## AMC 10 2020 Test A

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**Instructions**

- This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.
- You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer.
- No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the test if before 2006. No problems on the test will require the use of a calculator).
- Figures are not necessarily drawn to scale.
- You will have
**75 minutes**working time to complete the test.

What value of $x$ satisfies

\[x- \frac{3}{4} = \frac{5}{12} - \frac{1}{3}?\]$\textbf{(A)}\ -\dfrac{2}{3}\qquad\textbf{(B)}\ \dfrac{7}{36}\qquad\textbf{(C)}\ \dfrac{7}{12}\qquad\textbf{(D)}\ \dfrac{2}{3}\qquad\textbf{(E)}\ \dfrac{5}{6}$

$\textbf{E}$

Adding $\dfrac{3}{4}$ to both sides, $x= \dfrac{5}{12} - \dfrac{1}{3} + \dfrac{3}{4} = \dfrac{5}{12} - \dfrac{4}{12} + \dfrac{9}{12}=\dfrac{5}{6}$.

The numbers $3, 5, 7, a,$ and $b$ have an average (arithmetic mean) of $15$. What is the average of $a$ and $b$?

$\textbf{(A) } 0 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 30 \qquad\textbf{(D) } 45 \qquad\textbf{(E) } 60$

$\textbf{C}$

The average of the numbers $3, 5, 7, a,$ and $b$ is $$\dfrac{3+5+7+a+b}{5}=15\rightarrow a+b=60$$ Hence, the average of $a$ and $b$ is $\dfrac{a+b}2=\dfrac{60}{2}=30$.

Assuming $a\neq3$, $b\neq4$, and $c\neq5$, what is the value in simplest form of the following expression?

$$\frac{a-3}{5-c} \cdot \frac{b-4}{3-a} \cdot \frac{c-5}{4-b}$$

$\textbf{(A) } -1 \qquad \textbf{(B) } 1 \qquad \textbf{(C) } \dfrac{abc}{60} \qquad \textbf{(D) } \dfrac{1}{abc} - \dfrac{1}{60} \qquad \textbf{(E) } \dfrac{1}{60} - \dfrac{1}{abc}$

$\textbf{A}$

If $x\neq y,$ then $\dfrac{x-y}{y-x}=-1.$ We use this fact to simplify the original expression:\[\frac{\color{red}\overset{-1}{\cancel{a-3}}}{\color{blue}\underset{1}{\cancel{5-c}}} \cdot \frac{\color{green}\overset{-1}{\cancel{b-4}}}{\color{red}\underset{1}{\cancel{3-a}}} \cdot \frac{\color{blue}\overset{-1}{\cancel{c-5}}}{\color{green}\underset{1}{\cancel{4-b}}}=(-1)(-1)(-1)={-}1\]

A driver travels for $2$ hours at $60$ miles per hour, during which her car gets $30$ miles per gallon of gasoline. She is paid $\$0.50$ per mile, and her only expense is gasoline at $\$2.00$ per gallon. What is her net rate of pay, in dollars per hour, after this expense?

$\textbf{(A) }20 \qquad\textbf{(B) }22 \qquad\textbf{(C) }24 \qquad\textbf{(D) } 25\qquad\textbf{(E) } 26$

$\textbf{E}$

Since the driver travels $60$ miles per hour and gets $\$0.50$ per mile, she earns $\$30$ per hour of driving. For each hour she needs $2$ gallons of gasoline to drive for 60 miles. The price of gasoline is $\$2$ per gallon. So she spends $\$4$ per hour on gasoline. Subtracting the gas cost, her net rate of money earned per hour is $\$26$.

What is the sum of all real numbers $x$ for which $|x^2-12x+34|=2?$

$\textbf{(A) } 12 \qquad \textbf{(B) } 15 \qquad \textbf{(C) } 18 \qquad \textbf{(D) } 21 \qquad \textbf{(E) } 25$

$\textbf{C}$

We have $x^2-12x+34=2\ \text{or}\ -2$. When $x^2-12x+34=2$, we get $x=4$ and $x=8$. When $x^2-12x+34=-2$, we get $x=6$. So the answer is $4+8+6=18$.

How many $4$-digit positive integers (that is, integers between $1000$ and $9999$, inclusive) having only even digits are divisible by $5?$

$\textbf{(A) } 80 \qquad \textbf{(B) } 100 \qquad \textbf{(C) } 125 \qquad \textbf{(D) } 200 \qquad \textbf{(E) } 500$

$\textbf{B}$

The units digit, for all numbers divisible by 5, must be either $0$ or $5$. However, since all digits are even, the units digit must be $0$. The middle two digits can be 0, 2, 4, 6, or 8, but the thousands digit can only be 2, 4, 6, or 8 since it cannot be zero.

Now there is 1 choice for the units digit, 5 choices for each of the middle 2 digits, and 4 choices for the thousands digit, for a total of $4\times5\times5\times1 = 100$ numbers.

The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$-by-$5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?

$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$

$\textbf{C}$

Consider the five rows in the square. Each row must have the same sum of numbers, meaning that the sum of all the numbers in the square divided by $5$ is the sum of each row. The sum of the $25$ integers is $-10+-9+\cdots+14=11+12+13+14=50$. So the common sum is $\dfrac{50}{5}=10$.

What is the value of $$1+2+3-4+5+6+7-8+\cdots+197+198+199-200?$$

$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$

$\textbf{B}$

\begin{align*}

&1+2+3-4+5+6+7-8+\cdots+197+198+199-200\\

=\ &1+2+3+4+5+6+7+8+\cdots+197+198+199+200-2\times(4+8+\cdots+200)\\

=\ &1+2+3+4+5+6+7+8+\cdots+197+198+199+200-2\times4(1+2+\cdots+50)\\

=\ &\dfrac{(1+200)\times200}{2}-8\times\dfrac{(1+50)\times50}{2}\\

=\ &20100-10200\\

=\ &9900

\end{align*}

A single bench section at a school event can hold either $7$ adults or $11$ children. When $N$ bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of $N?$

$\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77$

$\textbf{B}$

The least common multiple of $7$ and $11$ is $77$. Hence, there must be $77$ adults and $77$ children. The total number of benches is $\dfrac{77}{7}+\dfrac{77}{11}=11+7=18$.

Seven cubes, whose volumes are $1$, $8$, $27$, $64$, $125$, $216$, and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?

$\textbf{(A) } 644 \qquad \textbf{(B) } 658 \qquad \textbf{(C) } 664 \qquad \textbf{(D) } 720 \qquad \textbf{(E) } 749$

$\textbf{B}$

The volume of each cube follows the pattern of $n^3$, for $n$ is between $1$ and $7$.

Ignoring overlap, the total surface area of these cubes is $$6( 1^2 + 2^2 + \cdots + 7^2 ) = 6\sum_{n=1}^{7} n^2 = 6 \left( \frac{7\times(7 + 1)\times(2 \times 7 + 1)}{6} \right) = 7 \times 8 \times 15 = 840$$ Then we need to subtract out the overlapped parts of these cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to $$2\sum_{n=1}^{6} n^2 = 182$$ Subtracting the overlapped surface area from the total surface area, we get $840 - 182 = 658$.

What is the median of the following list of $4040$ numbers$?$\[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\]

$\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$

$\textbf{C}$

If we sort the sequence in order, it is clearly that 2020 is not the 2020th number because some perfect squares less than 2020 will appear twice. The largest perfect square less than 2020 is $44^2=1936$. Hence, we see that the 2020th number is $2020-44=1976$, and the 2021st number is 1977. Then we get the median $$\dfrac{1976+1977}{2}=1976.5$$

Triangle $AMC$ is isosceles with $AM = AC$. Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$. What is the area of $\triangle AMC?$

$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$

$\textbf{C}$

Since quadrilateral $UVCM$ has perpendicular diagonals, its area is $\dfrac12MV\cdot CU=\dfrac12\times12\times12=72$. Note that $\triangle AUV$ has $\dfrac14$ the area of $\triangle AMC$ by similarity, so the area of $\triangle AUV$ is $[AUV]=\dfrac13[UVCM]=\dfrac13\times72=24$. Hence, the area of $\triangle AMC$ is $24+72=96$.

A frog sitting at the point $(1, 2)$ begins a sequence of jumps, where each jump is parallel to one of the coordinate axes and has length $1$, and the direction of each jump (up, down, right, or left) is chosen independently at random. The sequence ends when the frog reaches a side of the square with vertices $(0,0), (0,4), (4,4),$ and $(4,0)$. What is the probability that the sequence of jumps ends on a vertical side of the square$?$

$\textbf{(A) } \dfrac{1}{2} \qquad \textbf{(B) } \dfrac{5}{8} \qquad \textbf{(C) } \dfrac{2}{3} \qquad \textbf{(D) } \dfrac{3}{4} \qquad \textbf{(E) } \dfrac{7}{8}$

$\textbf{B}$

Drawing out the square, it's easy to see that if the frog goes to the left, it will immediately hit a vertical side of the square. Therefore, the probability of this happening is $\dfrac{1}{4}$.

If the frog goes to the right, it will be in the center of the square at $(2,2)$, and by symmetry (since the frog is equidistant from all sides of the square), the chance it will finally hit a vertical side of the square is $\dfrac{1}{2}$. The probability of this happening is $\dfrac{1}{4} \times \dfrac{1}{2} = \dfrac{1}{8}$.

If the frog goes either up or down, it will hit a diagonal of the square, and again, is equidistant relating to the two closer sides and also equidistant relating the two further sides. Hence, the probability for it to hit a vertical side is $\dfrac{1}{2}$. Because there's a $\dfrac{1}{2}$ chance of the frog going up or down, the total probability for this case is $\dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}$.

Summing up all the cases, the answer is $\dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{4} = \dfrac{5}{8}$.

We have another way to think about this question. Let $P_{(x,y)}$ denote the probability of the frog hitting a vertical edge when it is at $(x,y)$ in the beginning. Note that $P_{(1,2)}=P_{(3,2)}$ by reflective symmetry over the line $x=2$. Similarly, $P_{(1,1)}=P_{(1,3)}=P_{(3,1)}=P_{(3,3)}$, and $P_{(2,1)}=P_{(2,3)}$. Now we create equations for the probabilities at each of these points/states by considering the probability of going either up, down, left, or right from that point:\[P_{(1,2)}=\frac{1}{4}+\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}\]\[P_{(2,2)}=\frac{1}{2}P_{(1,2)}+\frac{1}{2}P_{(2,1)}\]\[P_{(1,1)}=\frac{1}{4}+\frac{1}{4}P_{(1,2)}+\frac{1}{4}P_{(2,1)}\]\[P_{(2,1)}=\frac{1}{2}P_{(1,1)}+\frac{1}{4}P_{(2,2)}\]We have a system of $4$ equations in $4$ variables, so we can solve for each of these probabilities. The answer is $P_{(1,2)}=\dfrac{5}{8}$.

Real numbers $x$ and $y$ satisfy $x + y = 4$ and $x \cdot y = -2$. What is the value of$$x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y?$$

$\textbf{(A)}\ 360\qquad\textbf{(B)}\ 400\qquad\textbf{(C)}\ 420\qquad\textbf{(D)}\ 440\qquad\textbf{(E)}\ 480$

$\textbf{D}$

\begin{align*}

x + \frac{x^3}{y^2} + \frac{y^3}{x^2} + y&=\dfrac{x^3y^2+x^5+y^5+x^2y^3}{x^2y^2}\\

&=\dfrac{x^3(x^2+y^2)+y^3(x^2+y^2)}{x^2y^2}\\

&=\dfrac{(x^2+y^2)(x^3+y^3)}{x^2y^2}\\

&=\dfrac{(x^2+y^2)(x+y)(x^2-xy+y^2)}{x^2y^2}

\end{align*}

Given that $x+y=4$ and $xy=-2$, we have $$(x+y)^2=x^2+2xy+y^2=x^2+2(-2)+y^2=16\rightarrow x^2+y^2=20$$ So the answer is $\dfrac{20\times4\times(20+2)}{(-2)^2}=440$.

A positive integer divisor of $12!$ is chosen at random. The probability that the divisor chosen is a perfect square can be expressed as $\dfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n$?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 12\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 23$

$\textbf{E}$

The prime factorization of $12!$ is $2^{10} \times 3^5 \times 5^2 \times 7 \times 11$. This yields a total of $11 \times 6 \times 3 \times 2 \times 2$ divisors of $12!.$ In order to produce a perfect square divisor, there must be an even exponent for each number in the prime factorization. Hence, we need to find a divisor of $4^5\times9^2\times25$. So the number of perfect square divisor is $6\times3\times2$. The probability that the divisor chosen is a perfect square is\[\frac{6\times 3\times 2}{11\times 6\times 3\times 2\times 2}=\frac{1}{22}\] Therefore, the answer is $m+n=1+22=23$.

A point is chosen at random within the square in the coordinate plane whose vertices are $(0, 0)$, $(2020, 0)$, $(2020, 2020),$ and $(0, 2020)$. The probability that the point is within $d$ units of a lattice point is $\dfrac{1}{2}$. (A point $(x, y)$ is a lattice point if $x$ and $y$ are both integers.) What is $d$ to the nearest tenth$?$

$\textbf{(A) } 0.3 \qquad \textbf{(B) } 0.4 \qquad \textbf{(C) } 0.5 \qquad \textbf{(D) } 0.6 \qquad \textbf{(E) } 0.7$

$\textbf{B}$

The diagram represents each unit square of the given $2020 \times 2020$ square. If we draw a quarter of a circle from each corner (where the lattice points are located), each with radius $d$, the area covered by the quarters should be $0.5$. Hence, we get \[4 \times \frac{1}{4} \pi d^2 = \frac{1}{2}\] Solving for $d$, we obtain $d = \dfrac{1}{\sqrt{2\pi}}$, where with $\pi \approx 3$, we get $d \approx \dfrac{1}{\sqrt{6}} \approx \dfrac{1}{2.5} =0.4$.

Define$$P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).$$How many integers $n$ are there such that $P(n)\leq 0$?

$\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100$

$\textbf{E}$

Notice that $P(x)$ is nonpositive when $x$ is between $100^2$ and $99^2, 98^2$ and $97^2$, $\cdots$, $2^2$ and $1^2$ (inclusive). So the number of such integers is \begin{align*}

&\left(100^2-99^2+1\right)+\left(98^2-97^2+1\right)+\cdots+\left(2^2-1^2+1\right)\\

=\ &[(100+99)(100-99) + 1] + [(98+97)(98-97)+1] + \cdots + [(2+1)(2-1)+1]\\

=\ &200 + 196 + 192 + \cdots + 4\\

=\ &4\times(1+2+\cdots + 50)\\

=\ &4 \times\frac{50 \times 51}{2}\\

=\ &5100

\end{align*}

Let $(a,b,c,d)$ be an ordered quadruple of not necessarily distinct integers, each one of them in the set ${0,1,2,3}.$ For how many such quadruples is it true that $a\cdot d-b\cdot c$ is odd? (For example, $(0,3,1,1)$ is one such quadruple, because $0\cdot 1-3\cdot 1 = -3$ is odd.)

$\textbf{(A) } 48 \qquad \textbf{(B) } 64 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 128 \qquad \textbf{(E) } 192$

$\textbf{C}$

There are two terms, $ad$ and $bc$, in the expression. We need exactly one term to be odd and one term to be even. By symmetry, we can set $ad$ to be odd and $bc$ to be even, then multiply the number of arrangements by $2.$ If $ad$ is odd, both $a$ and $d$ must be odd, so there are $2\times2=4$ possibilities for $ad.$ Consider $bc.$ Let us say that $b$ is even. Then there are $2\times4=8$ possibilities for $bc.$ However, $b$ can be odd, in which case we have $2\times2=4$ more possibilities for $bc.$ Thus, there are $12$ ways to choose $bc$ and $4$ ways to choose $ad.$ Considering symmetry, we have $2 \times 4 \times 12= 96$ arrangements for $ad-bc.$

As shown in the figure below, a regular dodecahedron (the polyhedron consisting of $12$ congruent regular pentagonal faces) floats in empty space with two horizontal faces. Note that there is a ring of five slanted faces adjacent to the top face, and a ring of five slanted faces adjacent to the bottom face. How many ways are there to move from the top face to the bottom face via a sequence of adjacent faces so that each face is visited at most once and moves are not permitted from the bottom ring to the top ring?

$\textbf{(A) } 125 \qquad \textbf{(B) } 250 \qquad \textbf{(C) } 405 \qquad \textbf{(D) } 640 \qquad \textbf{(E) } 810$

$\textbf{E}$

Since we start at the top face and end at the bottom face without moving from the bottom ring to the top ring or revisiting a face, our journey must consist of the top face, a series of faces in the top ring, a series of faces in the bottom ring, and the bottom face, in that order.

We have $5$ choices for which face we visit first on the top ring. From there, we have $9$ choices for how far around the top ring we go before moving down: $1,2,3,$ or $4$ faces around clockwise, $1,2,3,$ or $4$ faces around counterclockwise, or immediately going down to the bottom ring without visiting any other faces in the top ring.

We then have $2$ choices for which bottom ring face to visit first (since every top-ring face is adjacent to exactly $2$ bottom-ring faces) and then once again $9$ choices for how to travel around the bottom ring. We then proceed to the bottom face, completing the trip.

Multiplying together all the numbers of choices we have, we get $5 \times 9 \times 2 \times 9 = 810$.

Quadrilateral $ABCD$ satisfies $\angle ABC = \angle ACD = 90^{\circ}, AC = 20$, and $CD = 30$. Diagonals $\overline{AC}$ and $\overline{BD}$ intersect at point $E$, and $AE = 5$. What is the area of quadrilateral $ABCD$?

$\textbf{(A) } 330 \qquad\textbf{(B) } 340 \qquad\textbf{(C) } 350 \qquad\textbf{(D) } 360 \qquad\textbf{(E) } 370$

$\textbf{D}$

Since $AC=20$ and $CD=30$, we get the area of $\triangle ACD$: $[ACD]=300$. Now we need to find $[ABC]$ to get the area of the whole quadrilateral. Drop an altitude from $B$ to $AC$ and call the point of intersection $F$. Let $FE=x$. Since $AE=5$, then $AF=5-x$.

By dropping this altitude, we see two similar triangles, $\triangle BFE \sim \triangle DCE$. Since $EC$ is $20-5=15$, and $DC=30$, we get that $BF=2x$.

We also see that $\triangle ABF\sim\triangle BCF$. So $$\dfrac{BF}{CF}=\dfrac{AF}{BF}$$ or $$\dfrac{2x}{15+x}=\dfrac{5-x}{2x}$$ Hence, we get $x=3$. Then $BF=2\times3=6$, $[ABC]=60$. Thus $[ABCD]=[ACD]+[ABC]=300+60=360$.

There exists a unique strictly increasing sequence of nonnegative integers $a_1 < a_2 < … < a_k$ such that\[\frac{2^{289}+1}{2^{17}+1} = 2^{a_1} + 2^{a_2} + … + 2^{a_k}.\]What is $k?$

$\textbf{(A) } 117 \qquad \textbf{(B) } 136 \qquad \textbf{(C) } 137 \qquad \textbf{(D) } 273 \qquad \textbf{(E) } 306$

For how many positive integers $n \le 1000$ is

$$\left\lfloor \dfrac{998}{n} \right\rfloor+\left\lfloor \dfrac{999}{n} \right\rfloor+\left\lfloor \dfrac{1000}{n}\right \rfloor$$

not divisible by $3$? (Recall that $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.)

$\textbf{(A) } 22 \qquad\textbf{(B) } 23 \qquad\textbf{(C) } 24 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 26$

$\textbf{A}$

Clearly, $n=1$ fails because $998+999+1000$ is a multiple of 3. Except for the special case of $n=1$,\[\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor\]equals either $0$ or $1$.

If $\left\lfloor \dfrac{1000}{n} \right\rfloor - \left\lfloor \dfrac{998}{n} \right\rfloor=0$, this implies that $\left\lfloor \dfrac{998}{n} \right\rfloor = \left\lfloor \dfrac{999}{n} \right\rfloor = \left\lfloor \dfrac{1000}{n} \right\rfloor$, so their sum must be a multiple of $3$.

If $\left\lfloor \dfrac{1000}{n} \right\rfloor - \left\lfloor \dfrac{998}{n} \right\rfloor=1$, the sum of the three floor terms is $3 \left\lfloor \dfrac{999}{n} \right\rfloor \pm 1$, so it is never a multiple of $3$. Thus, we are looking for all $n \neq 1$ such that\[\left\lfloor \frac{1000}{n} \right\rfloor - \left\lfloor \frac{998}{n} \right\rfloor = 1\]This implies that either\[\left\lfloor \frac{998}{n} \right\rfloor + 1 = \left\lfloor \frac{999}{n} \right\rfloor\]or\[\left\lfloor \frac{999}{n} \right\rfloor + 1 = \left\lfloor \frac{1000}{n} \right\rfloor\] Let's analyze the first equation of these two. The first equation is equivalent to the statement that there is a positive integer $a$ such that\[\frac{998}{n} < a \leq \frac{999}{n} \implies 998 < an \leq 999 \implies an = 999 \implies a = \frac{999}{n} \implies n | 999\]Analogously, the second equation implies that\[n | 1000\]So this condition can only be satisfied if $n\neq 1$ is a factor of 999 or 1000. By prime factorization, we get $999=3^3\times37$ and $1000=2^3\times5^3$. Hence, we see that $999$ has $4\times2=8$ divisors and $1000$ has $4\times4=16$ divisors. Their only common factor is $1$. Since $n\neq1$, the number of possible values of $n$ is $(8-1)+(16-1)=22$.

Let $T$ be the triangle in the coordinate plane with vertices $\left(0,0\right)$, $\left(4,0\right)$, and $\left(0,3\right)$. Consider the following five isometries (rigid transformations) of the plane: rotations of $90^{\circ}$, $180^{\circ}$, and $270^{\circ}$ counterclockwise around the origin, reflection across the $x$-axis, and reflection across the $y$-axis. How many of the $125$ sequences of three of these transformations (not necessarily distinct) will return $T$ to its original position? (For example, a $180^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by a reflection across the $y$-axis will return $T$ to its original position, but a $90^{\circ}$ rotation, followed by a reflection across the $x$-axis, followed by another reflection across the $x$-axis will not return $T$ to its original position.)

$\textbf{(A) } 12\qquad\textbf{(B) } 15\qquad\textbf{(C) }17 \qquad\textbf{(D) }20 \qquad\textbf{(E) }25$

$\textbf{A}$

First, any combination of motions we can make must reflect $T$ an even number of times. This is because every time we reflect $T$, it changes orientation. Once $T$ has been flipped once, no combination of rotations will put it back in place because it is the mirror image; however, flipping it again changes it back to the original orientation. Since we are only allowed $3$ transformations and an even number of them must be reflections, we either reflect $T$ $0$ times or $2$ times.

$\textbf{Case 1: 0 reflections on $T$.}$

In this case, we must use $3$ rotations to return $T$ to its original position. This could be $90+90+180$ or $180+270+270$. Each of the combinations above has 3 ways to arrange the order. So we have 6 ways for case 1.

$\textbf{Case 2: 2 reflections on $T$.}$

In this case, we first eliminate the possibility of having two of the same reflection. Since two reflections across the x-axis maps $T$ back to itself, inserting a rotation before, between, or after these two reflections would change $T$'s final location, meaning that any combination involving two reflections across the x-axis would not map $T$ back to itself. The same applies to two reflections across the y-axis.

Therefore, we must use one reflection about the x-axis, one reflection about the y-axis, and one rotation. Since a reflection about the x-axis changes the sign of the y component, a reflection about the y-axis changes the sign of the x component, and a $180^\circ$ rotation changes both signs, these three transformation composed (in any order) will suffice. It is therefore only a question of arranging the three, giving us $3! = 6$ combinations for case 2.

In conclusion, the answer is $6+6= 12$.

Let $n$ be the least positive integer greater than $1000$ for which

$$\gcd(63, n+120) =21\quad \text{and} \quad \gcd(n+63, 120)=60.$$

What is the sum of the digits of $n$?

$\textbf{(A) } 12 \qquad\textbf{(B) } 15 \qquad\textbf{(C) } 18 \qquad\textbf{(D) } 21\qquad\textbf{(E) } 24$

$\textbf{C}$

Given that $\gcd(63, n+120) =21$ and $\gcd(n+63, 120)=60$, we have\[\gcd(63, n+120) = \gcd(63, n+120 + 63) = \gcd(63, n+183) = 21\]\[\gcd(n+63, 120) = \gcd(n+63 + 120, 120) = \gcd(n+183, 120) = 60\] Hence, we know $n+183=21a$, where integer $a$ is not a multiple of 3; and $n+183=60b$, where integer $b$ is not a multiple of 2. Since $\text{lcm}(21,60)=420$, we get $n+183=420c$, where integer $c$ is not divisible by 2 and 3. Apparently $c=1$ is too small because $n$ must be greater than 1000. Next, we try $c=5$, from which $n=420\times5-183=1917$. So the answer is $1+9+1+7=18$.

Jason rolls three fair standard six-sided dice. Then he looks at the rolls and chooses a subset of the dice (possibly empty, possibly all three dice) to reroll. After rerolling, he wins if and only if the sum of the numbers face up on the three dice is exactly $7$. Jason always plays to optimize his chances of winning. What is the probability that he chooses to reroll exactly two of the dice?

$\textbf{(A) } \dfrac{7}{36} \qquad\textbf{(B) } \dfrac{5}{24} \qquad\textbf{(C) } \dfrac{2}{9} \qquad\textbf{(D) } \dfrac{17}{72} \qquad\textbf{(E) } \dfrac{1}{4}$

$\textbf{A}$

We have 4 cases after the initial roll:

$\textbf{Case 1:}$ Jason rerolls exactly zero dice if and only if the sum of the three dice is $7,$ in which the probability of winning is always $1.$

$\textbf{Case 2:}$ If Jason rerolls exactly one die, then the sum of the two other dice must be $2,3,4,5,$ or $6.$ The probability of winning is always $\dfrac16,$ as exactly $1$ of the $6$ possible outcomes of the die rerolled results in a win.

$\textbf{Case 3:}$ If Jason rerolls exactly two dice, then the outcome of the remaining die must be $1,2,3,4,$ or $5.$ Applying casework to the remaining die produces the following table:

The probability of winning is at most $\dfrac{5}{36}$, which is less than the probability $\dfrac16$ in Case 2.

$\textbf{Case 4:}$ If Jason rerolls all three dice, then the probability of winning is always $\dfrac{\dbinom62}{6^3}=\dfrac{15}{216}=\dfrac{5}{72}.$

For the denominator, rolling three dice gives a total of $6^3=216$ possible outcomes.

For the numerator, this is the same as counting the ordered triples of positive integers $(a,b,c)$ for which $a+b+c=7.$ Suppose that $7$ balls are lined up in a row. There are $6$ gaps between the balls, and placing dividers in $2$ of the gaps separates the balls into $3$ piles. From left to right, the numbers of balls in the piles correspond to $a,b,$ and $c,$ respectively. There are $\dbinom62=15$ ways to place the dividers. Note that the dividers' positions and the ordered triples have one-to-one correspondence, and $1\leq a,b,c\leq6$ holds for all such ordered triples. So the numerator we are looking for is 15.

We see that the probability of winning in Case 4 is less than the first three cases in Case 3, but greater than the last two cases in Case 3. Furthermore, the probability of winning in Case 2 is greater than all cases in Case 3 and Case 4: $$\dfrac16>\dfrac5{36}>\dfrac4{36}>\dfrac3{36}>\dfrac5{72}>\dfrac2{36}>\dfrac1{36}$$

The optimal strategy is that:

1. If the sum of the three dice is 7, we reroll no dice.

2. If the sum of any two dice is less than 7, we reroll 1 die because rolling 1 die will give Jason a $\dfrac{1}{6}$ chance of winning.

3. The sum of any two dice is 7 or greater.

If the smallest die is 1, the other two dice must be 6, which has $3$ possible permutations. This case corresponds to the first case in Case 3, so we reroll 2 dice.

If the smallest die is 2, the other two dice can be 5 or 6. There are 3 ways to choose which die is the smallest one, 2 ways each to choose the value of the other dice, so there are $2 \times 2 \times 3 = 12$ possible permutations. This case corresponds to the second case in Case 3, so we reroll 2 dice.

If the smallest die is 3, the other two dice can be 4, 5, or 6. Similarly, there are 3 ways to choose which die is the smallest one, and 3 ways each to choose the value of the other dice, so there are $3 \times3 \times3 = 27$ possible permutations. This case corresponds to the third case in Case 3, so we reroll 2 dice.

If the smallest die is 4 or above, we reroll 3 dice because the possibility of winning $\dfrac{5}{72}>\dfrac{2}{36}>\dfrac1{36}$.

Therefore, the cases where 1, 2, or 3 are the smallest die covers all cases where 2 rerolls is optimal. There are $3+12+27 = 42$ permutations out of 216 total permutations, giving us a probability of $\dfrac{42}{216} = \dfrac{7}{36}$.