqrt{(a-b)(p-q)}$
| [
"Note the original pension as $k\\sqrt{x}$, where $x$ is the number of years served. Then, based on the problem statement, two equations can be set up. \n\n\n\\[k\\sqrt{x+a} = k\\sqrt{x} + p\\]\n\\[k\\sqrt{x+b} = k\\sqrt{x} + q\\]\n\n\nSquare the first equation to get\n\n\n\\[k^2x + ak^2 = k^2x + 2pk\\sqrt{x} + p^2.\\]\nBecause both sides have \\[k^2x\\], they cancel out. Similarly, the second equation will become $bk^2 = q^2 + 2qk\\sqrt{x}.$ Then, $a$ can be multiplied to the second equation and $b$ can be multiplied to the first equation so that the left side of both equations becomes $abk^2$. Finally, by setting the equations equal to each other, \\[bp^2 + 2bpk\\sqrt{x} = aq^2 + 2aqk\\sqrt{x}.\\] Isolate $k\\sqrt{x}$ to get $\\fbox{D} = \\frac{aq^2-bp^2}{2(bp-aq)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1970_AHSME_Problems/35.json | AHSME |
1980_AHSME_Problems | 20 | 0 | Probability | Multiple Choice | A box contains $2$ pennies, $4$ nickels, and $6$ dimes. Six coins are drawn without replacement,
with each coin having an equal probability of being chosen. What is the probability that the value of coins drawn is at least $50$ cents?
$\text{(A)} \ \frac{37}{924} \qquad \text{(B)} \ \frac{91}{924} \qquad \text{(C)} \ \frac{127}{924} \qquad \text{(D)}\ \frac{132}{924}\qquad \text{(E)}\ \text{none of these}$
| [
"We want the number of Successful Outcomes over the number of Total Outcomes. We want to calculate the total outcomes first. Since we have $12$ coins and we need to choose $6$, we have $\\binom{12}{6}$ = $924$ Total outcomes. For our successful outcomes, we can have $(1) 1$ penny and $5$ dimes, $2$ nickels and $4$ dimes, $1$ nickel and $5$ dimes, or $6$ dimes. \n\n\nFor the case of $1$ penny and $5$ dimes, there are $\\binom{6}{5}$ ways to choose the dimes and $2$ ways to choose the pennies. That is $6 \\cdot 2 = 12$ successful outcomes. For the case of $2$ nickels and $4$ dimes, we have $\\binom{6}{4}$ ways to choose the dimes and $\\binom{4}{2}$ ways to choose the nickels. We have $15 \\cdot 6$ = $90$ successful outcomes. For the case of $1$ nickel and $5$ dimes, we have $\\binom{4}{1} \\cdot \\binom{6}{5} = 24$. Lastly, we have $6$ dimes and $0$ nickels, and $0$ pennies, so we only have one case. Therefore, we have $\\dfrac {12 + 90 + 24 + 1}{924} = \\dfrac{127}{924}$ = $\\boxed{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/20.json | AHSME |
1980_AHSME_Problems | 16 | 0 | Geometry | Multiple Choice | Four of the eight vertices of a cube are the vertices of a regular tetrahedron. Find the ratio of the surface area of the cube to the surface area of the tetrahedron.
$\text{(A)} \ \sqrt 2 \qquad \text{(B)} \ \sqrt 3 \qquad \text{(C)} \ \sqrt{\frac{3}{2}} \qquad \text{(D)} \ \frac{2}{\sqrt{3}} \qquad \text{(E)} \ 2$
| [
"We assume the side length of the cube is $1$. The side length of the tetrahedron is $\\sqrt2$, so the surface area is $4\\times\\frac{2\\sqrt3}{4}=2\\sqrt3$. The surface area of the cube is $6\\times1\\times1=6$, so the ratio of the surface area of the cube to the surface area of the tetrahedron is $\\frac{6}{2\\sqrt3}=\\boxed{\\sqrt3}$.\n\n\n-aopspandy\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/16.json | AHSME |
1980_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | A positive number $x$ satisfies the inequality $\sqrt{x} < 2x$ if and only if
$\text{(A)} \ x > \frac{1}{4} \qquad \text{(B)} \ x > 2 \qquad \text{(C)} \ x > 4 \qquad \text{(D)} \ x < \frac{1}{4}\qquad \text{(E)} \ x < 4$
| [
"$\\sqrt{x}< 2x \\\\ x < 4x^2 \\\\ 0 < x(4x-1) \\\\ 0 < 4x-1 \\\\ 1 < 4x \\\\ x >\\frac{1}{4} \\\\ \\boxed{(A)}$\n\n\nNote: You can also draw a rough sketch.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/6.json | AHSME |
1980_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | Sides $AB,BC,CD$ and $DA$ of convex polygon $ABCD$ have lengths 3, 4, 12, and 13, respectively, and $\angle CBA$ is a right angle. The area of the quadrilateral is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); real r=degrees((12,5)), s=degrees((3,4)); pair D=origin, A=(13,0), C=D+12*dir(r), B=A+3*dir(180-(90-r+s)); draw(A--B--C--D--cycle); markscalefactor=0.05; draw(rightanglemark(A,B,C)); pair point=incenter(A,C,D); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$3$", A--B, dir(A--B)*dir(-90)); label("$4$", B--C, dir(B--C)*dir(-90)); label("$12$", C--D, dir(C--D)*dir(-90)); label("$13$", D--A, dir(D--A)*dir(-90));[/asy]
$\text{(A)} \ 32 \qquad \text{(B)} \ 36 \qquad \text{(C)} \ 39 \qquad \text{(D)} \ 42 \qquad \text{(E)} \ 48$
| [
"Connect C and A, and we have a 3-4-5 right triangle and 5-12-13 right triangle. The area of both is $\\frac{3\\cdot4}{2}+\\frac{5\\cdot12}{2}=36\\Rightarrow\\boxed{(B)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/7.json | AHSME |
1980_AHSME_Problems | 17 | 0 | Algebra | Multiple Choice | Given that $i^2=-1$, for how many integers $n$ is $(n+i)^4$ an integer?
$\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4$
| [
"$(n+i)^4=n^4+4in^3-6n^2-4in+1$, and this has to be an integer, so the sum of the imaginary parts must be $0$. \\[4in^3-4in=0\\] \\[4in^3=4in\\] \\[n^3=n\\]\nSince $n^3=n$, there are $\\boxed{3}$ solutions for $n$: $0$ and $\\pm1$.\n\n\n-aopspandy\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/17.json | AHSME |
1980_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | [asy] defaultpen(linewidth(0.7)+fontsize(10)); pair B=origin, C=(15,3), D=(5,1), A=7*dir(72)*dir(B--C), E=midpoint(A--C), F=intersectionpoint(A--D, B--E); draw(E--B--A--C--B^^A--D); label("$A$", A, dir(D--A)); label("$B$", B, dir(E--B)); label("$C$", C, dir(0)); label("$D$", D, SE); label("$E$", E, N); label("$F$", F, dir(80));[/asy]
In triangle $ABC$, $\measuredangle CBA=72^\circ$, $E$ is the midpoint of side $AC$,
and $D$ is a point on side $BC$ such that $2BD=DC$; $AD$ and $BE$ intersect at $F$.
The ratio of the area of triangle $BDF$ to the area of quadrilateral $FDCE$ is
$\text{(A)} \ \frac 15 \qquad \text{(B)} \ \frac 14 \qquad \text{(C)} \ \frac 13 \qquad \text{(D)}\ \frac{2}{5}\qquad \text{(E)}\ \text{none of these}$
| [
"We can use the principle of same height same area to solve this problem. \n$\\fbox{A}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/21.json | AHSME |
1980_AHSME_Problems | 10 | 0 | Algebra | Multiple Choice | The number of teeth in three meshed gears $A$, $B$, and $C$ are $x$, $y$, and $z$, respectively. (The teeth on all gears are the same size and regularly spaced.) The angular speeds, in revolutions per minutes of $A$, $B$, and $C$ are in the proportion
$\text{(A)} \ x: y: z ~~\text{(B)} \ z: y: x ~~ \text{(C)} \ y: z: x~~ \text{(D)} \ yz: xz: xy ~~ \text{(E)} \ xz: yx: zy$
| [
"The distance that each of the gears rotate is constant. Let us have the number of teeth per minute equal to $k$. The revolutions per minute are in ratio of:\n\\[\\frac{k}{x}:\\frac{k}{y}:\\frac{k}{z}\\]\n\\[yz:xz:xy.\\]\nTherefore, the answer is $\\fbox{D: yz:xz:xy}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/10.json | AHSME |
1980_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | Four balls of radius $1$ are mutually tangent, three resting on the floor and the fourth resting on the others.
A tetrahedron, each of whose edges have length $s$, is circumscribed around the balls. Then $s$ equals
$\text{(A)} \ 4\sqrt 2 \qquad \text{(B)} \ 4\sqrt 3 \qquad \text{(C)} \ 2\sqrt 6 \qquad \text{(D)}\ 1+2\sqrt 6\qquad \text{(E)}\ 2+2\sqrt 6$
| [
"$\\fbox{E}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/26.json | AHSME |
1980_AHSME_Problems | 30 | 0 | Number Theory | Multiple Choice | A six digit number (base 10) is squarish if it satisfies the following conditions:
(i) none of its digits are zero;
(ii) it is a perfect square; and
(iii) the first of two digits, the middle two digits and the last two digits of the number are all perfect squares when considered as two digit numbers.
How many squarish numbers are there?
$\text{(A)} \ 0 \qquad \text{(B)} \ 2 \qquad \text{(C)} \ 3 \qquad \text{(D)} \ 8 \qquad \text{(E)} \ 9$
| [
"N = a^2*10000 + b^2*100 + c^2*1\n\n\n\\begin{verbatim}\n = (a*100 + c)^2\n\\end{verbatim}\nwe get \n\n\n\\begin{verbatim}\n b^2 = 2*a*c\n\\end{verbatim}\nwhere \n4<=a,b,c<=9\n\n\n\n\n\n\n\\begin{verbatim}\n b^2=2*a*c, so \n\\end{verbatim}\n\\begin{verbatim}\n a=2*2, c=2*2*2, b=8\n a=2*3, c=3 (NO)\n a=2*2*2, c=3*3, b=3*4=12 (NO)\n\\end{verbatim}\nTherefore, there are two solutions (4,8,8) or (8,8,4)\n\n\nWwei.yu (talk)Wei\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/30.json | AHSME |
1980_AHSME_Problems | 27 | 0 | Algebra | Multiple Choice | The sum $\sqrt[3] {5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}$ equals
$\text{(A)} \ \frac 32 \qquad \text{(B)} \ \frac{\sqrt[3]{65}}{4} \qquad \text{(C)} \ \frac{1+\sqrt[6]{13}}{2} \qquad \text{(D)}\ \sqrt[3]{2}\qquad \text{(E)}\ \text{none of these}$
| [
"Lets set our original expression equal to $x$. So $\\sqrt[3] {5+2\\sqrt{13}}+\\sqrt[3]{5-2\\sqrt{13}} = x$. Cubing this gives us $x^3 = \\left(\\sqrt[3] {5+2\\sqrt{13}}+\\sqrt[3]{5-2\\sqrt{13}}\\right)^3 = 5 + 2\\sqrt{13} + 5 - 2\\sqrt{13} + 3\\left(\\sqrt[3] {5+2\\sqrt{13}}*\\sqrt[3]{5-2\\sqrt{13}}\\right)\\left(\\sqrt[3] {5+2\\sqrt{13}}+\\sqrt[3]{5-2\\sqrt{13}}\\right) = 10 - 9x$\nSo we have $x^3 + 9x - 10 = 0$. We can easily see that 1 is a root of this polynomial. By synthetic division, the new polynomial is $x^2 + x + 10$, which has no real roots. Thus $\\sqrt[3] {5+2\\sqrt{13}}+\\sqrt[3]{5-2\\sqrt{13}} = 1$. Since 1 is not A-D, our answer is $\\fbox{E}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/27.json | AHSME |
1980_AHSME_Problems | 1 | 0 | Algebra | Multiple Choice | The largest whole number such that seven times the number is less than 100 is
$\text{(A)} \ 12 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 14 \qquad \text{(D)} \ 15 \qquad \text{(E)} \ 16$
| [
"We want to find the smallest integer $x$ so that $7x < 100$. Dividing by 7 gets $x < 14\\dfrac{2}{7}$, so the answer is 14. $\\boxed{(C)}$\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/1.json | AHSME |
1980_AHSME_Problems | 11 | 0 | Algebra | Multiple Choice | If the sum of the first $10$ terms and the sum of the first $100$ terms of a given arithmetic progression are $100$ and $10$,
respectively, then the sum of first $110$ terms is:
$\text{(A)} \ 90 \qquad \text{(B)} \ -90 \qquad \text{(C)} \ 110 \qquad \text{(D)} \ -110 \qquad \text{(E)} \ -100$
| [
"Let $a$ be the first term of the sequence and let $d$ be the common difference of the sequence.\n\n\nSum of the first 10 terms: $\\frac{10}{2}(2a+9d)=100 \\Longleftrightarrow 2a+9d=20$\nSum of the first 100 terms: $\\frac{100}{2}(2a+99d)=10 \\Longleftrightarrow 2a+99d=\\frac{1}{5}$\n\n\nSolving the system, we get $d=-\\frac{11}{50}$, $a=\\frac{1099}{100}$. The sum of the first 110 terms is $\\frac{110}{2}(2a+109d)=55(-2)=-110$\n\n\nTherefore, $\\boxed{D}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/11.json | AHSME |
1980_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | The degree of $(x^2+1)^4 (x^3+1)^3$ as a polynomial in $x$ is
$\text{(A)} \ 5 \qquad \text{(B)} \ 7 \qquad \text{(C)} \ 12 \qquad \text{(D)} \ 17 \qquad \text{(E)} \ 72$
| [
"It becomes $(x^{8}+...)(x^{9}+...)$ with 8 being the degree of the first factor and 9 being the degree of the second factor, making the degree of the whole thing 17, or $\\boxed{(D)}$\n\n\n",
"First note that given a polynomial $P(x)$ and a polynomial $Q(x)$:\n\n\n\n\n$deg(P(x))^n = ndeg(P(x))$ and $deg(P(x)Q(x)) = deg(P(x))+deg(Q(x))$.\n\n\n\n\n\n\nWe let $x^2+1=P(x)$ and $x^3+1 = Q(x)$.\n\n\nHence $deg(P(x)) = 2$ and $deg(Q(x)) = 3$\n\n\nSo $deg(P(x))^4) = 4\\cdot2 = 8$ and $deg(Q(x)^3) = 3\\cdot3 = 9$\n\n\n\n\nNow we let $P(x)^4 = R(x)$ and $Q(x)^3 = S(x)$\n\n\nWe want to find $deg(R(x)S(x)) = deg(R(x))+deg(S(x)) = 9+8 = 17$. \n\n\nSo the answer is \\textbf{(D)} 17.\n\n\n\\begin{itemize}\n\\item Solution 2 by mihirb\n\\end{itemize}\n"
] | 2 | ./CreativeMath/AHSME/1980_AHSME_Problems/2.json | AHSME |
1980_AHSME_Problems | 28 | 0 | Algebra | Multiple Choice | The polynomial $x^{2n}+1+(x+1)^{2n}$ is not divisible by $x^2+x+1$ if $n$ equals
$\text{(A)} \ 17 \qquad \text{(B)} \ 20 \qquad \text{(C)} \ 21 \qquad \text{(D)} \ 64 \qquad \text{(E)} \ 65$
| [
"Let $h(x)=x^2+x+1$.\n\n\nThen we have\n\\[(x+1)^2n = (x^2+2x+1)^n = (h(x)+x)^n = g(x) \\cdot h(x) + x^n,\\]\nwhere $g(x)$ is $h^{n-1}(x) + nh^{n-2}(x) \\cdot x + ... + x^{n-1}$ (after expanding $(h(x)+x)^n$ according to the Binomial Theorem).\n\n\nNotice that\n\\[x^2n = x^2n+x^{2n-1}+x^{2n-2}+...x -x^{2n-1}-x^{2n-2}-x^{2n-3} +...\\]\n\n\n$x^n = x^n+x^{n-1}+x^{n-2} -x^{n-1}-x^{n-2}-x^{n-3} +....$\n\n\nTherefore, the left term from $x^2n$ is $x^{(2n-3u)}$\n\n\n\\begin{verbatim}\n the left term from $x^n$ is $x^{(n-3v)}$, \n\\end{verbatim}\nIf divisible by h(x), we need 2n-3u=1 and n-3v=2 or \n\n\n\\begin{verbatim}\n 2n-3u=2 and n-3v=1\n\\end{verbatim}\nThe solution will be n=1 or 2 mod(3). Therefore n=21 is impossible\n\n\n~~Wei\n\n\n",
"Notice that the roots of $w^2+w+1=0$ are also the third roots of unity (excluding $w=1$). This is fairly easy to prove: multiply both sides by $w-1$ and we get \\[(w-1)(w^2+w+1) = w^3 - 1 = 0.\\] These roots are $w = e^{i \\pi /3}$ and $w = e^{2i \\pi /3}$.\n\n\nNow we have \n\\begin{align*} w^{2n} + 1 + (w+1)^{2n} &= w^{2n} + 1 + (-w^2)^{2n} \\\\ &= w^{4n} + w^{2n} + 1\\\\ &= 0. \\end{align*}\nPlug in the roots of $w^2+w+1=0$. Note that\n\\[e^{2i \\pi /3} + 1 = -\\frac{1}{2} + \\frac{\\sqrt{3}}{2}i + 1 = \\frac{1}{2} + \\frac{\\sqrt{3}}{2}i = e^{i \\pi /3}.\\]\nHowever, this will not work if $n=3m$, so $n$ cannot be equal to $21$. Hence our answer is $\\textrm{(C)}$.\n\n\n",
"We start by noting that \\[x + 1 \\equiv -x^2 \\mod (x^2+x+1).\\]\nLet $n = 3k+r$, where $r \\in \\{ 0,1,2 \\}$.\n\n\nThus we have \\[x^{4n} + x^{2n} + 1 \\equiv x^{4r} + x^{2r} + 1 \\mod (x^3 -1).\\]\n\n\nWhen $r = 0$, \\[x^{4n} + x^{2n} + 1 \\equiv 3 \\mod (x^3 -1).\\]\nWhen $r = 1$, \\[x^{4n} + x^{2n} + 1 \\equiv x^2 + x + 1 \\mod (x^3 -1),\\] which will be divisible by $x^2+x+1$.\n\n\nWhen $r = 2$, \\[x^{4n} + x^{2n} + 1 \\equiv x^2 + x + 1 \\mod (x^3 -1),\\] which will also be divisible by $x^2+x+1$.\n\n\nThus $r \\ne 0$, so $n$ cannot be divisible by $3$, and the answer is $\\textrm{(C)}$.\n\n\n"
] | 3 | ./CreativeMath/AHSME/1980_AHSME_Problems/28.json | AHSME |
1980_AHSME_Problems | 12 | 0 | Geometry | Multiple Choice | The equations of $L_1$ and $L_2$ are $y=mx$ and $y=nx$, respectively. Suppose $L_1$ makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis ) as does $L_2$, and that $L_1$ has 4 times the slope of $L_2$. If $L_1$ is not horizontal, then $mn$ is
$\text{(A)} \ \frac{\sqrt{2}}{2} \qquad \text{(B)} \ -\frac{\sqrt{2}}{2} \qquad \text{(C)} \ 2 \qquad \text{(D)} \ -2 \qquad \text{(E)} \ \text{not uniquely determined}$
| [
"Solution by e_power_pi_times_i\n\n\n$4n = m$, as stated in the question. In the line $L_1$, draw a triangle with the coordinates $(0,0)$, $(1,0)$, and $(1,m)$. Then $m = \\tan(\\theta_1)$. Similarly, $n = \\tan(\\theta_2)$. Since $4n = m$ and $\\theta_1 = 2\\theta_2$, $\\tan(2\\theta_2) = 4\\tan(\\theta_2)$. Using the angle addition formula for tangents, $\\dfrac{2\\tan(\\theta_2)}{1-\\tan^2(\\theta_2)} = 4\\tan(\\theta_2)$. Solving, we have $\\tan(\\theta_2) = 0, \\dfrac{\\sqrt{2}}{2}$. But line $L_1$ is not horizontal, so therefore $(m,n) = (2\\sqrt{2},\\dfrac{\\sqrt{2}}{2})$. Looking at the answer choices, it seems the answer is $(2\\sqrt{2})(\\dfrac{\\sqrt{2}}{2}) = \\boxed{\\text{(C)} \\ 2}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/12.json | AHSME |
1980_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | For some real number $r$, the polynomial $8x^3-4x^2-42x+45$ is divisible by $(x-r)^2$. Which of the following numbers is closest to $r$?
$\text{(A)} \ 1.22 \qquad \text{(B)} \ 1.32 \qquad \text{(C)} \ 1.42 \qquad \text{(D)} \ 1.52 \qquad \text{(E)} \ 1.62$
| [
"Solution by e_power_pi_times_i\n\n\nDenote $s$ as the third solution. Then, by Vieta's, $2r+s = \\dfrac{1}{2}$, $r^2+2rs = -\\dfrac{21}{4}$, and $r^2s = -\\dfrac{45}{8}$. Multiplying the top equation by $2r$ and eliminating, we have $3r^2 = r+\\dfrac{21}{4}$. Combined with the fact that $s = \\dfrac{1}{2}-2r$, the third equation can be written as $(\\dfrac{r+\\dfrac{21}{4}}{3})(\\dfrac{1}{2}-2r) = -\\dfrac{45}{8}$, or $(4r+21)(4r-1) = 135$. Solving, we get $r = \\dfrac{3}{2}, -\\dfrac{13}{2}$. Plugging the solutions back in, we see that $-\\dfrac{13}{2}$ is an extraneous solution, and thus the answer is $\\boxed{\\text{(D)} \\ 1.52}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/24.json | AHSME |
1980_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | In the non-decreasing sequence of odd integers $\{a_1,a_2,a_3,\ldots \}=\{1,3,3,3,5,5,5,5,5,\ldots \}$ each odd positive integer $k$
appears $k$ times. It is a fact that there are integers $b, c$, and $d$ such that for all positive integers $n$,
$a_n=b\lfloor \sqrt{n+c} \rfloor +d$,
where $\lfloor x \rfloor$ denotes the largest integer not exceeding $x$. The sum $b+c+d$ equals
$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ 4$
| [
"Solution by e_power_pi_times_i\n\n\nBecause the set consists of odd numbers, and since $\\lfloor{}\\sqrt{n+c}\\rfloor{}$ is an integer and can be odd or even, $b = 2$ and $|a| = 1$. However, given that $\\lfloor{}\\sqrt{n+c}\\rfloor{}$ can be $0$, $a = 1$. Then, $a_1 = 1 = 2\\lfloor{}\\sqrt{1+c}\\rfloor{}+1$, and $\\lfloor{}\\sqrt{1+c}\\rfloor{}$ = 0, and $c = -1$ because $c$ is an integer. $b+c+d = 2+(-1)+1 = \\boxed{\\text{(C)}\\ 2}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/25.json | AHSME |
1980_AHSME_Problems | 13 | 0 | Algebra | Multiple Choice | A bug (of negligible size) starts at the origin on the coordinate plane. First, it moves one unit right to $(1,0)$. Then it makes a $90^\circ$ counterclockwise and travels $\frac 12$ a unit to $\left(1, \frac 12 \right)$. If it continues in this fashion, each time making a $90^\circ$ degree turn counterclockwise and traveling half as far as the previous move, to which of the following points will it come closest?
$\text{(A)} \ \left(\frac 23, \frac 23 \right) \qquad \text{(B)} \ \left( \frac 45, \frac 25 \right) \qquad \text{(C)} \ \left( \frac 23, \frac 45 \right) \qquad \text{(D)} \ \left(\frac 23, \frac 13 \right) \qquad \text{(E)} \ \left(\frac 25, \frac 45 \right)$
| [
"Writing out the change in $x$ coordinates and then in $y$ coordinates gives the infinite sum $1-\\frac{1}{4}+\\frac{1}{16}-\\dots$ and $\\frac{1}{2}-\\frac{1}{8}+\\dots$ respectively. Using the infinite geometric sum formula, we have $\\frac{1}{1+\\frac{1}{4}}=\\frac{4}{5}$ and $\\frac{\\frac{1}{2}}{1+\\frac{1}{4}}=\\frac{2}{5}$, thus the answer is $\\left( \\frac{4}{5}, \\frac{2}{5} \\right)$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/13.json | AHSME |
1980_AHSME_Problems | 29 | 0 | Number Theory | Multiple Choice | How many ordered triples (x,y,z) of integers satisfy the system of equations below?
\[\begin{array}{l} x^2-3xy+2y^2-z^2=31 \\ -x^2+6yz+2z^2=44 \\ x^2+xy+8z^2=100\\ \end{array}\]
$\text{(A)} \ 0 \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \\ \text{(D)}\ \text{a finite number greater than 2}\qquad\\ \text{(E)}\ \text{infinitely many}$
| [
"Sum of three equations, \n\n\n$x^2-2xy+2y^2+6yz+9z^2 = (x-y)^2+(y+3z)^2 = 175$\n\n\n(x,y,z) are integers, ie. $175 = a^2 + b^2$, \n\n\n$a^2$: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169\n$b^2$: 174, 171, 166, 159, 150, 139, 126, 111, 94, 75, 54, 31, 6\n\n\nso there is NO solution \n\n\nWwei.yu (talk) 22:09, 28 March 2020 (EDT)Wei\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/29.json | AHSME |
1980_AHSME_Problems | 3 | 0 | Algebra | Multiple Choice | If the ratio of $2x-y$ to $x+y$ is $\frac{2}{3}$, what is the ratio of $x$ to $y$?
$\text{(A)} \ \frac{1}{5} \qquad \text{(B)} \ \frac{4}{5} \qquad \text{(C)} \ 1 \qquad \text{(D)} \ \frac{6}{5} \qquad \text{(E)} \ \frac{5}{4}$
| [
"Cross multiplying gets \n\n\n$6x-3y=2x+2y\\\\4x=5y\\\\ \\dfrac{x}{y}=\\dfrac{5}{4}\\\\ \\boxed{(E)}$\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/3.json | AHSME |
1980_AHSME_Problems | 8 | 0 | Algebra | Multiple Choice | How many pairs $(a,b)$ of non-zero real numbers satisfy the equation
\[\frac{1}{a} + \frac{1}{b} = \frac{1}{a+b}\]
$\text{(A)} \ \text{none} \qquad \text{(B)} \ 1 \qquad \text{(C)} \ 2 \qquad \text{(D)} \ \text{one pair for each} ~b \neq 0$
$\text{(E)} \ \text{two pairs for each} ~b \neq 0$
| [
"We hope to simplify this expression into a quadratic in order to find the solutions. To do this, we find a common denominator to the LHS by multiplying by $ab$.\n\\[a+b=\\frac{ab}{a+b}\\]\n\\[a^2+2ab+b^2=ab\\]\n\\[a^2+ab+b^2=0.\\]\n\n\nBy the quadratic formula and checking the discriminant (imagining one of the variables to be constant), we see that this has no real solutions. Thus the answer is $\\boxed{\\text{(A)none}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/8.json | AHSME |
1980_AHSME_Problems | 22 | 0 | Algebra | Multiple Choice | For each real number $x$, let $f(x)$ be the minimum of the numbers $4x+1, x+2$, and $-2x+4$. Then the maximum value of $f(x)$ is
$\text{(A)} \ \frac{1}{3} \qquad \text{(B)} \ \frac{1}{2} \qquad \text{(C)} \ \frac{2}{3} \qquad \text{(D)} \ \frac{5}{2} \qquad \text{(E)}\ \frac{8}{3}$
| [
"The first two given functions intersect at $\\left(\\frac{1}{3},\\frac{7}{3}\\right)$, and last two at $\\left(\\frac{2}{3},\\frac{8}{3}\\right)$. Therefore \\[f(x)=\\left\\{ \\begin{matrix} 4x+1 & x<\\frac{1}{3} \\\\ x+2 & \\frac{1}{3}>x>\\frac{2}{3} \\\\ -2x+4 & x>\\frac{2}{3} \\end{matrix}\\right.\\]\nWhich attains a maximum at\n$\\boxed{(E)\\ \\frac{8}{3}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/22.json | AHSME |
1980_AHSME_Problems | 18 | 0 | Algebra | Multiple Choice | If $b>1$, $\sin x>0$, $\cos x>0$, and $\log_b \sin x = a$, then $\log_b \cos x$ equals
$\text{(A)} \ 2\log_b(1-b^{a/2}) ~~\text{(B)} \ \sqrt{1-a^2} ~~\text{(C)} \ b^{a^2} ~~\text{(D)} \ \frac 12 \log_b(1-b^{2a}) ~~\text{(E)} \ \text{none of these}$
| [
"\\[\\log_b \\sin x = a\\]\n\\[b^a=\\sin x\\]\n\\[\\log_b \\cos x=c\\]\n\\[b^c=\\cos x\\]\nSince $\\sin^2x+\\cos^2x=1$, \n\\[(b^c)^2+(b^a)^2=1\\]\n\\[b^{2c}+b^{2a}=1\\]\n\\[b^{2c}=1-b^{2a}\\]\n\\[\\log_b (1-b^{2a}) = 2c\\]\n\\[c=\\boxed{\\text{(D)} \\ \\frac 12 \\log_b(1-b^{2a})}\\]\n\n\n-aopspandy\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/18.json | AHSME |
1980_AHSME_Problems | 4 | 0 | Geometry | Multiple Choice | In the adjoining figure, CDE is an equilateral triangle and ABCD and DEFG are squares. The measure of $\angle GDA$ is
$\text{(A)} \ 90^\circ \qquad \text{(B)} \ 105^\circ \qquad \text{(C)} \ 120^\circ \qquad \text{(D)} \ 135^\circ \qquad \text{(E)} \ 150^\circ$
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair D=origin, C=D+dir(240), E=D+dir(300), F=E+dir(30), G=D+dir(30), A=D+dir(150), B=C+dir(150); draw(E--D--G--F--E--C--D--A--B--C); pair point=(0,0.5); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(-15)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$G$", G, dir(point--G));[/asy]
| [
"$m\\angle GDA=360^\\circ-90^\\circ-60^\\circ-90^\\circ=120^\\circ\\Rightarrow\\boxed{C}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/4.json | AHSME |
1980_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | If the function $f$ is defined by \[f(x)=\frac{cx}{2x+3} ,\quad x\neq -\frac{3}{2} ,\] satisfies $x=f(f(x))$ for all real numbers $x$ except $-\frac{3}{2}$, then $c$ is
$\text{(A)} \ -3 \qquad \text{(B)} \ - \frac{3}{2} \qquad \text{(C)} \ \frac{3}{2} \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}$
| [
"As $f(x)=cx/2x+3$, we can plug that into $f(f(x))$ and simplify to get $c^2x/2cx+6x+9 = x$\n. However, we have a restriction on x such that if $x=-3/2$ we have an undefined function. We can use this to our advantage. Plugging that value for x into $c^2x/2cx+6x+9 = x$ yields $c/2 = -3/2$, as the left hand side simplifies and the right hand side is simply the value we have chosen. This means that $c=-3 \\Rightarrow \\boxed{A}$.\n\n\n\n\n\n\n",
"Alternatively, after simplifying the function to $c^2x/2cx+6x+9 = x$, multiply both sides by $2cx+6x+9$ and divide by $x$ to yield $c^2=2cx+6x+9$. This can be factored to $x(2c+6) + (3+c)(3-c) = 0$. This means that both $2c+6$ and either one of $3+c$ or $3-c$ are equal to 0. $2c+6=0$ yields $c=-3$ and the other two yield $c=3,-3$. The clear solution is $c=-3 \\Rightarrow \\boxed{A}$\n\n\n\n\nThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. \n\n\n"
] | 2 | ./CreativeMath/AHSME/1980_AHSME_Problems/14.json | AHSME |
1980_AHSME_Problems | 15 | 0 | Algebra | Multiple Choice | A store prices an item in dollars and cents so that when 4% sales tax is added, no rounding is necessary because the result is exactly $n$ dollars where $n$ is a positive integer. The smallest value of $n$ is
$\text{(A)} \ 1 \qquad \text{(B)} \ 13 \qquad \text{(C)} \ 25 \qquad \text{(D)} \ 26 \qquad \text{(E)} \ 100$
| [
"Say that the price of the item in cents is $x$ (so $x$ is a positive integer as well). The sales tax would then be $\\frac{x}{25}$, so $n=\\frac{1}{100}\\left( x+\\frac{x}{25}\\right)=\\frac{26x}{2500}=\\frac{13x}{1250}$. \n\n\nSince $x$ is positive integer, the smallest possible integer value for $n=\\frac{13x}{1250}$ occurs when $x=1250$, which gives us the answer $\\fbox{\\text{(B)13}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/15.json | AHSME |
1980_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | If $AB$ and $CD$ are perpendicular diameters of circle $Q$, $P$ in $\overline{AQ}$, and $\measuredangle QPC = 60^\circ$, then the length of $PQ$ divided by the length of $AQ$ is
[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=(-1,0), B=(1,0), C=(0,1), D=(0,-1), Q=origin, P=(-0.5,0); draw(P--C--D^^A--B^^Circle(Q,1)); label("$A$", A, W); label("$B$", B, E); label("$C$", C, N); label("$D$", D, S); label("$P$", P, S); label("$Q$", Q, SE); label("$60^\circ$", P+0.0.5*dir(30), dir(30));[/asy]
$\text{(A)} \ \frac{\sqrt{3}}{2} \qquad \text{(B)} \ \frac{\sqrt{3}}{3} \qquad \text{(C)} \ \frac{\sqrt{2}}{2} \qquad \text{(D)} \ \frac12 \qquad \text{(E)} \ \frac23$
| [
"We find that $m\\angle PCQ=30^\\circ$. Because it is a $30^\\circ-60^\\circ-90^\\circ$ right triangle, we can let $PQ=x$, so $CQ=AQ=x\\sqrt{3}$. Thus, $\\frac{PQ}{AQ}=\\frac{x}{x\\sqrt{3}}=\\frac{\\sqrt{3}}{3}\\Rightarrow\\boxed{(B)}$.\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/5.json | AHSME |
1980_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | Let $C_1, C_2$ and $C_3$ be three parallel chords of a circle on the same side of the center.
The distance between $C_1$ and $C_2$ is the same as the distance between $C_2$ and $C_3$.
The lengths of the chords are $20, 16$, and $8$. The radius of the circle is
$\text{(A)} \ 12 \qquad \text{(B)} \ 4\sqrt{7} \qquad \text{(C)} \ \frac{5\sqrt{65}}{3} \qquad \text{(D)}\ \frac{5\sqrt{22}}{2}\qquad \text{(E)}\ \text{not uniquely determined}$
| [
"Let the center of the circle be on the origin with equation $x^2 + y^2 = r^2$. \nAs the chords are bisected by the x-axis their y-coordinates are $10, 8, 4$ respectively. Let the chord of length $10$ have x-coordinate $a$. Let $d$ be the common distance between chords. Thus, the coordinates of the top of the chords will be $(a, 10), (a+d, 8), (a+2d, 4)$ for the chords of length $20, 16$, and $8$ respectively. \nAs these points fall of the circle, we get three equations: \n$\\\\a^2 + 100 = r^2$\n$\\\\(a+d)^2 + 64 = r^2$\n$\\\\(a+2d)^2 + 16 = r^2$\nSubtracting the first equation from the second we get:\n$\\\\(a+d)^2 - a^2 - 36 = 0$\n$\\\\d(2a+d) = 36$\nSimilarly, by subtracting the first equation from the third we get:\n$\\\\(a+2d)^2 - a^2 - 84 = 0$\n$\\\\2d(2a+2d) = 84$\n$\\\\d(a+d) = 21$\nSubtracting these two equations gives us $ad = 15$. Expanding the second equation now gives us\n$\\\\a^2 + 2ad + d^2 + 64 = r^2$\n$\\\\a^2 + d^2 + 94 = r^2$\nSubtracting the first equation from this yields:\n$\\\\d^2 - 6 = 0$\n$\\\\d = \\sqrt{6}$\nCombining this with $ad = 15$ we get\n$\\\\\\sqrt{6}a = 15$\n$\\\\a = \\frac{15}{\\sqrt{6}} = \\frac{5\\sqrt{6}}{2}$\nPlugging this into the first equation finally us\n$\\\\(\\frac{5\\sqrt{6}}{2})^2 + 100 = r^2$\n$\\frac{150}{4} + 100 = \\frac{550}{4} = r^2$\n$\\\\r = \\sqrt{\\frac{550}{4}} = \\frac{5\\sqrt{22}}{2}$\n$\\fbox{D}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/19.json | AHSME |
1980_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | Line segments drawn from the vertex opposite the hypotenuse of a right triangle to the points trisecting the hypotenuse have lengths $\sin x$ and $\cos x$, where $x$ is a real number such that $0<x<\frac{\pi}{2}$. The length of the hypotenuse is
$\text{(A)} \ \frac{4}{3} \qquad \text{(B)} \ \frac{3}{2} \qquad \text{(C)} \ \frac{3\sqrt{5}}{5} \qquad \text{(D)}\ \frac{2\sqrt{5}}{3}\qquad \text{(E)}\ \text{not uniquely determined}$
| [
"Consider right triangle $ABC$ with hypotenuse $BC$. Let points $D$ and $E$ trisect $BC$. WLOG, let $AD=cos(x)$ and $AE=sin(x)$ (the proof works the other way around as well). \n\n\nApplying Stewart's theorem on $\\bigtriangleup ABC$ with point $D$, we obtain the equation\n\n\n\\[cos^{2}(x)\\cdot c=a^2 \\cdot \\frac{c}{3} + b^2 \\cdot \\frac{2c}{3} - \\frac{2c}{3}\\cdot \\frac{c}{3} \\cdot c\\]\n\n\nSimilarly using point $E$, we obtain\n\n\n\\[sin^{2}(x)\\cdot c=a^2 \\cdot \\frac{2c}{3} + b^2 \\cdot \\frac{c}{3} - \\frac{2c}{3}\\cdot \\frac{c}{3} \\cdot c\\]\n\n\nAdding these equations, we get\n\n\n\\[(sin^{2}(x) + cos^{2}(x)) \\cdot c=a^{2} c + b^{2} c - \\frac{4c^3}{9}\\]\n\n\n\\[c=(a^2 + b^2) c - \\frac{4c^3}{9}\\]\n\n\n\\[c=c^2 \\cdot c -\\frac{4c^3}{9}\\]\n\n\n\\[c=\\pm \\frac{3\\sqrt{5}}{5}, 0\\]\n\n\nAs the other 2 answers yield degenerate triangles, we see that the answer is \\[\\boxed{(C) \\frac{3\\sqrt{5}}{5}}\\]\n\n\n$\\fbox{}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/23.json | AHSME |
1980_AHSME_Problems | 9 | 0 | Geometry | Multiple Choice | A man walks $x$ miles due west, turns $150^\circ$ to his left and walks 3 miles in the new direction. If he finishes a a point $\sqrt{3}$ from his starting point, then $x$ is
$\text{(A)} \ \sqrt 3 \qquad \text{(B)} \ 2\sqrt{5} \qquad \text{(C)} \ \frac 32 \qquad \text{(D)} \ 3 \qquad \text{(E)} \ \text{not uniquely determined}$
| [
"Let us think about this. We only know that he ends up $\\sqrt{3}$ away from the origin. However, think about the locus of points $\\sqrt{3}$ away from the origin, a circle. However, his path could end on any part of the circle below the $x-$axis, so therefore, the answer is\n$\\fbox{E: not uniquely determined}.$\n\n\n(Note: Another way to do this is using law of cosines, which yields two solutions for x.)\n\n\n"
] | 1 | ./CreativeMath/AHSME/1980_AHSME_Problems/9.json | AHSME |
1995_AHSME_Problems | 20 | 0 | Probability | Multiple Choice | If $a,b$ and $c$ are three (not necessarily different) numbers chosen randomly and with replacement from the set $\{1,2,3,4,5 \}$, the probability that $ab + c$ is even is
$\mathrm{(A) \ \frac {2}{5} } \qquad \mathrm{(B) \ \frac {59}{125} } \qquad \mathrm{(C) \ \frac {1}{2} } \qquad \mathrm{(D) \ \frac {64}{125} } \qquad \mathrm{(E) \ \frac {3}{5} }$
| [
"The probability of $ab$ being odd is $\\left(\\frac 35\\right)^2 = \\frac{9}{25}$, so the probability of $ab$ being even is $1 - \\frac{9}{25} = \\frac {16}{25}$.\n\n\nThe probability of $c$ being odd is $3/5$ and being even is $2/5$\n\n\n$ab+c$ is even if $ab$ and $c$ are both odd, with probability $\\frac{9}{25} \\cdot \\frac{3}{5} = \\frac{27}{125}$; or $ab$ and $c$ are both even, with probability $\\frac{16}{25} \\cdot \\frac{2}{5} = \\frac{32}{125}$. Thus the total probability is $\\frac{59}{125} \\Rightarrow \\mathrm{(B)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/20.json | AHSME |
1995_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | Anita attends a baseball game in Atlanta and estimates that there are 50,000 fans in attendance. Bob attends a baseball game in Boston and estimates that there are 60,000 fans in attendance. A league official who knows the actual numbers attending the two games note that:
i. The actual attendance in Atlanta is within $10 \%$ of Anita's estimate.
ii. Bob's estimate is within $10 \%$ of the actual attendance in Boston.
To the nearest 1,000, the largest possible difference between the numbers attending the two games is
$\mathrm{(A) \ 10000 } \qquad \mathrm{(B) \ 11000 } \qquad \mathrm{(C) \ 20000 } \qquad \mathrm{(D) \ 21000 } \qquad \mathrm{(E) \ 22000 }$
| [
"Since the number of people at the game in Boston is certainly more than the number of fans in Atlanta, we need to compute the maximum of Bob's game minus the minimum of Anita's game. Note however that there is a slight different between conditions (i) and (ii); the attendence is \\textit{within} $\\pm 10 \\%$ from Anita's estimate of $50,000$, but Bob's estimate is \\textit{within} $\\pm 10 \\%$ of the actual attendance, or $0.9x \\le 60000 \\le 1.1x \\Longrightarrow x \\le \\frac{60000}{0.9}$. \n\n\nThe answer is $\\frac{60000}{0.9} - 50000 \\times 0.9 \\approx 22000 \\Longrightarrow \\mathrm{(E)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/16.json | AHSME |
1995_AHSME_Problems | 6 | 0 | Geometry | Multiple Choice | The figure shown can be folded into the shape of a cube. In the resulting cube, which of the lettered faces is opposite the face marked x?
[asy] defaultpen(linewidth(0.7)); path p=origin--(0,1)--(1,1)--(1,2)--(2,2)--(2,3); draw(p^^(2,3)--(4,3)^^shift(2,0)*p^^(2,0)--origin); draw(shift(1,0)*p, dashed); label("$x$", (0.3,0.5), E); label("$A$", (1.3,0.5), E); label("$B$", (1.3,1.5), E); label("$C$", (2.3,1.5), E); label("$D$", (2.3,2.5), E); label("$E$", (3.3,2.5), E);[/asy]
$\mathrm{(A) \ A } \qquad \mathrm{(B) \ B } \qquad \mathrm{(C) \ C } \qquad \mathrm{(D) \ D } \qquad \mathrm{(E) \ E }$
| [
"The marked side is the side with the $x.$ We imagine it folding up. First, we fold the $x$ upwards. Now we fold the $A$ upwards, and thus $x$ is touching $B$ on it's left side. We now fold $B$ up, and we realize that $x$ won't be touching $\\boxed{\\mathrm{(C)}}$ at all.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/6.json | AHSME |
1995_AHSME_Problems | 7 | 0 | Geometry | Multiple Choice | The radius of Earth at the equator is approximately 4000 miles. Suppose a jet flies once around Earth at a speed of 500 miles per hour relative to Earth. If the flight path is a neglibile height above the equator, then, among the following choices, the best estimate of the number of hours of flight is:
$\mathrm{(A) \ 8 } \qquad \mathrm{(B) \ 25 } \qquad \mathrm{(C) \ 50 } \qquad \mathrm{(D) \ 75 } \qquad \mathrm{(E) \ 100 }$
| [
"We want the number of hours that it takes the jet to fly the length of the circumference. $\\frac{8000\\pi}{500}=16\\pi$. The best estimate of that is $50\\Rightarrow \\mathrm{(C)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/7.json | AHSME |
1995_AHSME_Problems | 17 | 0 | Geometry | Multiple Choice | Given regular pentagon $ABCDE$, a circle can be drawn that is tangent to $\overline{DC}$ at $D$ and to $\overline{AB}$ at $A$. The number of degrees in minor arc $AD$ is
[asy]size(100); defaultpen(linewidth(0.7)); draw(rotate(18)*polygon(5)); real x=0.6180339887; draw(Circle((-x,0), 1)); int i; for(i=0; i<5; i=i+1) { dot(origin+1*dir(36+72*i)); } label("$B$", origin+1*dir(36+72*0), dir(origin--origin+1*dir(36+72*0))); label("$A$", origin+1*dir(36+72*1), dir(origin--origin+1*dir(36+72))); label("$E$", origin+1*dir(36+72*2), dir(origin--origin+1*dir(36+144))); label("$D$", origin+1*dir(36+72*3), dir(origin--origin+1*dir(36+72*3))); label("$C$", origin+1*dir(36+72*4), dir(origin--origin+1*dir(36+72*4))); [/asy]
$\mathrm{(A) \ 72 } \qquad \mathrm{(B) \ 108 } \qquad \mathrm{(C) \ 120 } \qquad \mathrm{(D) \ 135 } \qquad \mathrm{(E) \ 144 }$
| [
"Define major arc DA as $DA$, and minor arc DA as $da$. Extending DC and AB to meet at F, we see that $\\angle CFB=36=\\frac{DA-da}{2}$. We now have two equations: $DA-da=72$, and $DA+da=360$. Solving, $DA=216$ and $da=144\\Rightarrow \\mathrm{(E)}$.\nhello\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/17.json | AHSME |
1995_AHSME_Problems | 21 | 0 | Geometry | Multiple Choice | Two nonadjacent vertices of a rectangle are $(4,3)$ and $(-4,-3)$, and the coordinates of the other two vertices are integers. The number of such rectangles is
$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 3 } \qquad \mathrm{(D) \ 4 } \qquad \mathrm{(E) \ 5 }$
| [
"The center of the rectangle is $(0,0)$, and the distance from the center to a corner is $\\sqrt{4^2+3^2}=5$. The remaining two vertices of the rectangle must be another pair of points opposite each other on the circle of radius 5 centered at the origin. Let these points have the form $(\\pm x,\\pm y)$, where $x^2+y^2=25$. This equation has six pairs of integer solutions: $(\\pm 4, \\pm 3)$, $(\\pm 4, \\mp 3)$, $(\\pm 3, \\pm 4)$, $(\\pm 3, \\mp 4)$, $(\\pm 5, 0)$, and $(0, \\pm 5)$. The first pair of solutions are the endpoints of the given diagonal, and the other diagonal must span one of the other five pairs of points. $\\Rightarrow \\mathrm{(E)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/21.json | AHSME |
1995_AHSME_Problems | 10 | 0 | Geometry | Multiple Choice | The area of the triangle bounded by the lines $y = x, y = - x$ and $y = 6$ is
$\mathrm{(A) \ 12 } \qquad \mathrm{(B) \ 12\sqrt{2} } \qquad \mathrm{(C) \ 24 } \qquad \mathrm{(D) \ 24\sqrt{2} } \qquad \mathrm{(E) \ 36 }$
| [
"\\begin{center}\n[asy] defaultpen(fontsize(8)); draw((0,0)--(6,6)--(-6,6)--(0,0)); draw((0,-1)--(0,8)); draw((-7,0)--(7,0)); label(\"$(6,6)$\",(6,6), (1,1));label(\"$(-6,6)$\",(-6,6),(-1,1)); label(\"$y=x$\",(3,3),(1,-1));label(\"$y=-x$\",(-3,3),(-1,-0.5)); label(\"$6$\",(-3,6),(0,1));label(\"$6$\",(3,6),(0,1)); label(\"$6$\",(0,3),(-1,0)); [/asy]\\end{center}\nThe height of the triangle is $y = 6$, and the width of the triangle is $|x_1| + |x_2| = 2y = 12$. Thus the area of the triangle is $\\frac 12 \\cdot 6 \\cdot 12 = 36\\ \\mathrm{(E)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/10.json | AHSME |
1995_AHSME_Problems | 26 | 0 | Geometry | Multiple Choice | In the figure, $\overline{AB}$ and $\overline{CD}$ are diameters of the circle with center $O$, $\overline{AB} \perp \overline{CD}$, and chord $\overline{DF}$ intersects $\overline{AB}$ at $E$. If $DE = 6$ and $EF = 2$, then the area of the circle is
[asy]size(120); defaultpen(linewidth(0.7)); pair O=origin, A=(-5,0), B=(5,0), C=(0,5), D=(0,-5), F=5*dir(40), E=intersectionpoint(A--B, F--D); draw(Circle(O, 5)); draw(A--B^^C--D--F); dot(O^^A^^B^^C^^D^^E^^F); markscalefactor=0.05; draw(rightanglemark(B, O, D)); label("$A$", A, dir(O--A)); label("$B$", B, dir(O--B)); label("$C$", C, dir(O--C)); label("$D$", D, dir(O--D)); label("$F$", F, dir(O--F)); label("$O$", O, NW); label("$E$", E, SE);[/asy]
$\mathrm{(A) \ 23 \pi } \qquad \mathrm{(B) \ \frac {47}{2} \pi } \qquad \mathrm{(C) \ 24 \pi } \qquad \mathrm{(D) \ \frac {49}{2} \pi } \qquad \mathrm{(E) \ 25 \pi }$
| [
"\\textbf{Solution 1}\n\n\nLet the radius of the circle be $r$ and let $x=\\overline{OE}$. \n\n\nBy the Pythagorean Theorem, $OD^2+OE^2=DE^2 \\Rightarrow r^2+x^2=6^2=36$. \n\n\nBy Power of a point, $AE \\cdot EB = DE \\cdot EF \\Rightarrow (r+x)(r-x)=r^2-x^2=6\\cdot2=12$. \n\n\nAdding these equations yields $2r^2=48 \\Rightarrow r^2 = 24$. \n\n\nThus, the area of the circle is $\\pi r^2 = 24\\pi \\Rightarrow C$.\n\n\n\\textbf{Solution 2}\n\n\nLet the radius of the circle be $r$.\n\n\nWe can see that $\\triangle CFD$ has a right angle at $F$ and that $\\triangle EOD$ has a right angle at $O$. \n\n\nBoth triangles also share $\\angle ODE$, so $\\triangle CFD$ and $\\triangle EOD$ are similar.\n\n\nThis means that $\\frac {DE}{OD}=\\frac {CD}{FD}$.\n\n\nSo, $\\frac {6}{r}=\\frac {2r}{8}$. Simplifying, $r^2 = 24$.\n\n\nThis means the area is $24\\pi \\Rightarrow C$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/26.json | AHSME |
1995_AHSME_Problems | 30 | 0 | Geometry | Multiple Choice | A large cube is formed by stacking 27 unit cubes. A plane is perpendicular to one of the internal diagonals of the large cube and bisects that diagonal. The number of unit cubes that the plane intersects is
$\mathrm{(A) \ 16 } \qquad \mathrm{(B) \ 17 } \qquad \mathrm{(C) \ 18 } \qquad \mathrm{(D) \ 19 } \qquad \mathrm{(E) \ 20 }$
| [
"Place one corner of the cube at the origin of the coordinate system so that its sides are parallel to the axes.\n\n\nNow consider the diagonal from $(0,0,0)$ to $(3,3,3)$. The midpoint of this diagonal is at $\\left(\\frac 32,\\frac 32,\\frac 32\\right)$. The plane that passes through this point and is orthogonal to the diagonal has the equation $x+y+z=\\frac 92$.\n\n\nThe unit cube with opposite corners at $(x,y,z)$ and $(x+1,y+1,z+1)$ is intersected by this plane if and only if $x+y+z < \\frac 92 < (x+1)+(y+1)+(z+1)=(x+y+z)+3$. Therefore the cube is intersected by this plane if and only if $x+y+z\\in\\{2,3,4\\}$.\n\n\nThere are six cubes such that $x+y+z=2$: permutations of $(1,1,0)$ and $(2,0,0)$. \n\nSymmetrically, there are six cubes such that $x+y+z=4$. \n\nFinally, there are seven cubes such that $x+y+z=3$: permutations of $(2,1,0)$ and the central cube $(1,1,1)$.\n\n\nThat gives a total of $\\boxed{19}$ intersected cubes.\n\n\nNote that there are only 8 cubes that are not intersected by our plane: 4 in each of the two opposite corners that were connected by the original diagonal.\n\n\n",
"Place the cube so that its space diagonal is perpendicular to the ground. The space diagonal has length of $3\\sqrt{3}$, the altitude of the top vertex of the newly placed cube is $3\\sqrt{3}$. The plane perpendicular and bisecting the space diagonal is now parallel to the ground and also bisecting the space diagonal into $\\frac{3\\sqrt{3}}{2}$, so that the height of the plane is $\\frac{3\\sqrt{3}}{2}$.\n\n\nBy symmetry, the space diagonal is trisected by the pyramid at the top of the cube and the pyramid at the bottom of the cube.\n\n\nWe can prove that the space diagonal is trisected. Let the altitude of the pyramid at the top of the cube be $h$. The base of the pyramid is an equilateral triangle with side length of $\\sqrt{3^2+3^2}=3\\sqrt{2}$. The height of the triangle is $\\frac{ \\sqrt{3} }{2} \\cdot 3\\sqrt{2}$. The distance of the center of the triangle to the vertex is $\\frac23 \\cdot \\frac{ \\sqrt{3} }{2} \\cdot 3\\sqrt{2} = \\sqrt {6}$. Therefore, $h = \\sqrt{3^2-(\\sqrt {6})^2} = \\sqrt{9-6}=\\sqrt{3}$. The altitude of the pyramid at the bottom of the cube is also $h$. The altitude in the middle is $3\\sqrt{3}-\\sqrt{3}-\\sqrt{3}=\\sqrt{3}$.\n\n\nThe altitude of the vertex at the top is $3\\sqrt{3}$. The altitude of the second highest $3$ vertices are all $2\\sqrt{3}$. The altitude of the third highest $3$ vertices are all $\\sqrt{3}$. The altitude of the bottom-most vertex is $0$.\n\n\nBy scale, for the unit cube, place the cube so that its space diagonal is perpendicular to the ground. The altitude of the vertex at the top is $\\sqrt{3}$. The altitude of the second highest $3$ vertices are all $\\frac{2\\sqrt{3}}{3}$. The altitude of the third highest $3$ vertices are all $\\frac{\\sqrt{3}}{3}$. The altitude of the bottom-most vertex is $0$.\n\n\nThe length of the space diagonal of a unit cube is $\\sqrt{3}$. The highest vertex of the bottom-most unit cube has an altitude of $\\sqrt{3}$. As $\\sqrt{3} < \\frac{3\\sqrt{3}}{2}$, therefore, the plane will not pass through the unit cube at the bottom.\n\n\nFor the next $3$ cubes from the bottom, the altitude of their highest vertex is $\\frac{\\sqrt{3}}{3} + \\sqrt{3} = \\frac{4\\sqrt{3}}{3}$. As $\\frac{4\\sqrt{3}}{3} < \\frac{3\\sqrt{3}}{2}$, therefore, the plane will not pass through the next $3$ unit cubes.\n\n\nFor the next $3$ cubes from the bottom, the altitude of their highest vertex is $\\frac{2\\sqrt{3}}{3} + \\frac{\\sqrt{3}}{3} = \\frac{5\\sqrt{3}}{3}$. As $\\frac{5\\sqrt{3}}{3} > \\frac{3\\sqrt{3}}{2}$, therefore, the plane will pass through the next $3$ unit cubes.\n\n\nSo at the bottom half of the cube, there are only $1+3 = 4$ unit cubes that the plane does not passes through. By symmetry, the plane will not pass through $4$ unit cubes at the top half of the cube.\n\n\nThus, the plane does not pass through $4+4 = 8$ unit cubes, it passes through $27-8=\\boxed{\\textbf{(D) } 19}$ unit cubes.\n\n\n~isabelchen\n\n\n"
] | 2 | ./CreativeMath/AHSME/1995_AHSME_Problems/30.json | AHSME |
1995_AHSME_Problems | 27 | 0 | Number Theory | Multiple Choice | Consider the triangular array of numbers with 0,1,2,3,... along the sides and interior numbers obtained by adding the two adjacent numbers in the previous row. Rows 1 through 6 are shown.
\[\begin{tabular}{ccccccccccc} & & & & & 0 & & & & & \\ & & & & 1 & & 1 & & & & \\ & & & 2 & & 2 & & 2 & & & \\ & & 3 & & 4 & & 4 & & 3 & & \\ & 4 & & 7 & & 8 & & 7 & & 4 & \\ 5 & & 11 & & 15 & & 15 & & 11 & & 5 \end{tabular}\]
Let $f(n)$ denote the sum of the numbers in row $n$. What is the remainder when $f(100)$ is divided by 100?
$\mathrm{(A) \ 12 } \qquad \mathrm{(B) \ 30 } \qquad \mathrm{(C) \ 50 } \qquad \mathrm{(D) \ 62 } \qquad \mathrm{(E) \ 74 }$
| [
"Note that if we re-draw the table with an additional diagonal row on each side, the table is actually just two of Pascal's Triangles, except translated and summed. \n\\[\\begin{tabular}{ccccccccccccccc} & & & & & 1 & & 0 & & 1 & & & & \\\\ & & & & 1 & & 1 & & 1 & & 1 & & & \\\\ & & & 1 & & 2 & & 2 & & 2 & & 1 & & \\\\ & & 1 & & 3 & & 4 & & 4 & & 3 & & 1 & \\\\ & 1 & & 4 & & 7 & & 8 & & 7 & & 4 & & 1 \\\\ 1 & & 5 & & 11 & & 15 & & 15 & & 11 & & 5 & & 1 \\end{tabular}\\]\n\n\nThe sum of a row of Pascal's triangle is $2^{n-1}$; the sum of two of each of these rows, subtracting away the $2$ ones we included, yields $f(n) = 2^n - 2$. Now, $f(100) = 2^{100} - 2 \\equiv 2 \\pmod{4}$ and $f(100) = 2^{100} - 2 \\equiv 2^{20 \\cdot 5} - 2 \\equiv -1 \\pmod{25}$, and by the Chinese Remainder Theorem, we have $f(100) \\equiv 74 \\pmod{100} \\Longrightarrow \\mathrm{(E)}$.\n\n\n\\subsubsection{Remark (Chinese Remainder Theorem)}\nSolution 1 needs to calculate the last $2$ digits of $2^{100}$, meaning to solve $2^{100} (\\bmod{ 100})$. $100 = 4 \\cdot 25$, we are going to get $n \\bmod{ 100}$ from $n \\bmod{ 4}$ and $n \\bmod{ 25}$.\n\n\nBy Chinese Remainder Theorem, the general solution of the system of $2$ linear congruences is:\n\n\n\\begin{verbatim}\n$n \\equiv r_1 (\\bmod { \\quad m_1})$, $n \\equiv r_2 (\\bmod { \\quad m_2})$, $(m_1, m_2) = 1$\nFind $k_1$ and $k_2$ such that $k_1 m_1 \\equiv 1 (\\bmod{ \\quad m_2})$, $k_2 m_2 \\equiv 1 (\\bmod{ \\quad m_1})$\nThen $n \\equiv k_2 m_2 r_1 + k_1 m_1 r_2 (\\bmod{ \\quad m_1 m_2})$\n\\end{verbatim}\nIn this problem, $2^{100} \\equiv (2^{10})^{10} \\equiv 1024^{10} \\equiv 24^{10} \\equiv (-1)^{10} \\equiv 1 (\\bmod{ \\quad 25})$, $2^{100} \\equiv 0 (\\bmod{ \\quad 4})$:\n\n\n\\begin{verbatim}\n$n \\equiv 1 (\\bmod{ \\quad 25})$, $n \\equiv 0 (\\bmod{ \\quad 4})$, $(25, 4) = 1$\n$1 \\cdot 25 \\equiv 1 (\\bmod{ \\quad 4})$, $19 \\cdot 4 \\equiv 1 (\\bmod{ \\quad 25})$\nThen $n \\equiv 19 \\cdot 4 \\cdot 1 + 1 \\cdot 25 \\cdot 0 \\equiv 76 (\\bmod{ \\quad 100})$\n\\end{verbatim}\n~isabelchen\n\n\n",
"We sum the first few rows: $0, 2, 6, 14, 30, 62$. They are each two less than a power of $2$, so we try to prove it:\n\n\nLet the sum of row $n$ be $S_n$. To generate the next row, we add consecutive numbers. So we double the row, subtract twice the end numbers, then add twice the end numbers and add two. That makes $S_{n+1}=2S_n-2(n-1)+2(n-1)+2=2S_n+2$. If $S_n$ is two less than a power of 2, then it is in the form $2^x-2$. $S_{n+1}=2^{x+1}-4+2=2^{x+1}-2$. \n\n\nSince the first row is two less than a power of 2, all the rest are. Since the sum of the elements of row 1 is $2^1-2$, the sum of the numbers in row $n$ is $2^n-2$. Thus, using Modular arithmetic, $f(100)=2^{100}-2 \\bmod{100}$. $2^{10}=1024$, so $2^{100}-2\\equiv 24^{10}-2\\equiv (2^3 \\cdot 3)^{10} - 2$ $\\equiv 1024^3 \\cdot 81 \\cdot 81 \\cdot 9 - 2 \\equiv 24^3 \\cdot 19^2 \\cdot 9 - 2$ $\\equiv 74\\bmod{100} \\Rightarrow \\mathrm{(E)}$.\n\n\n",
"We derive the recurrence $S_{n+1}=2S_n + 2$ as above. Without guessing the form of the solution at this point we can easily solve this recurrence. Note that one can easily get rid of the \"$+2$\" as follows: Let $S_n=T_n-2$. Then $S_{n+1}=T_{n+1}-2$ and $2S_n+2 = 2(T_n-2)+2 = 2T_n-2$. Therefore $T_{n+1}=2T_n$. This obviously solves to $T_n=2^{n-1} T_1$. As $S_1=0$, we have $T_1=2$. Therefore $T_n=2^n$ and consecutively $S_n=2^n-2$.\n\n\n"
] | 3 | ./CreativeMath/AHSME/1995_AHSME_Problems/27.json | AHSME |
1995_AHSME_Problems | 1 | 0 | Arithmetic | Multiple Choice | Kim earned scores of 87,83, and 88 on her first three mathematics examinations. If Kim receives a score of 90 on the fourth exam, then her average will
$\mathrm{(A) \ \text{remain the same} } \qquad \mathrm{(B) \ \text{increase by 1} } \qquad \mathrm{(C) \ \text{increase by 2} } \qquad \mathrm{(D) \ \text{increase by 3} } \qquad \mathrm{(E) \ \text{increase by 4} }$
| [
"The average of the first three test scores is $\\frac{88+83+87}{3}=86$. The average of all four exams is $\\frac{87+83+88+90}{4}=87$. It increased by one point. $\\mathrm{(B)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/1.json | AHSME |
1995_AHSME_Problems | 11 | 0 | Counting | Multiple Choice | How many base 10 four-digit numbers, $N = \underline{a} \underline{b} \underline{c} \underline{d}$, satisfy all three of the following conditions?
(i) $4,000 \leq N < 6,000;$ (ii) $N$ is a multiple of 5; (iii) $3 \leq b < c \leq 6$.
$\mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 18 } \qquad \mathrm{(C) \ 24 } \qquad \mathrm{(D) \ 36 } \qquad \mathrm{(E) \ 48 }$
| [
"\\begin{itemize}\n\\item For condition (i), the restriction is put on $a$; $N<4000$ if $a<4$, and $N \\ge 6$ if $a \\ge 6$. Therefore, $a=4,5$.\n\n\\item For condition (ii), the restriction is put on $d$; it must be a multiple of $5$. Therefore, $d=0,5$.\n\n\\item For condition (iii), the restriction is put on $b$ and $c$. The possible ordered pairs of $b$ and $c$ are $(3,4)$, $(3,5)$, $(3,6)$, $(4,5), (4,6),$ and $(5,6),$ and there are $6$ of them. Alternatively, we are picking from the four digits 3, 4, 5, 6, and for every combination of two, there is exactly one way to arrange them in increasing order, so we have $\\binom{4}{2} = 6$ choices for $b$ and $c$ when we consider them together.\n\\end{itemize}\nMultiplying the possibilities for each restriction, $2 \\cdot 2 \\cdot 6=24\\Rightarrow \\mathrm{(C)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/11.json | AHSME |
1995_AHSME_Problems | 2 | 0 | Algebra | Multiple Choice | If $\sqrt {2 + \sqrt {x}} = 3$, then $x =$
$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ \sqrt{7} } \qquad \mathrm{(C) \ 7 } \qquad \mathrm{(D) \ 49 } \qquad \mathrm{(E) \ 121 }$
| [
"$2 + \\sqrt{x} = 9 \\Longrightarrow \\sqrt{x} = 7 \\Longrightarrow x = 49 \\Rightarrow \\mathrm{(D)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/2.json | AHSME |
1995_AHSME_Problems | 12 | 0 | Algebra | Multiple Choice | Let $f$ be a linear function with the properties that $f(1) \leq f(2), f(3) \geq f(4),$ and $f(5) = 5$. Which of the following is true?
$\mathrm{(A) \ f(0) < 0 } \qquad \mathrm{(B) \ f(0) = 0 } \qquad \mathrm{(C) \ f(1) < f(0) < f( - 1) } \qquad \mathrm{(D) \ f(0) = 5 } \qquad \mathrm{(E) \ f(0) > 5 }$
| [
"A linear function has the property that $f(a)\\leq f(b)$ either for all $a<b$, or for all $b<a$. Since $f(3)\\geq f(4)$, $f(1)\\geq f(2)$. Since $f(1)\\leq f(2)$, $f(1)=f(2)$. And if $f(a)=f(b)$ for $a\\neq b$, then $f(x)$ is a constant function. Since $f(5)=5$, $f(0)=5\\Rightarrow \\mathrm{(D)}$\n\n\n",
"If $f$ is a linear function, the statement $f(1) \\leq f(2)$ states that the slope of the line $\\frac{f(2) - f(1)}{2 - 1} = f(2) - f(1)$ is nonnegative: it is either positive or zero.\n\n\nSimilarly, the statement $f(3) \\geq f(4)$ states that the slope of the line $\\frac{f(4) - f(3)}{4 - 3} = f(4) - f(3)$ is nonpositive: it is either negative or zero.\n\n\nSince the slope of a linear function can only have one value, it must be zero, and thus the function is a constant. The statement $f(5) = 5$ tells us that the value of the constant is $5$, and thus that $f(x) = 5$. This leads to $f(0)=5\\Rightarrow \\mathrm{(D)}$\n\n\n",
"It should be very clear that $f(1)<f(2)$ and $f(3)>f(4)$ is contradictory because of the fact that linear functions are monotonic. The only thing that makes sense is $f(1)=f(2)$, and $f(3)=f(4)$. This means that $f(x)$ has a slope of $0$. So $f(x)=5$. So $f(0)=5$. Select $\\boxed{D}$. \n\n\n~hastapasta\n\n\n"
] | 3 | ./CreativeMath/AHSME/1995_AHSME_Problems/12.json | AHSME |
1995_AHSME_Problems | 24 | 0 | Algebra | Multiple Choice | There exist positive integers $A,B$ and $C$, with no common factor greater than $1$, such that
\[A \log_{200} 5 + B \log_{200} 2 = C.\]
What is $A + B + C$?
$\mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$
| [
"\\[A \\log_{200} 5 + B \\log_{200} 2 = C\\]\n\n\nSimplifying and taking the logarithms away,\n\n\n\\[5^A \\cdot 2^B=200^C=2^{3C} \\cdot 5^{2C}\\]\n\n\nTherefore, $A=2C$ and $B=3C$. Since $A, B,$ and $C$ are relatively prime, $C=1$, $B=3$, $A=2$. $A+B+C=6 \\Rightarrow \\mathrm{(A)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/24.json | AHSME |
1995_AHSME_Problems | 25 | 0 | Algebra | Multiple Choice | A list of five positive integers has mean $12$ and range $18$. The mode and median are both $8$. How many different values are possible for the second largest element of the list?
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ 12 }$
| [
"Let $a$ be the smallest element, so $a+18$ is the largest element. Since the mode is $8$, at least two of the five numbers must be $8$. The last number we denote as $b$.\n\n\nThen their average is $\\frac{a + (8) + (8) + b + (a+18)}5 = 12 \\Longrightarrow 2a + b = 26$. Clearly $a \\le 8$. Also we have $b \\le a + 18 \\Longrightarrow 26-2a \\le a + 18 \\Longrightarrow 8/3 < 3 \\le a$. Thus there are a maximum of $6$ values of $a$ which corresponds to $6$ values of $b$; listing shows that all such values work. The answer is $\\boxed{B}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/25.json | AHSME |
1995_AHSME_Problems | 13 | 0 | Arithmetic | Multiple Choice | The addition below is incorrect. The display can be made correct by changing one digit $d$, wherever it occurs, to another digit $e$. Find the sum of $d$ and $e$.
$\begin{tabular}{ccccccc} & 7 & 4 & 2 & 5 & 8 & 6 \\ + & 8 & 2 & 9 & 4 & 3 & 0 \\ \hline 1 & 2 & 1 & 2 & 0 & 1 & 6 \end{tabular}$
$\mathrm{(A) \ 4 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 10 } \qquad \mathrm{(E) \ \text{more than 10} }$
| [
"If we change $0$, the units column would be incorrect. \n\n\nIf we change $1$, then the leading $1$ in the sum would be incorrect.\n\n\nHowever, looking at the $2$ in the hundred-thousands column, it would be possible to change the $2$ to either a $5$ (no carry) or a $6$ (carry) to create a correct statement. \n\n\nChanging the $2$ to a $5$ would give $745586 + 859430$ on top, which equals $1605016$. This does not match up to the bottom.\n\n\nChanging the $2$ to a $6$ gives $746586 + 869430$ on top, which has a sum of $1616016$. This is the number on the bottom if the $2$s were changed to $6$s.\n\n\nThus $d=2$ and $e=6$. so $d+e= 8 \\boxed{\\mathrm{ (C)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/13.json | AHSME |
1995_AHSME_Problems | 29 | 0 | Counting | Multiple Choice | For how many three-element sets of distinct positive integers $\{a,b,c\}$ is it true that $a \times b \times c = 2310$?
$\mathrm{(A) \ 32 } \qquad \mathrm{(B) \ 36 } \qquad \mathrm{(C) \ 40 } \qquad \mathrm{(D) \ 43 } \qquad \mathrm{(E) \ 45 }$
| [
"$2310 = 2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11$. We wish to figure out the number of ways to distribute these prime factors amongst 3 different integers, without over counting triples which are simply permutations of one another.\n\n\nWe can account for permutations by assuming WLOG that $a$ contains the prime factor 2. Thus, there are $3^4$ ways to position the other 4 prime numbers. Note that, with the exception of when all of the prime factors belong to $a$, we have over counted each case twice, as for when we put certain prime factors into $b$ and the rest into $c$, we count the exact same case when we put those prime factors which were in $b$ into $c$.\n\n\nThus, our total number of cases is $\\frac{3^4 - 1}{2} = 40 \\Rightarrow \\boxed{C}.$\n\n\n",
"The prime factorization of $2310$ is $2310 = 2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11.$ Therefore, we have the equation \\[abc = 2310 = 2 \\cdot 3 \\cdot 5 \\cdot 7 \\cdot 11,\\]where $a, b, c$ must be distinct positive integers and order does not matter. There are $3$ ways to assign each prime number on the right-hand side to one of the variables $a, b, c,$ which gives $3^5 = 243$ solutions for $(a, b, c).$ However, three of these solutions have two $1$s and one $2310,$ which contradicts the fact that $a, b, c$ must be distinct. Because each prime factor appears only once, all other solutions have $a, b, c$ distinct. Correcting for this, we get $243 - 3 = 240$ ordered triples $(a, b, c)$ where $a, b, c$ are all distinct.\n\n\nFinally, since order does not matter, we must divide by $3!,$ the number of ways to order $a, b, c.$ This gives the final answer, \\[\\frac{240}{3!} = \\frac{240}{6} = \\boxed{40}.\\]\n\n\n"
] | 2 | ./CreativeMath/AHSME/1995_AHSME_Problems/29.json | AHSME |
1995_AHSME_Problems | 3 | 0 | Arithmetic | Multiple Choice | The total in-store price for an appliance is $\textdollar 99.99$. A television commercial advertises the same product for three easy payments of $\textdollar 29.98$ and a one-time shipping and handling charge of $\textdollar 9.98$. How many cents are saved by buying the appliance from the television advertiser?
$\mathrm{(A) \ 6 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 8 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$
| [
"IMPORTANT NOTICE: The original problem statement had \"how much is saved\". However, because this made little sense when the calculations were done, the problem statement was changed to \"how many cents\".\n\n\nWe see that 3 payments of $\\textdollar 29.98$ will be a total cost of $3\\cdot(30-.02)=90-.06$\n\n\nAdding this to $\\textdollar9.98$ we have a total of $99.98-.06$\n\n\nClearly, this differs from $\\textdollar 99.99$ by $7$ cents. Thus, the answer is $\\fbox{\\text{(B)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/3.json | AHSME |
1995_AHSME_Problems | 8 | 0 | Geometry | Multiple Choice | In $\triangle ABC$, $\angle C = 90^\circ, AC = 6$ and $BC = 8$. Points $D$ and $E$ are on $\overline{AB}$ and $\overline{BC}$, respectively, and $\angle BED = 90^\circ$. If $DE = 4$, then $BD =$
$\mathrm{(A) \ 5 } \qquad \mathrm{(B) \ \frac {16}{3} } \qquad \mathrm{(C) \ \frac {20}{3} } \qquad \mathrm{(D) \ \frac {15}{2} } \qquad \mathrm{(E) \ 8 }$
| [
"[asy] size(120); defaultpen(0.7); pair A = (0,6), B = (8,0), C= (0,0), D = (8/3,4), E = (8/3,0), F = (0, 3), G = (38/15,1.6); draw(A--B--E--D--E--B--C--A--B--cycle); label(\"\\(A\\)\",A,W); label(\"\\(B\\)\",B,E); label(\"\\(C\\)\",C,SW); label(\"\\(D\\)\",D,NE); label(\"\\(E\\)\",E,S); label(\"\\(6\\)\",F,W); label(\"\\(4\\)\",G,NW); [/asy]\n\n\n$\\triangle BAC$ is a $6-8-10$ right triangle with hypotenuse $AB = 10$.\n\n\n$\\triangle BDE$ is similar to $\\triangle BAC$ by angle-angle similarity since $E=C = 90^\\circ$ and $B=B$, and thus $\\frac{BD}{BA} = \\frac{DE}{AC}$.\n\n\nSolving the above for $BD$, we get $BD=\\frac{BA\\cdot DE}{AC} = 10\\cdot \\dfrac{4}{6}=\\dfrac{20}{3}\\Rightarrow \\boxed{\\mathrm{(C)}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/8.json | AHSME |
1995_AHSME_Problems | 22 | 0 | Geometry | Multiple Choice | A pentagon is formed by cutting a triangular corner from a rectangular piece of paper. The five sides of the pentagon have lengths $13, 19, 20, 25$ and $31$, although this is not necessarily their order around the pentagon. The area of the pentagon is
$\mathrm{(A) \ 459 } \qquad \mathrm{(B) \ 600 } \qquad \mathrm{(C) \ 680 } \qquad \mathrm{(D) \ 720 } \qquad \mathrm{(E) \ 745 }$
| [
"\\begin{center}\n[asy] defaultpen(linewidth(0.7)); draw((0,0)--(31,0)--(31,25)--(12,25)--(0,20)--cycle); draw((0,20)--(0,25)--(12,25)--cycle,linetype(\"4 4\")); [/asy]\\end{center}\nSince the pentagon is cut from a rectangle, the cut-off triangle must be right. Since all of the lengths given are integers, it follows that this triangle is a Pythagorean Triple. We know that $31$ and either $25,\\, 20$ must be the dimensions of the rectangle, since they are the largest lengths. With some trial-and-error, if we assign the shortest side, $13$, to be the hypotenuse of the triangle, we see the $5-12-13$ triple. Indeed this works, by placing the $31$ side opposite from the $19$ side and the $25$ side opposite from the $20$ side, leaving the cutaway side to be, as before, $13$.\n\n\nTo find the area of the pentagon, we subtract the area of the triangle from that of the big rectangle: $31\\cdot 25-\\frac{12\\cdot5}{2}=775-30=745\\Longrightarrow \\boxed{\\mathrm{(E)}745}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/22.json | AHSME |
1995_AHSME_Problems | 18 | 0 | Geometry | Multiple Choice | Two rays with common endpoint $O$ forms a $30^\circ$ angle. Point $A$ lies on one ray, point $B$ on the other ray, and $AB = 1$. The maximum possible length of $OB$ is
$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ \frac {1 + \sqrt {3}}{\sqrt 2} } \qquad \mathrm{(C) \ \sqrt{3} } \qquad \mathrm{(D) \ 2 } \qquad \mathrm{(E) \ \frac{4}{\sqrt{3}} }$
| [
"Triangle $OAB$ has the property that $\\angle O=30^{\\circ}$ and $AB=1$. \n\n\nFrom the Law of Sines, $\\frac{\\sin{\\angle A}}{OB}=\\frac{\\sin{\\angle O}}{AB}$. \n\n\nSince $\\sin 30^\\circ = \\frac{1}{2}$, we have:\n\n\n$\\frac{\\sin{\\angle A}}{OB}=\\frac{\\frac{1}{2}}{1}$\n\n\n$2\\sin{\\angle A}=OB$. \n\n\nThe maximum of $\\sin{\\angle A}$ is $1$ when $\\angle A = 90^\\circ$, so the maximum of $OB$ is $2\\Rightarrow \\mathrm{(D)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/18.json | AHSME |
1995_AHSME_Problems | 4 | 0 | Algebra | Multiple Choice | If $M$ is $30 \%$ of $Q$, $Q$ is $20 \%$ of $P$, and $N$ is $50 \%$ of $P$, then $\frac {M}{N} =$
$\mathrm{(A) \ \frac {3}{250} } \qquad \mathrm{(B) \ \frac {3}{25} } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ \frac {6}{5} } \qquad \mathrm{(E) \ \frac {4}{3} }$
| [
"We are given: $M=\\frac{3Q}{10}$, $Q=\\frac{P}{5}$, $N=\\frac{P}{2}$. We want M in terms of N, so we substitute N into everything:\n\n\n$\\frac{2}{5}N=\\frac{P}{5}=Q$\n\n\n$M=\\frac{3N}{25}$\n\n\n$\\frac{M}{N}=\\frac{3}{25} \\Rightarrow \\mathrm{(B)}$\n\n\n",
"Alternatively, picking an arbitrary value for $Q$ of $100$, we find that $M = 30\\% \\cdot 100 = 0.30 \\cdot 100 = 30$.\n\n\nWe find that $Q = 20\\% \\cdot P$, meaning $100 = 0.2\\cdot P$, giving $P = \\frac{100}{0.2} = \\frac{1000}{2} = 500$.\n\n\nFinally, since $N$ is $50\\%$ of $P$, we have $N = 50\\% \\cdot 500 = 0.5 \\cdot 500 = 250$.\n\n\nThus, $M = 30$ and $N = 250$, so their ratio $\\frac{M}{N} = \\frac{30}{250} = \\frac{3}{25} \\Rightarrow \\mathrm{(B)}$\n\n\nThis method does not prove that the answer must be constant, but it proves that if the answer is a constant, it must be $B$.\n\n\n"
] | 2 | ./CreativeMath/AHSME/1995_AHSME_Problems/4.json | AHSME |
1995_AHSME_Problems | 14 | 0 | Algebra | Multiple Choice | If $f(x) = ax^4 - bx^2 + x + 5$ and $f( - 3) = 2$, then $f(3) =$
$\mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }$
| [
"$f(-x) = a(-x)^4 - b(-x)^2 - x + 5$\n\n\n$f(-x) = ax^4 - bx^2 - x + 5$\n\n\n$f(-x) = (ax^4 - bx^2 + x + 5) - 2x$\n\n\n$f(-x) = f(x) - 2x$. \n\n\nThus $f(3) = f(-3)-2(-3) = 8 \\Rightarrow \\mathrm{(E)}$.\n\n\n",
"If $f(-3) = 2$, then $81a - 9b + -3 + 5 = 2$. Simplifying, we get $81a - 9b = 0$.\n\n\nGetting an expression for $f(3)$, we find $f(3) = 81a - 9b + 3 + 5$. Since the first two terms sum up to zero, we get $f(3) = 8$, which is answer $\\mathrm{(E)}$\n\n\n\n\n\n\n",
"Substituting $x = -3$, we get\n\\[f(-3) = 81a - 9b - 3 + 5 = 81a - 9b + 2.\\]\nBut $f(-3) = 2$, so $81a - 9b + 2 = 2$, which means $81a - 9b = 0$. Then\n\\[f(3) = 81a - 9b + 3 + 5 = 0 + 3 + 5 = \\boxed{8}.\\]\n\n\n"
] | 3 | ./CreativeMath/AHSME/1995_AHSME_Problems/14.json | AHSME |
1995_AHSME_Problems | 15 | 0 | Number Theory | Multiple Choice | Five points on a circle are numbered 1,2,3,4, and 5 in clockwise order. A bug jumps in a clockwise direction from one point to another around the circle; if it is on an odd-numbered point, it moves one point, and if it is on an even-numbered point, it moves two points. If the bug begins on point 5, after 1995 jumps it will be on point
$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \ 2 } \qquad \mathrm{(C) \ 3 } \qquad \mathrm{(D) \ 4 } \qquad \mathrm{(E) \ 5 }$
| [
"Let us see how the bug moves.\n\n\nFirst, we see that if it starts at point 5, it moves to point 1.\n\n\nAt point 1, it moves to point 2.\n\n\nAt point 2, since 2 is even, it moves to point 4.\n\n\nThen at point 4, it moves to point 1.\n\n\nWe can see that at this point, the bug will cycle between 1, 2, and 4\n\n\nMore specifically, we can see that all numbers congruent to 0 (mod 3) will have the bug on point 4 on that step number.\n\n\nThus, we can conclude that the answer is $\\fbox{\\text{(D)}}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/15.json | AHSME |
1995_AHSME_Problems | 5 | 0 | Geometry | Multiple Choice | A rectangular field is 300 feet wide and 400 feet long. Random sampling indicates that there are, on the average, three ants per square inch through out the field. [12 inches = 1 foot.] Of the following, the number that most closely approximates the number of ants in the field is
$\mathrm{(A) \ \text{500 thousand} } \qquad \mathrm{(B) \ \text{5 million} } \qquad \mathrm{(C) \ \text{50 million} } \qquad \mathrm{(D) \ \text{500 million} } \qquad \mathrm{(E) \ \text{5 billion} }$
| [
"The rectangular field is $300 \\text{ feet} \\cdot \\frac{12 \\text{ inches}}{1 \\text{ foot}} = 3600 \\text{ inches}$ wide and $400 \\text{ feet} \\cdot \\frac{12 \\text{ inches}}{1 \\text{ foot}} = 4800 \\text{ inches}$ inches long.\n\n\nIt has an area of $3.6\\cdot 10^3 \\cdot 4.8 \\cdot 10^3 = 17.28 \\cdot 10^6$ square inches. \n\n\nWe multiply by $3$ to account for the ants to get approximately $50 \\cdot 10^6$, which is $50$ million $\\Rightarrow \\mathrm{(C)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/5.json | AHSME |
1995_AHSME_Problems | 19 | 0 | Geometry | Multiple Choice | Equilateral triangle $DEF$ is inscribed in equilateral triangle $ABC$ such that $\overline{DE} \perp \overline{BC}$. The ratio of the area of $\triangle DEF$ to the area of $\triangle ABC$ is
$\mathrm{(A) \ \frac {1}{6} } \qquad \mathrm{(B) \ \frac {1}{4} } \qquad \mathrm{(C) \ \frac {1}{3} } \qquad \mathrm{(D) \ \frac {2}{5} } \qquad \mathrm{(E) \ \frac {1}{2} }$
| [
"[asy]draw((0,0)--(3,0)); draw((0,0)--(1.5, 2.59807621)); draw((3,0)--(1.5, 2.59807621)); draw((1,0)--(1, 1.73205081)); draw((1,0)--(2.5, 0.866025404)); draw((1, 1.73205081)--(2.5, 0.866025404));[/asy]\n\n\nLet's take one of the smaller right triangles. Without loss of generality, let the smaller leg be $1$. Since the triangle is a 30-60-90 right triangle, then the other leg is $\\sqrt{3}$ and the hypotenuse is $2$. The side length of the bigger triangle is $1 + 2 = 3$ and the side length of the smaller triangle is $\\sqrt{3}$. The ratio of the areas of two similar triangles is the square of the ratio of two corresponding side lengths, so the ratio of the area of triangle DEF to the area of triangle ABC is $\\left(\\frac{\\sqrt{3}}{3}\\right)^2 = \\dfrac{1}{3} \\rightarrow (C)$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/19.json | AHSME |
1995_AHSME_Problems | 23 | 0 | Geometry | Multiple Choice | The sides of a triangle have lengths $11,15,$ and $k$, where $k$ is an integer. For how many values of $k$ is the triangle obtuse?
$\mathrm{(A) \ 5 } \qquad \mathrm{(B) \ 7 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 13 } \qquad \mathrm{(E) \ 14 }$
| [
"By the Law of Cosines, a triangle is obtuse if the sum of the squares of two of the sides of the triangles is less than the square of the third. The largest angle is either opposite side $15$ or side $k$. If $15$ is the largest side, \n\n\n\\[15^2 >11^2 + k^2 \\Longrightarrow k < \\sqrt{104}\\]\n\n\nBy the Triangle Inequality we also have that $k > 4$, so $k$ can be $5, 6, 7, \\ldots , 10$, or $6$ values.\n\n\nIf $k$ is the largest side, \n\n\n\\[k^2 >11^2 + 15^2 \\Longrightarrow k > \\sqrt{346}\\]\n\n\nCombining with the Triangle Inequality $19 \\le k < 26$, or $7$ values. These total $13\\ \\mathrm{(D)}$ values of $k$. \n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/23.json | AHSME |
1995_AHSME_Problems | 9 | 0 | Counting | Multiple Choice | Consider the figure consisting of a square, its diagonals, and the segments joining the midpoints of opposite sides. The total number of triangles of any size in the figure is
$\mathrm{(A) \ 10 } \qquad \mathrm{(B) \ 12 } \qquad \mathrm{(C) \ 14 } \qquad \mathrm{(D) \ 16 } \qquad \mathrm{(E) \ 18 }$
| [
"[asy]draw((0,0)--(2,2)); draw((0,2)--(2,0)); draw((0,0)--(0,2)); draw((0,2)--(2,2)); draw((2,2)--(2,0)); draw((0,0)--(2,0)); draw((1,0)--(1,2)); draw((0,1)--(2,1));[/asy]\n\n\nThere are 8 little triangles, 4 triangles with twice the area, and 4 triangles with four times the area of the smaller triangles. $8+4+4=16\\Rightarrow \\mathrm{(D)}$\n\n\n"
] | 1 | ./CreativeMath/AHSME/1995_AHSME_Problems/9.json | AHSME |
1977_AHSME_Problems | 20 | 0 | Counting | Multiple Choice | $\begin{tabular}{ccccccccccccc}& & & & & & C & & & & & &\\ & & & & & C & O & C & & & & &\\ & & & & C & O & N & O & C & & & &\\ & & & C & O & N & T & N & O & C & & &\\ & & C & O & N & T & E & T & N & O & C & &\\ & C & O & N & T & E & S & E & T & N & O & C &\\ C & O & N & T & E & S & T & S & E & T & N & O & C \end{tabular}$
For how many paths consisting of a sequence of horizontal and/or vertical line segments, with each segment connecting a pair of adjacent letters in the diagram above, is the word CONTEST spelled out as the path is traversed from beginning to end?
$\textbf{(A) }63\qquad \textbf{(B) }128\qquad \textbf{(C) }129\qquad \textbf{(D) }255\qquad \textbf{(E) }\text{none of these}$
| [
"The path will either start on the left side, start on the right side, or go down the middle. Say it starts on the left side. Instead of counting it starting from the $C$, let's start at the T and work backwards. Starting at the $T$, there are $2$ options to choose the $S$. Then, there are $2$ options to chose the $E$, $T$, $N$, $O$, and $C$, for a total of $2^6=64$ possibilities. But we need to subtract the case where it does down the middle, which makes it $63$ possibilities. The right side also has $63$ possibilities by symmetry. There is $1$ way for it to go down the middle, for a total of $63+63+1=127$ paths, so the answer is $\\textbf{(E) }\\text{none of these}$.\n\n\n~alexanderruan\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/20.json | AHSME |
1977_AHSME_Problems | 16 | 0 | Algebra | Multiple Choice | If $i^2 = -1$, then the sum \[\cos{45^\circ} + i\cos{135^\circ} + \cdots + i^n\cos{(45 + 90n)^\circ} + \cdots + i^{40}\cos{3645^\circ}\]
equals
$\text{(A)}\ \frac{\sqrt{2}}{2} \qquad \text{(B)}\ -10i\sqrt{2} \qquad \text{(C)}\ \frac{21\sqrt{2}}{2} \qquad\\ \text{(D)}\ \frac{\sqrt{2}}{2}(21 - 20i) \qquad \text{(E)}\ \frac{\sqrt{2}}{2}(21 + 20i)$
| [
"Solution by e_power_pi_times_i\n\n\nNotice that the sequence repeats every $4$ terms. These $4$ terms are $\\cos(45^\\circ)$, $i\\cos(135^\\circ)$, $-\\cos(225^\\circ)$, and $-i\\cos(315^\\circ)$. This is because $\\cos(\\theta) = \\cos(360^\\circ - \\theta)$. Also $\\cos(\\theta) = -\\cos(180^\\circ - \\theta)$, so the $4$ terms are just $\\cos(45^\\circ)$, $-i\\cos(45^\\circ)$, $\\cos(45^\\circ)$, and $-i\\cos(45^\\circ)$, which adds up to $\\sqrt{2}-i\\sqrt{2}$. The sequence repeats $10$ times, and an extra $i^{40}\\cos(3645^\\circ)$. The sum is $10(\\sqrt{2}-i\\sqrt{2})+\\dfrac{\\sqrt{2}}{2} = \\dfrac{21\\sqrt{2}}{2} - 10i\\sqrt{2} = \\boxed{\\text{(D) }\\dfrac{\\sqrt{2}}{2}(21-20i)}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/16.json | AHSME |
1977_AHSME_Problems | 6 | 0 | Algebra | Multiple Choice | If $x, y$ and $2x + \frac{y}{2}$ are not zero, then
$\left( 2x + \frac{y}{2} \right)^{-1} \left[(2x)^{-1} + \left( \frac{y}{2} \right)^{-1} \right]$
equals
$\textbf{(A) }1\qquad \textbf{(B) }xy^{-1}\qquad \textbf{(C) }x^{-1}y\qquad \textbf{(D) }(xy)^{-1}\qquad \textbf{(E) }\text{none of these}$
| [
"We can write $\\left( 2x+ \\frac{y}{2} \\right)^{-1} = \\left( \\frac{4x+y}{2} \\right)^{-1} = \\frac{2}{4x+y}$.\n\n\nThen, the expression simplifies to \\[\\left( 2x + \\frac{y}{2} \\right)^{-1} \\left[(2x)^{-1} + \\left( \\frac{y}{2} \\right)^{-1} \\right] \\Rightarrow \\frac{2}{4x+y} \\left( \\frac{1}{2x} + \\frac{2}{y} \\right) \\Rightarrow \\frac{2}{4x+y} \\left( \\frac{y+4x}{2xy} \\right) \\Rightarrow \\frac{1}{xy}.\\]\n\n\n\n\nThus, our answer is $\\textbf{(D) }(xy)^{-1}$. ~jiang147369\n\n\n\n\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/6.json | AHSME |
1977_AHSME_Problems | 7 | 0 | Algebra | Multiple Choice | If $t = \frac{1}{1 - \sqrt[4]{2}}$, then $t$ equals
$\text{(A)}\ (1-\sqrt[4]{2})(2-\sqrt{2})\qquad \text{(B)}\ (1-\sqrt[4]{2})(1+\sqrt{2})\qquad \text{(C)}\ (1+\sqrt[4]{2})(1-\sqrt{2}) \qquad \\ \text{(D)}\ (1+\sqrt[4]{2})(1+\sqrt{2})\qquad \text{(E)}-(1+\sqrt[4]{2})(1+\sqrt{2})$
| [
"Solution by e_power_pi_times_i\n\n\n$t = \\dfrac{1}{1-\\sqrt[4]{2}} = (\\dfrac{1}{1-\\sqrt[4]{2}})(\\dfrac{1+\\sqrt[4]{2}}{1+\\sqrt[4]{2}}) = (\\dfrac{1+\\sqrt[4]{2}}{1-\\sqrt{2}})(\\dfrac{1+\\sqrt{2}}{1+\\sqrt{2}}) = \\dfrac{(1+\\sqrt[4]{2})(1+\\sqrt{2})}{-1} = \\text{(E)}-(1+\\sqrt[4]{2})(1+\\sqrt{2})$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/7.json | AHSME |
1977_AHSME_Problems | 17 | 0 | Probability | Multiple Choice | Three fair dice are tossed at random (i.e., all faces have the same probability of coming up).
What is the probability that the three numbers turned up can be arranged to form an arithmetic progression with common difference one?
$\textbf{(A) }\frac{1}{6}\qquad \textbf{(B) }\frac{1}{9}\qquad \textbf{(C) }\frac{1}{27}\qquad \textbf{(D) }\frac{1}{54}\qquad \textbf{(E) }\frac{7}{36}$
| [
"Solution by e_power_pi_times_i\n\n\n\n\nThe denominator of the probability is $6^3 = 216$. We are asked to find the probability that the three numbers are consecutive. The three numbers can be arranged in $6$ ways, and there are $4$ triplets of consecutive numbers from $1-6$. The probability is $\\dfrac{6\\cdot4}{216} = \\dfrac{24}{216} = \\boxed{\\text{(B) }\\dfrac{1}{9}}$.\n\n\n"
] | 1 | ./CreativeMath/AHSME/1977_AHSME_Problems/17.json | AHSME |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.