problem stringlengths 15 4.7k | solution stringlengths 2 11.9k | answer stringclasses 51
values | problem_type stringclasses 8
values | question_type stringclasses 4
values | problem_is_valid stringclasses 1
value | solution_is_valid stringclasses 1
value | source stringclasses 6
values | synthetic bool 1
class |
|---|---|---|---|---|---|---|---|---|
In the [obtuse triangle](https://artofproblemsolving.com/wiki/index.php/Obtuse_triangle) $ABC$ with $\angle C>90^\circ$, $AM=MB$, $MD\perp BC$, and $EC\perp BC$ ($D$ is on $BC$, $E$ is on $AB$, and $M$ is on $EB$). If the [area](https://artofproblemsolving.com/wiki/index.php/Area) of $\triangle ABC$ is $24$, then the a... | [1984AHSME26.png](https://artofproblemsolving.com/wiki/index.php/File:1984AHSME26.png)
We let side $BC$ have length $a$, $AB$ have length $c$, and $\angle ABC$ have angle measure $\beta$. We then have that
\[[ABC]=24=\frac{AB\cdot BC\sin{\angle ABC}}{2}=\frac{ac\sin{\beta}}{2}\]
Now I shall find the lengths of $BD$ and... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The number of distinct pairs of [integers](https://artofproblemsolving.com/wiki/index.php/Integers) $(x, y)$ such that $0<x<y$ and $\sqrt{1984}=\sqrt{x}+\sqrt{y}$ is
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ }4 \qquad \mathrm{(E) \ } 7$ | We can simplify $\sqrt{1984}$ to $8\sqrt{31}$. Therefore, the only solutions are $a\sqrt{31}+b\sqrt{31}$ such that $a+b=8$ and $0<a<b$. The only solutions to this are $a=1, b=7; a=2, b=6; a=3, b=5$. Each of these gives distinct pairs of $(x, y)$, so there are $3$ pairs, $\boxed{\text{C}}$. | 3 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
The largest [integer](https://artofproblemsolving.com/wiki/index.php/Integer) $n$ for which $n^{200}<5^{300}$ is
$\mathrm{(A) \ }8 \qquad \mathrm{(B) \ }9 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ }11 \qquad \mathrm{(E) \ } 12$ | Since both sides are positive, we can take the $100th$ root of both sides to find the largest integer $n$ such that $n^2<5^3$. Fortunately, this is simple to evaluate: $5^3=125$, and the largest [square](https://artofproblemsolving.com/wiki/index.php/Perfect_square) less than $125$ is $11^2=121$, so the largest $n$ is ... | 11 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Exactly three of the interior angles of a convex [polygon](https://artofproblemsolving.com/wiki/index.php/Polygon) are obtuse. What is the maximum number of sides of such a polygon?
$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }5 \qquad \mathrm{(C) \ } 6 \qquad \mathrm{(D) \ } 7 \qquad \mathrm{(E) \ }8$ | All angle measures are in degrees.
The sum of the interior angle measures of an $n$-gon is $180(n-2)=180n-360$. Let the three obtuse angle measures be $o_1, o_2,$ and $o_3$, and the $n-3$ acute angle measures be $a_1, a_2, a_3, \cdots$.
Since $90<o_i<180$, $3(90)=270<o_1+o_2+o_3<3(180)=540$.
Similarly, since $0<a_i... | 6 | Geometry | MCQ | Yes | Yes | amc_aime | false |
If $A=20^\circ$ and $B=25^\circ$, then the value of $(1+\tan A)(1+\tan B)$ is
$\mathrm{(A)\ } \sqrt{3} \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ } 1+\sqrt{2} \qquad \mathrm{(D) \ } 2(\tan A+\tan B) \qquad \mathrm{(E) \ }\text{none of these}$ | Solution 1
First, let's leave everything in variables and see if we can simplify $(1+\tan A)(1+\tan B)$.
We can write everything in terms of sine and cosine to get $\left(\frac{\cos A}{\cos A}+\frac{\sin A}{\cos A}\right)\left(\frac{\cos B}{\cos B}+\frac{\sin B}{\cos B}\right)=\frac{(\sin A+\cos A)(\sin B+\cos B)}{\co... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A wooden [cube](https://artofproblemsolving.com/wiki/index.php/Cube) with edge length $n$ units (where $n$ is an integer $>2$) is painted black all over. By slices parallel to its faces, the cube is cut into $n^3$ smaller cubes each of unit length. If the number of smaller cubes with just one face painted black is equa... | Notice that if we remove the outer layer of unit cubes from the entire cube, we're left with a smaller cube of side length $n-2$. Notice also that this contains all of the unpainted cubes and nothing else, so there are $(n-2)^3$ unpainted cubes. Also notice that if we take one face of the cube and remove the outer edge... | 8 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
How many integers $x$ satisfy the equation $(x^2-x-1)^{x+2}=1?$
$\mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ } 5 \qquad \mathrm{(E) \ }\text{none of these}$ | Notice that any power of $1$ is $1$, so $x^2-x-1=1$ would give valid solutions.
$x^2-x-2=0$
$(x-2)(x+1)=0$
$x=2, -1$
Also, $-1$ to an even power also gives $1$, so we check $x^2-x-1=-1$
$x^2-x=0$
$x(x-1)=0$
$x=0, 1$
However, $x=1$ gives an odd power of $-1$, so this is discarded. Finally, notice that anything to the $... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In a circle with center $O$, $AD$ is a [diameter](https://artofproblemsolving.com/wiki/index.php/Diameter), $ABC$ is a [chord](https://artofproblemsolving.com/wiki/index.php/Chord), $BO=5$ and $\angle ABO= \ \stackrel{\frown}{CD} \ =60^\circ$. Then the length of $BC$ is
$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }3+\sqrt{... | Since $\angle CAD$ is inscribed and intersects an arc of length $60^\circ$, $\angle CAD=30^\circ$. Thus, $\triangle ABO$ is a $30-60-90$ right triangle. Thus, $AO=BO\sqrt{3}=5\sqrt{3}$ and $AB=2BO=10$. Since $AO$ and $DO$ are both radii, $DO=AO=5\sqrt{3}$ and $AD=10\sqrt{3}$. Since $\angle ACD$ is inscribed in a semici... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Consider a sequence $x_1, x_2, x_3, \cdots$ defined by
$x_1=\sqrt[3]{3}$
$x_2=(\sqrt[3]{3})^{\sqrt[3]{3}}$
and in general
$x_n=(x_{n-1})^{\sqrt[3]{3}}$ for $n>1$.
What is the smallest value of $n$ for which $x_n$ is an [integer](https://artofproblemsolving.com/wiki/index.php/Integer)?
$\mathrm{(A)\ } 2 \qquad \mathrm{(... | First, we will use induction to prove that $x_n=\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-1}\right)}$
We see that $x_1=\sqrt[3]{3}=\sqrt[3]{3}^{\left(\sqrt[3]{3}^0\right)}$. This is our base case.
Now, we have $x_n=(x_{n-1})^{\sqrt[3]{3}}=\left(\sqrt[3]{3}^{\left(\sqrt[3]{3}^{n-2}\right)}\right)^{\sqrt[3]{3}}=\sqrt[3]{3}^{\lef... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In right $\triangle ABC$ with legs $5$ and $12$, arcs of circles are drawn, one with center $A$ and radius $12$, the other with center $B$ and radius $5$. They intersect the [hypotenuse](https://artofproblemsolving.com/wiki/index.php/Hypotenuse) in $M$ and $N$. Then, $MN$ has length
$\mathrm{(A)\ } 2 \qquad \mathrm{(B... | First of all, from the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem), $AB=\sqrt{AC^2+BC^2}=\sqrt{12^2+5^2}=\sqrt{144+25}=\sqrt{169}=13$. Also, since $AM$ and $AC$ are radii of the same circle, $AM=AC=12$. Therefore, $MB=AB-AM=13-12=1$. Also, since $BN$ and $BC$ are radii of t... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Let $\lfloor x \rfloor$ be the greatest integer less than or equal to $x$. Then the number of real solutions to $4x^2-40\lfloor x \rfloor +51=0$ is
$\mathrm{(A)\ } 0 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \ } 2 \qquad \mathrm{(D) \ } 3 \qquad \mathrm{(E) \ }4$ | We can rearrange the equation into $4x^2=40\lfloor x \rfloor-51$. Obviously, the RHS is an integer, so $4x^2=n$ for some integer $n$. We can therefore make the substitution $x=\frac{\sqrt{n}}{2}$ to get
\[40\left\lfloor \frac{\sqrt{n}}{2}\right\rfloor-51=n\]
(We'll try the case where $x=-\frac{\sqrt{n}}{2}$ later.) Now... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
John scores $93$ on this year's AHSME. Had the old scoring system still been in effect, he would score only $84$ for the same answers.
How many questions does he leave unanswered? (In the new scoring system that year, one receives $5$ points for each correct answer,
$0$ points for each wrong answer, and $2$ points for... | Let $c$, $w$, and $u$ be the number of correct, wrong, and unanswered questions respectively. From the old scoring system, we have $30+4c-w=84$, from the new scoring system we have $5c+2u=93$, and since there are $30$ problems in the AHSME, $c+w+u=30$. Solving the simultaneous equations yields $u=9$, which is $\boxed{B... | 9 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
A parabola $y = ax^{2} + bx + c$ has vertex $(4,2)$. If $(2,0)$ is on the parabola, then $abc$ equals
$\textbf{(A)}\ -12\qquad \textbf{(B)}\ -6\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 6\qquad \textbf{(E)}\ 12$ | Consider the quadratic in completed square form: it must be $y=a(x-4)^{2}+2$. Now substitute $x=2$ and $y=0$ to give $a=-\frac{1}{2}$. Now expanding gives $y=-\frac{1}{2}x^{2}+4x-6$, so the product is $-\frac{1}{2} \cdot 4 \cdot -6 = 3 \cdot 4 = 12$, which is $\boxed{E}$. | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In $\triangle ABC, AB = 8, BC = 7, CA = 6$ and side $BC$ is extended, as shown in the figure, to a point $P$ so that $\triangle PAB$
is similar to $\triangle PCA$. The length of $PC$ is
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 8\qquad \textbf{(C)}\ 9\qquad \textbf{(D)}\ 10\qquad \textbf{(E)}\ 11$ | Since we are given that $\triangle{PAB}\sim\triangle{PCA}$, we have $\frac{PC}{PA}=\frac{6}{8}=\frac{PA}{PC+7}$.
Solving for $PA$ in $\frac{PC}{PA}=\frac{6}{8}=\frac{3}{4}$ gives us $PA=\frac{4PC}{3}$.
We also have $\frac{PA}{PC+7}=\frac{3}{4}$. Substituting $PA$ in for our expression yields $\frac{\frac{4PC}{3}}{PC+... | 9 | Geometry | MCQ | Yes | Yes | amc_aime | false |
A plane intersects a right circular cylinder of radius $1$ forming an ellipse.
If the major axis of the ellipse is $50\%$ longer than the minor axis, the length of the major axis is
$\textbf{(A)}\ 1\qquad \textbf{(B)}\ \frac{3}{2}\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ \frac{9}{4}\qquad \textbf{(E)}\ 3$ | We note that we can draw the minor axis to see that because the minor axis is the minimum distance between two opposite points on the ellipse, we can draw a line through two opposite points of the cylinder, and so the minor axis is $2(1) = 2$. Therefore, our answer is $2(1.5) = 3$, and so our answer is $\boxed{E}$. | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Let $p(x) = x^{2} + bx + c$, where $b$ and $c$ are integers.
If $p(x)$ is a factor of both $x^{4} + 6x^{2} + 25$ and $3x^{4} + 4x^{2} + 28x + 5$, what is $p(1)$?
$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 1\qquad \textbf{(C)}\ 2\qquad \textbf{(D)}\ 4\qquad \textbf{(E)}\ 8$ | $p(x)$ must be a factor of $3(x^4+6x^2+25)-(3x^4+4x^2+28x+5)=14x^2-28x+70=14(x^2-2x+5)$.
Therefore $p(x)=x^2 -2x+5$ and $p(1)=4$.
The answer is $\fbox{(D) 4}$. | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
$ABCDE$ is a regular pentagon. $AP, AQ$ and $AR$ are the perpendiculars dropped from $A$ onto $CD, CB$ extended and $DE$ extended,
respectively. Let $O$ be the center of the pentagon. If $OP = 1$, then $AO + AQ + AR$ equals
$\textbf{(A)}\ 3\qquad \textbf{(B)}\ 1 + \sqrt{5}\qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 2 ... | To solve the problem, we compute the area of regular pentagon $ABCDE$ in two different ways. First, we can divide regular pentagon $ABCDE$ into five congruent triangles.
If $s$ is the side length of the regular pentagon, then each of the triangles $AOB$, $BOC$, $COD$, $DOE$, and $EOA$ has base $s$ and height 1, so the... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Two of the altitudes of the scalene triangle $ABC$ have length $4$ and $12$.
If the length of the third altitude is also an integer, what is the biggest it can be?
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ \text{none of these}$ | Assume we have a scalene triangle $ABC$. Arbitrarily, let $12$ be the height to base $AB$ and $4$ be the height to base $AC$. Due to area equivalences, the base $AC$ must be three times the length of $AB$.
Let the base $AB$ be $x$, thus making $AC = 3x$. Thus, setting the final height to base $BC$ to $h$, we note that... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many ordered triples $(a, b, c)$ of non-zero real numbers have the property that each number is the product of the other two?
$\textbf{(A)}\ 1 \qquad \textbf{(B)}\ 2 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 4 \qquad \textbf{(E)}\ 5$ | We have $ab = c$, $bc = a$, and $ca = b$, so multiplying these three equations together gives $a^{2}b^{2}c^{2} = abc \implies abc(abc-1)=0$, and as $a$, $b$, and $c$ are all non-zero, we cannot have $abc = 0$, so we must have $abc = 1$. Now substituting $bc = a$ gives $a(bc) = 1 \implies a^2 = 1 \implies a = \pm 1$. If... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A triangular corner with side lengths $DB=EB=1$ is cut from equilateral triangle ABC of side length $3$.
The perimeter of the remaining quadrilateral is
$\text{(A)} \ 6 \qquad \text{(B)} \ 6\frac{1}{2} \qquad \text{(C)} \ 7 \qquad \text{(D)} \ 7\frac{1}{2} \qquad \text{(E)} \ 8$ | $\triangle DBE$ is similar to $\triangle ABC$ by AA, so $\overline{DE}$ = 1 by similarity, and $\overline{CE} = \overline{AD} = 2$, by subtraction. Thus the perimeter is $3+2+2+1 = 8$, or $\boxed{E}$. -slackroadia | 8 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Evaluate
$\log_{10}(\tan 1^{\circ})+\log_{10}(\tan 2^{\circ})+\log_{10}(\tan 3^{\circ})+\cdots+\log_{10}(\tan 88^{\circ})+\log_{10}(\tan 89^{\circ}).$
$\textbf{(A)}\ 0 \qquad \textbf{(B)}\ \frac{1}{2}\log_{10}(\frac{\sqrt{3}}{2}) \qquad \textbf{(C)}\ \frac{1}{2}\log_{10}2\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ \text... | Because $\tan x \tan (90^\circ - x) = \tan x \cot x = 1$, $\tan 45^\circ = 1$, and $\log a + \log b = \log {ab}$, the answer is $\log_{10} {\tan 1^\circ \tan 2^\circ \dots \tan 89^\circ} = \log_{10} 1 = 0.$ $\boxed{\textbf{(A)}}.$ | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A cube of cheese $C=\{(x, y, z)| 0 \le x, y, z \le 1\}$ is cut along the planes $x=y, y=z$ and $z=x$. How many pieces are there?
(No cheese is moved until all three cuts are made.)
$\textbf{(A)}\ 5 \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 7 \qquad \textbf{(D)}\ 8 \qquad \textbf{(E)}\ 9$ | The cut $x = y$ separates the cube into points with $x y$, and analogous results apply for the other cuts. Thus, which piece a particular point is in depends only on the relative sizes of its coordinates $x$, $y$, and $z$ - for example, all points with the ordering $x < y < z$ are in the same piece. Thus, as there are... | 6 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
How many primes less than $100$ have $7$ as the ones digit? (Assume the usual base ten representation)
$\text{(A)} \ 4 \qquad \text{(B)} \ 5 \qquad \text{(C)} \ 6 \qquad \text{(D)} \ 7 \qquad \text{(E)} \ 8$ | List out all numbers that have 7 as the ones digit less than 100: ${7, 17, 27, 37, 47, 57, 67, 77, 87, 97}$. Only $7, 17,37, 47,67,$ and $97$ are prime. Thus, it is $\boxed{C}$. -slackroadia | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
$\frac{2^1+2^0+2^{-1}}{2^{-2}+2^{-3}+2^{-4}}$ equals
$\text{(A)} \ 6 \qquad \text{(B)} \ 8 \qquad \text{(C)} \ \frac{31}{2} \qquad \text{(D)} \ 24 \qquad \text{(E)} \ 512$ | We can factorise to give $\frac{2^1 + 2^0 + 2^{-1}}{2^{-3}(2^1 + 2^0 + 2^{-1})} = \frac{1}{2^{-3}} = 8$, which is $\boxed{\text{B}}$. | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A student recorded the exact percentage frequency distribution for a set of measurements, as shown below.
However, the student neglected to indicate $N$, the total number of measurements. What is the smallest possible value of $N$?
\[\begin{tabular}{c c}\text{measured value}&\text{percent frequency}\\ \hline 0 & 12.5\... | Note that $12.5\% = \frac{1}{8}$, $25\% = \frac{1}{4}$, and $50\% = \frac{1}{2}$. Thus, since the frequencies must be integers, $N$ must be divisible by $2$, $4$, and $8$ (so that $\frac{N}{8}$ etc. are integers), or in other words, $N$ is divisible by $8$. Thus the smallest possible value of $N$ is the smallest positi... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
If $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^3 + bx^2 + 1$, then $b$ is
$\textbf{(A)}\ -2\qquad \textbf{(B)}\ -1\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ 1\qquad \textbf{(E)}\ 2$ | Using polynomial division, we find that the remainder is $(2a+b)x+(a+b+1)$, so for the condition to hold, we need this remainder to be $0$. This gives $2a+b=0$ and $a+b+1=0$, so $b=-2a$ and $a-2a+1=0 \implies a=1 \implies b=-2$, which is $\boxed{\text{A}}.$ | -2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In one of the adjoining figures a square of side $2$ is dissected into four pieces so that $E$ and $F$ are the midpoints
of opposite sides and $AG$ is perpendicular to $BF$. These four pieces can then be reassembled into a rectangle as shown
in the second figure. The ratio of height to base, $XY / YZ$, in this rectang... | Within $WXYZ$, the parallelogram piece has vertical side $BF = \sqrt{1^2 + 2^2} = \sqrt{5}$, and diagonal side $FD = 1$. Thus the triangle in the bottom-right hand corner (the one with horizontal side $YZ$) must have hypotenuse $1$, and the only such triangle in the original figure is $\triangle AFG$, so we deduce $YZ ... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
For how many integers $x$ does a triangle with side lengths $10, 24$ and $x$ have all its angles acute?
$\textbf{(A)}\ 4\qquad \textbf{(B)}\ 5\qquad \textbf{(C)}\ 6\qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ \text{more than } 7$ | We first notice that the sides $10$ and $24$, can be part of $2$ different right triangles, one with sides $10,24,26$, and the other with a leg
somewhere between $21$ and $22$. We now notice that if $x$ is less than or equal to $21$, one of the angles is obtuse, and that the same is the same for
any value of $x$ above... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
If $b$ and $c$ are constants and $(x + 2)(x + b) = x^2 + cx + 6$, then $c$ is
$\textbf{(A)}\ -5\qquad \textbf{(B)}\ -3\qquad \textbf{(C)}\ -1\qquad \textbf{(D)}\ 3\qquad \textbf{(E)}\ 5$ | We first start out by expanding the left side of the equation, $(x+2)(x+b)=x^{2}+bx+2x+2b=x^2+(2+b)x+2b=x^2+cx+6$. We know the constant
terms have to be equal so we have $2b=6$, so $b=3$. Plugging $b=3$ back in yields $x^2+(2+3)x+6=x^2+cx+6$. Thus, $c=5 \implies \boxed{\text{E}}$. | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
$(-1)^{5^{2}}+1^{2^{5}}=$
$\textrm{(A)}\ -7\qquad\textrm{(B)}\ -2\qquad\textrm{(C)}\ 0\qquad\textrm{(D)}\ 1\qquad\textrm{(E)}\ 57$ | $(-1)^{5^2} + 1^{2^5} = (-1)^{25}+1^{32}=-1+1=0$ thus the answer is C. | 0 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Suppose that 7 boys and 13 girls line up in a row. Let $S$ be the number of places in the row where a boy and a girl are standing next to each other. For example, for the row $\text{GBBGGGBGBGGGBGBGGBGG}$ we have that $S=12$. The average value of $S$ (if all possible orders of these 20 people are considered) is closest... | We approach this problem using Linearity of Expectation. Consider a pair of two people standing next to each other. Ignoring all other people, the probability that a boy is standing on the left position and a girl is standing on the right position is $\frac7{20}\cdot\frac{13}{19}$. Similarly, if a girl is standing on t... | 9 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
In the figure, $ABCD$ is an isosceles trapezoid with side lengths $AD=BC=5$, $AB=4$, and $DC=10$. The point $C$ is on $\overline{DF}$ and $B$ is the midpoint of hypotenuse $\overline{DE}$ in right triangle $DEF$. Then $CF=$
$\textrm{(A)}\ 3.25\qquad\textrm{(B)}\ 3.5\qquad\textrm{(C)}\ 3.75\qquad\textrm{(D)}\ 4.0\qquad... | Drop perpendiculars from $A$ and $B$; then the triangle $DBY$ is similar to $DEF$ but with corresponding sides of half the length.
$XY=AB=4$ and $DX=YC=3$, hence $DY=7\implies DF=14\implies CF=\boxed{4.0}$. | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
If $a,b>0$ and the triangle in the first quadrant bounded by the coordinate axes and the graph of $ax+by=6$ has area 6, then $ab=$
$\mathrm{(A) \ 3 } \qquad \mathrm{(B) \ 6 } \qquad \mathrm{(C) \ 12 } \qquad \mathrm{(D) \ 108 } \qquad \mathrm{(E) \ 432 }$ | Setting $y=0$ we have that the $x-$intercept of the line is $x= \frac{6}{a}$. Similarly setting $x=0$ we find the $y-$intercept to be $y= \frac{6}{b}$. Then $\frac{18}{ab}=\frac{1}{2}\cdot\frac{6}{a}\cdot\frac{6}{b}$ so that $\frac{18}{ab} = 6$, simplifying we would get $ab=3$. Hence the answer is $\fbox{A}$.
-$\LaTeX$... | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
For how many integers $n$ between 1 and 100 does $x^2+x-n$ factor into the product of two linear factors with integer coefficients?
$\mathrm{(A) \ 0 } \qquad \mathrm{(B) \ 1 } \qquad \mathrm{(C) \ 2 } \qquad \mathrm{(D) \ 9 } \qquad \mathrm{(E) \ 10 }$ | For $x^2+x-n$ to factor into a product of two linear factors, we must have $x^2+x-n = (x + a)(x + b)$, where $a$ and $b$ are integers.
By expansion of the product of the linear factors and comparison to the original quadratic,
$ab = -n$
$a + b = 1$.
The only way for this to work if $n$ is a positive integer is if $a = ... | 9 | Algebra | MCQ | Yes | Yes | amc_aime | false |
How many positive integers less than $50$ have an odd number of positive integer divisors?
$\text{(A) } 3\quad \text{(B) } 5\quad \text{(C) } 7\quad \text{(D) } 9\quad \text{(E) } 11$ | Divisors come in pairs, unless there is an integer square root, so we just need the perfect squares below $50$. There are $7$, so $\fbox{C}$ | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
If the six solutions of $x^6=-64$ are written in the form $a+bi$, where $a$ and $b$ are real, then the product of those solutions with $a>0$ is
$\text{(A) } -2\quad \text{(B) } 0\quad \text{(C) } 2i\quad \text{(D) } 4\quad \text{(E) } 16$ | This equation is $r^6e^{6\theta i}=2^6e^{(\pi\pm 2k\pi) i}$. Solving in the usual way, $r=2$ and $\theta\in\{\pm 30^\circ,\pm 90^\circ,\pm 150^\circ\}$.
Thus there are only two solutions with positive real part, and they are conjugates, so their product is $r^2=4$ which is $\fbox{D}$ | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Ten people form a circle. Each picks a number and tells it to the two neighbors adjacent to them in the circle. Then each person computes and announces the average of the numbers of their two neighbors. The figure shows the average announced by each person (not the original number the person picked.)
The number pick... | For $i\in\{1,2,3,\ldots,10\},$ suppose Person $i$ picks the number $a_i$ and announces the number $i.$ We wish to find $a_6.$
Taking the indices modulo $10,$ we are given that $\frac{a_{i-1}+a_{i+1}}{2}=i,$ from which $a_{i-1}+a_{i+1}=2i.$
We have ten equations: five with odd-numbered indices and five with even-numbere... | 1 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $R_n=\tfrac{1}{2}(a^n+b^n)$ where $a=3+2\sqrt{2}$ and $b=3-2\sqrt{2}$, and $n=0,1,2,\cdots,$ then $R_{12345}$ is an integer. Its units digit is
$\text{(A) } 1\quad \text{(B) } 3\quad \text{(C) } 5\quad \text{(D) } 7\quad \text{(E) } 9$ | $(a+b)R_n=\tfrac12(a^{n+1}+b^{n+1})+ab\cdot\tfrac12(a^{n-1}+b^{n-1})=R_{n+1}+abR_{n-1}$
but $a+b=6$ and $ab=1$, so this means that $R_{n+1}=6R_n-R_{n-1}$. Since $R_0=1$ and $R_1=3$, all terms are integers and we can continue the recurrence $\rm{mod}\ 10$ to get the repeating sequence $1,3,7,9,7,3$. The number $12345$ i... | 9 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
If for any three distinct numbers $a$, $b$, and $c$ we define $f(a,b,c)=\frac{c+a}{c-b}$, then $f(1,-2,-3)$ is
$\textbf {(A) } -2 \qquad \textbf {(B) } -\frac{2}{5} \qquad \textbf {(C) } -\frac{1}{4} \qquad \textbf {(D) } \frac{2}{5} \qquad \textbf {(E) } 2$ | If we plug in $1$ as $a$, $-2$ as $b$, and $-3$ as $c$ in the expression $\frac{c+a}{c-b}$, then we get $\frac{-3+1}{-3-(-2)}=\frac{-2}{-1}=2$, which is choice $\boxed{\textbf{E}}$. | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
A positive integer $N$ is a palindrome if the integer obtained by reversing the sequence of digits of $N$ is equal to $N$. The year 1991 is the only year in the current century with the following 2 properties:
(a) It is a palindrome
(b) It factors as a product of a 2-digit prime palindrome and a 3-digit prime palindrom... | Solution by e_power_pi_times_i
Notice that all four-digit palindromes are divisible by $11$, so that is our two-digit prime. Because the other factor is a three-digit number, we are looking at palindromes between $1100$ and $2000$, which also means that the last digit of the three-digit number is $1$. Checking through ... | 4 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
An $n$-digit positive integer is cute if its $n$ digits are an arrangement of the set $\{1,2,...,n\}$ and its first
$k$ digits form an integer that is divisible by $k$, for $k = 1,2,...,n$. For example, $321$ is a cute $3$-digit integer because $1$ divides $3$, $2$ divides $32$, and $3$ divides $321$. How many cute ... | $\fbox{C}$ Let the number be $abcdef$. We know $1$ will always divide $a$. $5$ must divide $abcde$, so $e$ must be $5$ or $0$, but we can only use the digits $1$ to $6$, so $e = 5$. $4$ must divide $abcd$, so it must divide $cd$ (the test for divisibility by 4 is that the last two digits form a number divisible by 4), ... | 2 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
$(4^{-1}-3^{-1})^{-1}=$
(A) $-12$ (B) $-1$ (C) $\frac{1}{12}$ (D) $1$ (E) $12$ | $\fbox{A}$ This is $\frac{1}{\frac{1}{4} - \frac{1}{3}} = \frac{1}{\frac{-1}{12}} = -12.$ | -12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The number of positive integers $k$ for which the equation
\[kx-12=3k\]
has an integer solution for $x$ is
$\text{(A) } 3\quad \text{(B) } 4\quad \text{(C) } 5\quad \text{(D) } 6\quad \text{(E) } 7$ | $\fbox{D}$
$kx -12 = 3k$
$-12=3k-kx$
$-12=k(3-x)$
$\frac{-12}{k}=3-x$
Positive factors of $-12$:
$1,2,3,4,6,12$
6 factors, each of which have an integer solution for $x$ in $\frac{-12}{k}=3-x$ | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $y=mx+b$ be the image when the line $x-3y+11=0$ is reflected across the $x$-axis. The value of $m+b$ is
$\text{(A) -6} \quad \text{(B) } -5\quad \text{(C) } -4\quad \text{(D) } -3\quad \text{(E) } -2$ | $\fbox{C}$
First we want to put this is slope-intercept form, so we get $y=\dfrac{1}{3}x+\dfrac{11}{3}$. When we reflect a line across the x-axis, the line's x-intercept stays the same, while the y-intercept and the slope are additive inverses of the original line. Since $m+b$ is the sum of the slope and the y-intercep... | -4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
How many pairs of positive integers (a,b) with $a+b\le 100$ satisfy the equation
\[\frac{a+b^{-1}}{a^{-1}+b}=13?\]
$\text{(A) } 1\quad \text{(B) } 5\quad \text{(C) } 7\quad \text{(D) } 9\quad \text{(E) } 13$ | $\fbox{C}$ We can rewrite the left-hand side as $\frac{a^2b+a}{b+ab^2}$ by multiplying the numerator and denominator by $ab$. Now we can multiply the top part of the fraction by the denominator to give $a^2b+a = 13b + 13ab^2 \implies ab(a-13b)=13b-a = -(a-13b) \implies (ab+1)(a-13b) = 0$. Now, as $a$ and $b$ are positi... | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If
\[\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\]
for three positive numbers $x,y$ and $z$, all different, then $\frac{x}{y}=$
$\text{(A) } \frac{1}{2}\quad \text{(B) } \frac{3}{5}\quad \text{(C) } \frac{2}{3}\quad \text{(D) } \frac{5}{3}\quad \text{(E) } 2$ | $\fbox{E}$ We have $\frac{x+y}{z} = \frac{x}{y} \implies xy+y^2=xz$ and $\frac{y}{x-z} = \frac{x}{y} \implies y^2=x^2-xz \implies x^2-y^2=xz$. Equating the two expressions for $xz$ gives $xy+y^2=x^2-y^2 \implies x^2-xy-2y^2=0 \implies (x+y)(x-2y)=0$, so as $x+y$ cannot be $0$ for positive $x$ and $y$, we must have $x-2... | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The 2-digit integers from 19 to 92 are written consecutively to form the integer $N=192021\cdots9192$. Suppose that $3^k$ is the highest power of 3 that is a factor of $N$. What is $k$?
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) more than } 3$ | Solution 1
We can determine if our number is divisible by $3$ or $9$ by summing the digits. Looking at the one's place, we can start out with $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ and continue cycling though the numbers from $0$ through $9$. For each one of these cycles, we add $0 + 1 + ... + 9 = 45$. This is divisible by $9$... | 1 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Let $ABCD$ be a parallelogram of area $10$ with $AB=3$ and $BC=5$. Locate $E,F$ and $G$ on segments $\overline{AB},\overline{BC}$ and $\overline{AD}$, respectively, with $AE=BF=AG=2$. Let the line through $G$ parallel to $\overline{EF}$ intersect $\overline{CD}$ at $H$. The area of quadrilateral $EFHG$ is
$\text{(A) } ... | $\fbox{C}$ Use vectors. Place an origin at $A$, with $B = p, D = q, C = p + q$. We know that $\|p \times q\|=10$, and also $E=\frac{2}{3}p, F=p+\frac{2}{5}q, G = \frac{2}{5}q$, and now we can find the area of $EFHG$ by dividing it into two triangles and using cross-products (the expressions simplify using the fact that... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Let $i=\sqrt{-1}$. The product of the real parts of the roots of $z^2-z=5-5i$ is
$\text{(A) } -25\quad \text{(B) } -6\quad \text{(C) } -5\quad \text{(D) } \frac{1}{4}\quad \text{(E) } 25$ | Applying the quadratic formula gives \[z=\frac{-1\pm\sqrt{21-20i}}{2}\]
Let \[\sqrt{21-20i}=a+bi\] where $a$ and $b$ are real. Squaring both sides and equating real and imaginary parts gives \[a^2-b^2=21\] \[2ab=-20\] Substituting $b=-\frac{10}{a}$, letting $a^2=n$ and solving gives $n=25, -4$, from which we see that $... | -6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
For integers $a,b,$ and $c$ define $\fbox{a,b,c}$ to mean $a^b-b^c+c^a$. Then $\fbox{1,-1,2}$ equals:
$\text{(A) } -4\quad \text{(B) } -2\quad \text{(C) } 0\quad \text{(D) } 2\quad \text{(E) } 4$ | Plug in the values for $a,b,c$ and you get $1^{-1} - (-1)^2 + 2^1 \Rightarrow 1-1+2 \Rightarrow \fbox{2}$
$\fbox{D}$ | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Consider the non-decreasing sequence of positive integers
\[1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots\]
in which the $n^{th}$ positive integer appears $n$ times. The remainder when the $1993^{rd}$ term is divided by $5$ is
$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) } 4$ | The sequence of 1's ends at position 1, and the sequence of 2's ends at position 1+2, and the sequence of $n$'s ends at position $1+2+\dots+n$.
Therefore we want to find the smallest integer $n$ that satisfies $\frac{n(n+1)}{2}\geq 1993$.
By trial and error, the value of $n$ is $63$, and $63 \div 5$ has a remainder of ... | 3 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
How many ordered pairs $(m,n)$ of positive integers are solutions to
\[\frac{4}{m}+\frac{2}{n}=1?\]
$\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } \text{more than }6$ | Multiply both sides by $mn$ to clear the denominator. Moving all the terms to the right hand side, the equation becomes $0 = mn-4n-2m$. Adding 8 to both sides allows us to factor the equation as follows: $(m-4)(n-2) = 8$. Since the problem only wants integer pairs $(m,n)$, the pairs are given by the factors of 8, which... | 4 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The sides of $\triangle ABC$ have lengths $6,8,$ and $10$. A circle with center $P$ and radius $1$ rolls around the inside of $\triangle ABC$, always remaining tangent to at least one side of the triangle. When $P$ first returns to its original position, through what distance has $P$ traveled?
$\text{(A) } 10\quad \tex... | Start by considering the triangle traced by $P$ as the circle moves around the triangle. It turns out this triangle is similar to the $6-8-10$ triangle (Proof: Realize that the slope of the line made while the circle is on $AC$ is the same as line $AC$ and that it makes a right angle when the circle switches from being... | 12 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Last year a bicycle cost $160 and a cycling helmet $40. This year the cost of the bicycle increased by $5\%$, and the cost of the helmet increased by $10\%$. The percent increase in the combined cost of the bicycle and the helmet is:
$\text{(A) } 6\%\quad \text{(B) } 7\%\quad \text{(C) } 7.5\%\quad \text{(D) } 8\%\quad... | Since the bicycle originally cost $160, the new cost will be 160 * 5% = 160 * 1/20 = 8, and 160 + 8 = 168. The new cost of the helmet will be 40 * 10% = 40 * 1/10 = 4, and 40 + 4 = 44. 160 + 40 = 200, and 168 + 44 = 212, so the total percent increase is 12/200 = 6%, or A. | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Triangle $ABC$ is inscribed in a circle, and $\angle B = \angle C = 4\angle A$. If $B$ and $C$ are adjacent vertices of a regular polygon of $n$ sides inscribed in this circle, then $n=$
$\textbf{(A)}\ 5 \qquad\textbf{(B)}\ 7 \qquad\textbf{(C)}\ 9 \qquad\textbf{(D)}\ 15 \qquad\textbf{(E)}\ 18$ | We solve for $\angle A$ as follows: \[4\angle A+4\angle A+\angle A=180\implies 9\angle A=180\implies \angle A=20.\] That means that minor arc $\widehat{BC}$ has measure $40^\circ$. We can fit a maximum of $\frac{360}{40}=\boxed{\textbf{(C) }9}$ of these arcs in the circle.
--Solution by [TheMaskedMagician](https://arto... | 9 | Geometry | MCQ | Yes | Yes | amc_aime | false |
A sample consisting of five observations has an arithmetic mean of $10$ and a median of $12$. The smallest value that the range (largest observation minus smallest) can assume for such a sample is
$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 5 \qquad\textbf{(D)}\ 7 \qquad\textbf{(E)}\ 10$ | The minimum range occurs in the set $\{7,7,12,12,12\}$, so the answer is $\boxed{\textbf{(C)}\ 5}$ | 5 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $x$ and $y$ are non-zero real numbers such that
\[|x|+y=3 \qquad \text{and} \qquad |x|y+x^3=0,\]
then the integer nearest to $x-y$ is
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ 3 \qquad\textbf{(E)}\ 5$ | We have two cases to consider: $x$ is positive or $x$ is negative. If $x$ is positive, we have $x+y=3$ and $xy+x^3=0$
Solving for $y$ in the top equation gives us $3-x$. Plugging this in gives us: $x^3-x^2+3x=0$. Since we're told $x$ is not zero, we can divide by $x$, giving us: $x^2-x+3=0$
The discriminant of this is... | -3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
In the $xy$-plane, the segment with endpoints $(-5,0)$ and $(25,0)$ is the diameter of a circle. If the point $(x,15)$ is on the circle, then $x=$
$\textbf{(A)}\ 10 \qquad\textbf{(B)}\ 12.5 \qquad\textbf{(C)}\ 15\qquad\textbf{(D)}\ 17.5 \qquad\textbf{(E)}\ 20$ | We see that the center of this circle is at $\left(\frac{-5+25}{2},0\right)=(10,0)$. The radius is $\frac{30}{2}=15$. So the equation of this circle is \[(x-10)^2+y^2=225.\] Substituting $y=15$ yields $(x-10)^2=0$ so $x=\boxed{\textbf{(A) }10}$.
--Solution by [TheMaskedMagician](https://artofproblemsolving.comhttp://ww... | 10 | Geometry | MCQ | Yes | Yes | amc_aime | false |
In the sequence
\[..., a, b, c, d, 0, 1, 1, 2, 3, 5, 8,...\]
each term is the sum of the two terms to its left. Find $a$.
$\textbf{(A)}\ -3 \qquad\textbf{(B)}\ -1 \qquad\textbf{(C)}\ 0 \qquad\textbf{(D)}\ 1 \qquad\textbf{(E)}\ 3$ | We work backwards to find $a$.
\[d+0=1\implies d=1\]
\[c+1=0\implies c=-1\]
\[b+(-1)=1\implies b=2\]
\[a+2=-1\implies a=\boxed{\textbf{(A)}-3.}\]
--Solution by [TheMaskedMagician](https://artofproblemsolving.comhttp://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685) | -3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
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 \... | If we change $0$, the units column would be incorrect.
If we change $1$, then the leading $1$ in the sum would be incorrect.
However, 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.
Changing the $2$... | 8 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
If $f(x) = ax^4 - bx^2 + x + 5$ and $f( - 3) = 2$, then $f(3) =$
$\mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }$ | $f(-x) = a(-x)^4 - b(-x)^2 - x + 5$
$f(-x) = ax^4 - bx^2 - x + 5$
$f(-x) = (ax^4 - bx^2 + x + 5) - 2x$
$f(-x) = f(x) - 2x$.
Thus $f(3) = f(-3)-2(-3) = 8 \Rightarrow \mathrm{(E)}$. | 8 | Algebra | MCQ | Yes | Yes | amc_aime | false |
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) \ \... | Triangle $OAB$ has the property that $\angle O=30^{\circ}$ and $AB=1$.
From the [Law of Sines](https://artofproblemsolving.com/wiki/index.php/Law_of_Sines), $\frac{\sin{\angle A}}{OB}=\frac{\sin{\angle O}}{AB}$.
Since $\sin 30^\circ = \frac{1}{2}$, we have:
$\frac{\sin{\angle A}}{OB}=\frac{\frac{1}{2}}{1}$
$2\sin{\an... | 2 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Two nonadjacent vertices of a [rectangle](https://artofproblemsolving.com/wiki/index.php/Rectangle) are $(4,3)$ and $(-4,-3)$, and the [coordinates](https://artofproblemsolving.com/wiki/index.php/Coordinate) of the other two vertices are integers. The number of such rectangles is
$\mathrm{(A) \ 1 } \qquad \mathrm{(B) \... | 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 $x^2+y^2=25$.... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
There exist positive integers $A,B$ and $C$, with no [common factor](https://artofproblemsolving.com/wiki/index.php/Greatest_common_divisor) 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) ... | \[A \log_{200} 5 + B \log_{200} 2 = C\]
Simplifying and taking the [logarithms](https://artofproblemsolving.com/wiki/index.php/Logarithm) away,
\[5^A \cdot 2^B=200^C=2^{3C} \cdot 5^{2C}\]
Therefore, $A=2C$ and $B=3C$. Since $A, B,$ and $C$ are relatively prime, $C=1$, $B=3$, $A=2$. $A+B+C=6 \Rightarrow \mathrm{(A)}$ | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
The addition below is incorrect. What is the largest digit that can be changed to make the addition correct?
$\begin{tabular}{rr}&\ \texttt{6 4 1}\\ &\texttt{8 5 2}\\ &+\texttt{9 7 3}\\ \hline &\texttt{2 4 5 6}\end{tabular}$
$\text{(A)}\ 4\qquad\text{(B)}\ 5\qquad\text{(C)}\ 6\qquad\text{(D)}\ 7\qquad\text{(E)}\ 8$ | Doing the addition as is, we get $641 + 852 + 973 = 2466$. This number is $10$ larger than the desired sum of $2456$. Therefore, we must make one of the three numbers $10$ smaller.
We may either change $641 \rightarrow 631$, $852 \rightarrow 842$, or $973 \rightarrow 963$. Either change results in a valid sum. The ... | 7 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
A function $f$ from the integers to the integers is defined as follows:
\[f(n) =\begin{cases}n+3 &\text{if n is odd}\\ \ n/2 &\text{if n is even}\end{cases}\]
Suppose $k$ is odd and $f(f(f(k))) = 27$. What is the sum of the digits of $k$?
$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 1... | First iteration
To get $f(k) = 27$, you could either have $f(27 - 3)$ and add $3$, or $f(27\cdot 2)$ and divide by $2$.
If you had the former, you would have $f(24)$, and the function's rule would have you divide. Thus, $k=54$ is the only number for which $f(k) = 27$.
Second iteration
Going out one step, if you have ... | 6 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Each day Walter gets $3$ dollars for doing his chores or $5$ dollars for doing them exceptionally well. After $10$ days of doing his chores daily, Walter has received a total of $36$ dollars. On how many days did Walter do them exceptionally well?
$\text{(A)}\ 3\qquad\text{(B)}\ 4\qquad\text{(C)}\ 5\qquad\text{(D)}\ 6\... | If Walter had done his chores for $10$ days without doing any of them well, he would have earned $3 \cdot 10 = 30$ dollars. He got $6$ dollars more than this.
He gets a $5 - 3 = 2$ dollar bonus every day he does his chores well. Thus, he did his chores exceptionally well $\frac{6}{2} = 3$ days, and the answer is $\bo... | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $\texttt{a}$ and $\texttt{b}$ are digits for which
$\begin{array}{ccc}& 2 & a\\ \times & b & 3\\ \hline & 6 & 9\\ 9 & 2\\ \hline 9 & 8 & 9\end{array}$
then $\texttt{a+b =}$
$\mathrm{(A)\ } 3 \qquad \mathrm{(B) \ }4 \qquad \mathrm{(C) \ } 7 \qquad \mathrm{(D) \ } 9 \qquad \mathrm{(E) \ }12$ | From the units digit calculation, we see that the units digit of $a\times 3$ is $9$. Since $0 \le a \le 9$ and $a$ is an integer, the only value of $a$ that works is is $a=3$. As a double-check, that does work, since $23 \times 3 = 69$, which is the first line of the multiplication.
The second line of the multiplicat... | 7 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
How many two-digit positive integers $N$ have the property that the sum of $N$ and the number obtained by reversing the order of the digits of is a perfect square?
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$ | Let $N = 10t + u$, where $t$ is the tens digit and $u$ is the units digit.
The condition of the problem is that $10t + u + 10u + t$ is a perfect square.
Simplifying and factoring, we want $11(t+u)$ to be a perfect square.
Thus, $t+u$ must at least be a multiple of $11$, and since $t$ and $u$ are digits, the only multip... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
A line $x=k$ intersects the graph of $y=\log_5 x$ and the graph of $y=\log_5 (x + 4)$. The distance between the points of intersection is $0.5$. Given that $k = a + \sqrt{b}$, where $a$ and $b$ are integers, what is $a+b$?
$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}... | Since the line $x=k$ is vertical, we are only concerned with vertical distance.
In other words, we want to find the value of $k$ for which the distance $|\log_5 x - \log_5 (x+4)| = \frac{1}{2}$
Since $\log_5 x$ is a strictly increasing function, we have:
$\log_5 (x + 4) - \log_5 x = \frac{1}{2}$
$\log_5 (\frac{x+4}{x})... | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Ashley, Betty, Carlos, Dick, and Elgin went shopping. Each had a whole number of dollars to spend, and together they had $56$ dollars. The absolute difference between the amounts Ashley and Betty had to spend was $19$ dollars. The absolute difference between the amounts Betty and Carlos had was $7$ dollars, between Car... | Working backwards, if $6 \le E \le 10$, then $6 \pm 11 \le A \le 10 \pm 11$. Since $A$ is a positive integer, $17 \le A \le 21$.
Since $17 \le A \le 21$, we know that $17 \pm 19 \le B \le 21 \pm 19$. But if $B=36$, which is the smallest possible "plus" value, then $E + A + B = 6 + 17 + 36 = 59$, which is too much m... | 10 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
Let $ABCD$ be a parallelogram and let $\overrightarrow{AA^\prime}$, $\overrightarrow{BB^\prime}$, $\overrightarrow{CC^\prime}$, and $\overrightarrow{DD^\prime}$ be parallel rays in space on the same side of the plane determined by $ABCD$. If $AA^{\prime} = 10$, $BB^{\prime}= 8$, $CC^\prime = 18$, and $DD^\prime = 22$ a... | Let $ABCD$ be a unit square with $A(0,0,0)$, $B(0,1,0)$, $C(1,1,0)$, and $D(1,0,0)$. Assume that the rays go in the +z direction. In this case, $A^\prime(0,0,10)$, $B^\prime(0,1,8)$, $C^\prime(1,1,18)$, and $D^\prime(1,0,22)$. Finding the midpoints of $A^\prime C^\prime$ and $B^\prime D^\prime$ gives $M(\frac{1}{2}... | 1 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$, angle $APB$ is twice angle $ACB$, and $\overline{AC}$ intersects $\overline{BP}$ at point $D$. If $PB = 3$ and $PD= 2$, then $AD\cdot CD =$
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 6\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ ... | The product of two lengths with a common point brings to mind the [Power of a Point Theorem](https://artofproblemsolving.com/wiki/index.php/Power_of_a_Point_Theorem).
Since $PA = PB$, we can make a circle with radius $PA$ that is centered on $P$, and both $A$ and $B$ will be on that circle. Since $\angle APB = \wideha... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many ordered triples of integers $(a,b,c)$ satisfy $|a+b|+c = 19$ and $ab+|c| = 97$?
$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 10\qquad\textbf{(E)}\ 12$ | [WLOG](https://artofproblemsolving.com/wiki/index.php/WLOG), let $a \ge 0$, and let $a \ge b$. We can say this because if we have one solution $(a,b) = (a_0, b_0)$ with $a_0 \ge 0$ and $a_0 > b_0$, we really have the four solutions $(a_0, b_0), (-a_0, -b_0), (b_0, a_0), (-b_0, -a_0)$ by the symmetry of the original pr... | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Call a positive real number special if it has a decimal representation that consists entirely of digits $0$ and $7$. For example, $\frac{700}{99}= 7.\overline{07}= 7.070707\cdots$ and $77.007$ are special numbers. What is the smallest $n$ such that $1$ can be written as a sum of $n$ special numbers?
$\textbf{(A)}\ 7\qq... | Define a super-special number to be a number whose decimal expansion only consists of $0$'s and $1$'s. The problem is equivalent to finding the number of super-special numbers necessary to add up to $\frac{1}{7}=0.142857142857\hdots$. This can be done in $8$ numbers if we take
\[0.111111\hdots, 0.011111\hdots, 0.010111... | 8 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
If $x$, $y$, and $z$ are real numbers such that
$(x-3)^2 + (y-4)^2 + (z-5)^2 = 0$,
then $x + y + z =$
$\mathrm{(A)\ } -12 \qquad \mathrm{(B) \ }0 \qquad \mathrm{(C) \ } 8 \qquad \mathrm{(D) \ } 12 \qquad \mathrm{(E) \ }50$ | If the sum of three squared expressions is zero, then each expression itself must be zero, since $a^2 \ge 0$ with the equality iff $a=0$.
In this case, $x-3=0$, $y-4=0$, and $z-5=0$. Adding the three equations and moving the constant to the right gives $x + y + z = 12$, and the answer is $\boxed{D}$. | 12 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Mientka Publishing Company prices its bestseller Where's Walter? as follows:
$C(n) =\left\{\begin{matrix}12n, &\text{if }1\le n\le 24\\ 11n, &\text{if }25\le n\le 48\\ 10n, &\text{if }49\le n\end{matrix}\right.$
where $n$ is the number of books ordered, and $C(n)$ is the cost in dollars of $n$ books. Notice that $25$ ... | Clearly, the areas of concern are where the piecewise function shifts value.
Since $C(25) = 11\cdot 25 = 275$, we want to find the least value of $n$ for which $C(n) > 275$.
If $n \le 24$, then $C(n) = 12n$, so for $C(n) > 275$, $12n > 275$, which is equivalent to $n > 22.91$. Thus, both $n=23$ and $n=24$ will be mo... | 6 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Let $R$ be a rectangle. How many circles in the plane of $R$ have a diameter both of whose endpoints are vertices of $R$?
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }5 \qquad \mathrm{(E) \ }6$ | There are $6$ pairs of vertices of $R$. However, both diagonals determine the same circle, therefore the answer is $\boxed{5}$.
[asy] size(200); defaultpen(0.8); pair A=(0,0), B=(5,0), C=(5,2), D=(0,2); draw ( A--B--C--D--cycle ); draw( circle( (A+B)/2, length((A-B)/2) ), red ); draw( circle( (C+B)/2, length((C-B)/2... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
How many different prime numbers are factors of $N$ if
$\log_2 ( \log_3 ( \log_5 (\log_ 7 N))) = 11?$
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }3 \qquad \mathrm{(D) \ } 4\qquad \mathrm{(E) \ }7$ | Re-writing as exponents, we have $\log_3 ( \log_5 (\log_ 7 N)) = 2^{11}$, and so forth, such that $N = 7^{5^{3^{2^{11}}}}$, which only has $7$ as a prime factor $\mathbf{(A)}$. | 1 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
How many triangles have area $10$ and vertices at $(-5,0),(5,0)$ and $(5\cos \theta, 5\sin \theta)$ for some angle $\theta$?
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }2 \qquad \mathrm{(C) \ }4 \qquad \mathrm{(D) \ }6 \qquad \mathrm{(E) \ } 8$ | The triangle can be seen as having the base on the $x$ axis and height $|5\sin\theta|$. The length of the base is $10$, thus the height must be $2$. The equation $|\sin\theta| = \frac 25$ has $\boxed{4}$ solutions, one in each quadrant.
Visually, the set of points of the form $(5\cos \theta, 5\sin \theta)$ is a circ... | 4 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Three cards, each with a positive integer written on it, are lying face-down on a table. Casey, Stacy, and Tracy are told that
(a) the numbers are all different,
(b) they sum to $13$, and
(c) they are in increasing order, left to right.
First, Casey looks at the number on the leftmost card and says, "I don't have enou... | Initially, there are the following possibilities for the numbers on the cards: $(1,2,10)$, $(1,3,9)$, $(1,4,8)$, $(1,5,7)$, $(2,3,8)$, $(2,4,7)$, $(2,5,6)$, and $(3,4,6)$.
If Casey saw the number $3$, she would have known the other two numbers. As she does not, we eliminated the possibility $(3,4,6)$.
At this moment, i... | 4 | Logic and Puzzles | MCQ | Yes | Yes | amc_aime | false |
A point $(x,y)$ in the plane is called a lattice point if both $x$ and $y$ are integers. The area of the largest square that contains exactly three lattice points in its interior is closest to
$\mathrm{(A) \ } 4.0 \qquad \mathrm{(B) \ } 4.2 \qquad \mathrm{(C) \ } 4.5 \qquad \mathrm{(D) \ } 5.0 \qquad \mathrm{(E) \ } 5... | [asy] real e = 0.1; dot((0,-1)); dot((1,-1)); dot((-1,0)); dot((0,0)); dot((1,0)); dot((2,0)); dot((-1,1)); dot((0,1)); dot((1,1)); dot((0,2)); dot((-1,-1)); dot((2,2)); dot((1,2)); dot((2,1)); dot((2,-1)); dot((-1,2)); draw((0.8, -1.4+e)--(1.8-e, 0.6)--(-0.2, 1.6-e)--(-1.2+e, -0.4)--cycle); [/asy]
The best square's s... | 5 | Geometry | MCQ | Yes | Yes | amc_aime | false |
For each positive integer $n$, let
$a_n = \frac{(n+9)!}{(n-1)!}$
Let $k$ denote the smallest positive integer for which the rightmost nonzero digit of $a_k$ is odd. The rightmost nonzero digit of $a_k$ is
$\mathrm{(A) \ }1 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ }5 \qquad \mathrm{(D) \ } 7 \qquad \mathrm{(E) \ }... | We have $a_n = n(n+1)\dots (n+9)$.
The value $a_n$ can be written as $2^{x_n} 5^{y_n} r_n$, where $r_n$ is not divisible by 2 and 5. The number of trailing zeroes is $z_n = \min(x_n,y_n)$. The last non-zero digit is the last digit of $2^{x_n-z_n} 5^{y_n-z_n} r_n$.
Clearly, the last non-zero digit is even iff $x_n - z_n... | 9 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Define $[a,b,c]$ to mean $\frac {a+b}c$, where $c \neq 0$. What is the value of
$\left[[60,30,90],[2,1,3],[10,5,15]\right]?$
$\mathrm{(A) \ }0 \qquad \mathrm{(B) \ }0.5 \qquad \mathrm{(C) \ }1 \qquad \mathrm{(D) \ }1.5 \qquad \mathrm{(E) \ }2$ | Note that $[ta,tb,tc] = \frac{ta+tb}{tc} = \frac{t(a+b)}{tc} = \frac{a+b}{c} = [a,b,c]$.
Thus $[60,30,90] = [2,1,3] = [10,5,15] = \frac{2+1}{3} = 1$, and $[1,1,1] = \frac{1+1}{1} = 2 \Longrightarrow \mathbf{(E)}$. | 2 | Algebra | MCQ | Yes | Yes | amc_aime | false |
If $2^{1998}-2^{1997}-2^{1996}+2^{1995} = k \cdot 2^{1995},$ what is the value of $k$?
$\mathrm{(A) \ } 1 \qquad \mathrm{(B) \ } 2 \qquad \mathrm{(C) \ } 3 \qquad \mathrm{(D) \ } 4 \qquad \mathrm{(E) \ } 5$ | $2^{1998} - 2^{1997} - 2^{1996} + 2^{1995}$
Factor out $2^{1995}$:
$2^{1995}(2^{3} - 2^{2} - 2^{1} + 1)$
Simplify:
$2^{1995}\cdot (8 - 4 - 2 + 1)$
$2^{1995}\cdot (3)$
By comparing the answer with the original equation, $k=3$, and the answer is $\text{(C)}.$ | 3 | Algebra | MCQ | Yes | Yes | amc_aime | false |
Four girls — Mary, Alina, Tina, and Hanna — sang songs in a concert as trios, with one girl sitting out each time. Hanna sang $7$ songs, which was more than any other girl, and Mary sang $4$ songs, which was fewer than any other girl. How many songs did these trios sing?
$\textbf{(A)}\ 7 \qquad \textbf{(B)}\ 8 \qquad ... | Alina and Tina must sing more than $4$, but less than $7$, songs. Therefore, Alina sang $5$ or $6$ songs, and Tina sang $5$ or $6$ songs, with $4$ possible combinations.
However, since every song is a trio, if you add up all the numbers of songs a person sang for all four singers, it must be divisible by $3$. Thus, ... | 7 | Combinatorics | MCQ | Yes | Yes | amc_aime | false |
Consider all [triangles](https://artofproblemsolving.com/wiki/index.php/Triangle) $ABC$ satisfying in the following conditions: $AB = AC$, $D$ is a point on $\overline{AC}$ for which $\overline{BD} \perp \overline{AC}$, $AC$ and $CD$ are integers, and $BD^{2} = 57$. Among all such triangles, the smallest possible valu... | Thus $AD = AC - CD$ and $AB = AC$ are integers. By the [Pythagorean Theorem](https://artofproblemsolving.com/wiki/index.php/Pythagorean_Theorem),
\[AD^2 + 57 = AB^2 \Longrightarrow 1 \cdot 57 = 3 \cdot 19 = (AB - AD)(AB + AD).\]
Thus $AC = AB = \frac {1 + 57}{2} = 29$ or $\frac {3 + 19}{2} = 11 \Longrightarrow \mathrm... | 11 | Geometry | MCQ | Yes | Yes | amc_aime | false |
The graphs of $y = -|x-a| + b$ and $y = |x-c| + d$ intersect at points $(2,5)$ and $(8,3)$. Find $a+c$.
$\mathrm{(A) \ } 7 \qquad \mathrm{(B) \ } 8 \qquad \mathrm{(C) \ } 10 \qquad \mathrm{(D) \ } 13\qquad \mathrm{(E) \ } 18$ | Each of the graphs consists of two orthogonal half-lines. In the first graph both point downwards at a $45^\circ$ angle, in the second graph they point upwards. One can easily find out that the only way how to get these graphs to intersect in two points is the one depicted below:
Obviously, the maximum of the first g... | 10 | Algebra | MCQ | Yes | Yes | amc_aime | false |
There are unique integers $a_{2},a_{3},a_{4},a_{5},a_{6},a_{7}$ such that
\[\frac {5}{7} = \frac {a_{2}}{2!} + \frac {a_{3}}{3!} + \frac {a_{4}}{4!} + \frac {a_{5}}{5!} + \frac {a_{6}}{6!} + \frac {a_{7}}{7!}\]
where $0\leq a_{i} < i$ for $i = 2,3,\ldots,7$. Find $a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7}$.
$\text... | Multiply out the $7!$ to get
\[5 \cdot 6! = (3 \cdot 4 \cdots 7)a_2 + (4 \cdots 7)a_3 + (5 \cdot 6 \cdot 7)a_4 + 42a_5 + 7a_6 + a_7 .\]
By [Wilson's Theorem](https://artofproblemsolving.com/wiki/index.php/Wilson%27s_Theorem) (or by straightforward division), $a_7 + 7(a_6 + 6a_5 + \cdots) \equiv 5 \cdot 6! \equiv -5 \e... | 9 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
Let $x_1, x_2, \ldots , x_n$ be a sequence of integers such that
(i) $-1 \le x_i \le 2$ for $i = 1,2, \ldots n$
(ii) $x_1 + \cdots + x_n = 19$; and
(iii) $x_1^2 + x_2^2 + \cdots + x_n^2 = 99$.
Let $m$ and $M$ be the minimal and maximal possible values of $x_1^3 + \cdots + x_n^3$, respectively. Then $\frac Mm =$
$\mathr... | Clearly, we can ignore the possibility that some $x_i$ are zero, as adding/removing such variables does not change the truth value of any condition, nor does it change the value of the sum of cubes. Thus we'll only consider $x_i\in\{-1,1,2\}$.
Also, order of the $x_i$ does not matter, so we are only interested in the ... | 7 | Algebra | MCQ | Yes | Yes | amc_aime | false |
What is the sum of the digits of the decimal form of the product $2^{1999}\cdot 5^{2001}$?
$\textbf{(A)}\ 2\qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ 5 \qquad \textbf{(D)}\ 7\qquad \textbf{(E)}\ 10$ | $2^{1999}\cdot5^{2001}=2^{1999}\cdot5^{1999}\cdot5^{2}=25\cdot10^{1999}$, a number with the digits "25" followed by 1999 zeros. The sum of the digits in the decimal form would be $2+5=7$, thus making the answer $\boxed{\text{D}}$. | 7 | Number Theory | MCQ | Yes | Yes | amc_aime | false |
What is the largest number of acute angles that a convex hexagon can have?
$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4\qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6$ | The sum of the interior angles of a hexagon is $720$ degrees. In a convex polygon, each angle must be strictly less than $180$ degrees.
Six acute angles can only sum to less than $90\cdot 6 = 540$ degrees, so six acute angles could not form a hexagon.
Five acute angles and one obtuse angle can only sum to less than ... | 3 | Geometry | MCQ | Yes | Yes | amc_aime | false |
Three numbers, $a_1, a_2, a_3$, are drawn randomly and without replacement from the set $\{1, 2, 3,\ldots, 1000\}$. Three other numbers, $b_1, b_2, b_3$, are then drawn randomly and without replacement from the remaining set of $997$ numbers. Let $p$ be the probability that, after suitable rotation, a brick of dimensio... | There is a total of $P(1000,6)$ possible ordered $6$-tuples $(a_1,a_2,a_3,b_1,b_2,b_3).$
There are $C(1000,6)$ possible sets $\{a_1,a_2,a_3,b_1,b_2,b_3\}.$ We have five valid cases for the increasing order of these six elements:
$aaabbb$
$aababb$
$aabbab$
$abaabb$
$ababab$
Note that the $a$'s are different from each... | 5 | Combinatorics | math-word-problem | Yes | Yes | amc_aime | false |
Find the least positive integer $n$ such that no matter how $10^{n}$ is expressed as the product of any two positive integers, at least one of these two integers contains the digit $0$. | If a factor of $10^{n}$ has a $2$ and a $5$ in its [prime factorization](https://artofproblemsolving.com/wiki/index.php/Prime_factorization), then that factor will end in a $0$. Therefore, we have left to consider the case when the two factors have the $2$s and the $5$s separated, so we need to find the first power of ... | 8 | Number Theory | math-word-problem | Yes | Yes | amc_aime | false |
Suppose that $x,$ $y,$ and $z$ are three positive numbers that satisfy the equations $xyz = 1,$ $x + \frac {1}{z} = 5,$ and $y + \frac {1}{x} = 29.$ Then $z + \frac {1}{y} = \frac {m}{n},$ where $m$ and $n$ are [relatively prime](https://artofproblemsolving.com/wiki/index.php/Relatively_prime) positive integers. Find $... | We can rewrite $xyz=1$ as $\frac{1}{z}=xy$.
Substituting into one of the given equations, we have
\[x+xy=5\]
\[x(1+y)=5\]
\[\frac{1}{x}=\frac{1+y}{5}.\]
We can substitute back into $y+\frac{1}{x}=29$ to obtain
\[y+\frac{1+y}{5}=29\]
\[5y+1+y=145\]
\[y=24.\]
We can then substitute once again to get
\[x=\frac15\]
\[z=\f... | 5 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
The number
$\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}$
can be written as $\frac mn$ where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | Solution 1
$\frac 2{\log_4{2000^6}} + \frac 3{\log_5{2000^6}}$
$=\frac{\log_4{16}}{\log_4{2000^6}}+\frac{\log_5{125}}{\log_5{2000^6}}$
$=\frac{\log{16}}{\log{2000^6}}+\frac{\log{125}}{\log{2000^6}}$
$=\frac{\log{2000}}{\log{2000^6}}$
$=\frac{\log{2000}}{6\log{2000}}$
$=\frac{1}{6}$
Therefore, $m+n=1+6=\boxed{007}$
Sol... | 7 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
Find the least positive integer $n$ such that $\frac 1{\sin 45^\circ\sin 46^\circ}+\frac 1{\sin 47^\circ\sin 48^\circ}+\cdots+\frac 1{\sin 133^\circ\sin 134^\circ}=\frac 1{\sin n^\circ}.$ | We apply the identity
\begin{align*} \frac{1}{\sin n \sin (n+1)} &= \frac{1}{\sin 1} \cdot \frac{\sin (n+1) \cos n - \sin n \cos (n+1)}{\sin n \sin (n+1)} \\ &= \frac{1}{\sin 1} \cdot \left(\frac{\cos n}{\sin n} - \frac{\cos (n+1)}{\sin (n+1)}\right) \\ &= \frac{1}{\sin 1} \cdot \left(\cot n - \cot (n+1)\right). \end{... | 1 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
Given that $z$ is a complex number such that $z+\frac 1z=2\cos 3^\circ$, find the least integer that is greater than $z^{2000}+\frac 1{z^{2000}}$. | Using the quadratic equation on $z^2 - (2 \cos 3 )z + 1 = 0$, we have $z = \frac{2\cos 3 \pm \sqrt{4\cos^2 3 - 4}}{2} = \cos 3 \pm i\sin 3 = \text{cis}\,3^{\circ}$.
There are other ways we can come to this conclusion. Note that if $z$ is on the [unit circle](https://artofproblemsolving.com/wiki/index.php/Unit_circle) i... | 0 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
A [sphere](https://artofproblemsolving.com/wiki/index.php/Sphere) is inscribed in the [tetrahedron](https://artofproblemsolving.com/wiki/index.php/Tetrahedron) whose vertices are $A = (6,0,0), B = (0,4,0), C = (0,0,2),$ and $D = (0,0,0).$ The [radius](https://artofproblemsolving.com/wiki/index.php/Radius) of the spher... | The center $I$ of the insphere must be located at $(r,r,r)$ where $r$ is the sphere's radius.
$I$ must also be a distance $r$ from the plane $ABC$
The signed distance between a plane and a point $I$ can be calculated as $\frac{(I-G) \cdot P}{|P|}$, where G is any point on the plane, and P is a vector perpendicular to A... | 5 | Geometry | math-word-problem | Yes | Yes | amc_aime | false |
The solutions to the system of equations
$\log_{225}x+\log_{64}y=4$
$\log_{x}225-\log_{y}64=1$
are $(x_1,y_1)$ and $(x_2,y_2)$. Find $\log_{30}\left(x_1y_1x_2y_2\right)$. | Let $A=\log_{225}x$ and let $B=\log_{64}y$.
From the first equation: $A+B=4 \Rightarrow B = 4-A$.
Plugging this into the second equation yields $\frac{1}{A}-\frac{1}{B}=\frac{1}{A}-\frac{1}{4-A}=1 \Rightarrow A = 3\pm\sqrt{5}$ and thus, $B=1\pm\sqrt{5}$.
So, $\log_{225}(x_1x_2)=\log_{225}(x_1)+\log_{225}(x_2)=(3+\sq... | 12 | Algebra | math-word-problem | Yes | Yes | amc_aime | false |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.