Answer: E

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The right side is $\dfrac{5}{12}-\dfrac13=\dfrac{5}{12}-\dfrac4{12}=\dfrac1{12}$. Thus $x=\dfrac34+\dfrac1{12}=\dfrac9{12}+\dfrac1{12}=\dfrac56$. Thus, E is the correct answer.

Answer: C

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The five numbers have total sum $5\cdot15=75$. Since $3+5+7=15$, we have $a+b=75-15=60$, so the average of $a$ and $b$ is $30$. Thus, C is the correct answer.

Answer: A

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Rewrite the denominator factors as $5-c=-(c-5)$, $3-a=-(a-3)$, and $4-b=-(b-4)$. The expression becomes $\dfrac{(a-3)(b-4)(c-5)}{-(a-3)(b-4)(c-5)}=-1$. Thus, A is the correct answer.

Answer: E

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The driver travels $2\cdot60=120$ miles, so she is paid $120\cdot \$0.50=\$60$. The trip uses $120/30=4$ gallons of gasoline, costing $4\cdot\$2=\$8$. Her net pay is $60-8=52$ dollars over $2$ hours, or $26$ dollars per hour. Thus, E is the correct answer.

Answer: C

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The equation means $x^2-12x+34=2$ or $x^2-12x+34=-2$. The first gives $x^2-12x+32=0$, with roots $4$ and $8$. The second gives $(x-6)^2=0$, with root $6$. The sum of all real solutions is $4+8+6=18$. Thus, C is the correct answer.

Answer: B

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The last digit must be $0$, because the number is divisible by $5$ and all digits are even. The thousands digit can be $2,4,6,$ or $8$, and each of the hundreds and tens digits has $5$ choices. Thus there are $4\cdot5\cdot5=100$ such integers. Thus, B is the correct answer.

Answer: C

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The sum of the integers from $-10$ to $14$ is $25\cdot\dfrac{-10+14}{2}=50$. If every row has common sum $S$, then the five row sums add to $50$, so $5S=50$ and $S=10$. Thus, C is the correct answer.

Answer: B

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Group the terms in blocks of four: $(1+2+3-4)+(5+6+7-8)+\cdots+(197+198+199-200)$. The $j$th block is $(4j-3)+(4j-2)+(4j-1)-4j=8j-6$.

There are $50$ blocks, so the sum is $\sum_{j=1}^{50}(8j-6)=8\cdot\dfrac{50\cdot51}{2}-6\cdot50=9900$. Thus, B is the correct answer.

Answer: B

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If the equal number of adults and children is $P$, then the adults use $P/7$ bench sections and the children use $P/11$ bench sections. Thus $N=P\left(\dfrac17+\dfrac1{11}\right)=\dfrac{18P}{77}$.

The least positive integer occurs when $P=77$, giving $N=18$. Thus, B is the correct answer.

Answer: B

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The cube side lengths are $1,2,3,4,5,6,7$, stacked from largest on bottom to smallest on top. The sum of the surface areas of the separate cubes is $6(1^2+2^2+\cdots+7^2)=6\cdot140=840$.

Each contact hides two square faces, with areas $1^2,2^2,\ldots,6^2$. Subtracting these hidden faces gives $840-2(1^2+2^2+\cdots+6^2)=840-182=658$. Thus, B is the correct answer.

Answer: C

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For a number near the median, the sorted list includes all ordinary integers up to that number and all squares up to that number. Since $44^2=1936$ and $45^2=2025$, there are $44$ squares not exceeding any number from $1936$ through $2020$.

At $1975$, there are $1975+44=2019$ list entries at most $1975$. At $1976$, there are $2020$ entries at most $1976$, so the $2020$th entry is $1976$, and the next is $1977$. The median is $1976.5$. Thus, C is the correct answer.

Answer: C

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Let the centroid be the origin. Since a centroid divides each median in a $2:1$ ratio, we may place median $MV$ horizontally with $M=(8,0)$ and $V=(-4,0)$, and median $CU$ vertically with $C=(0,8)$ and $U=(0,-4)$.

Because $U$ is the midpoint of $AM$, we get $A=2U-M=(-8,-8)$. The area of $\triangle AMC$ is $\dfrac12 |(16,8)\times(8,16)|=\dfrac12(256-64)=96$. Thus, C is the correct answer.

Answer: B

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Let $p(x,y)$ be the probability of eventually hitting a vertical side first from point $(x,y)$. By symmetry, set $a=p(1,1)=p(1,3)$, $b=p(2,1)=p(2,3)$, $c=p(1,2)$, and $d=p(2,2)$.

The averaging equations are $a=\dfrac{1+b+c}{4}$, $b=\dfrac{2a+d}{4}$, $c=\dfrac{1+2a+d}{4}$, and $d=\dfrac{b+c}{2}$. Solving gives $c=\dfrac58$, which is the desired probability from $(1,2)$. Thus, B is the correct answer.

Answer: D

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Let $S_k=x^k+y^k$. Since $x+y=4$ and $xy=-2$, the numbers $x$ and $y$ satisfy $t^2-4t-2=0$, so $S_k=4S_{k-1}+2S_{k-2}$.

Using $S_0=2$ and $S_1=4$, we get $S_2=20$, $S_3=88$, $S_4=392$, and $S_5=1744$. The expression is $x+y+\dfrac{x^5+y^5}{x^2y^2}=4+\dfrac{1744}{4}=440$. Thus, D is the correct answer.

Answer: E

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The prime factorization of $12!$ is $2^{10}3^5 5^2 7^1 11^1$. Therefore $12!$ has $(10+1)(5+1)(2+1)(1+1)(1+1)=792$ positive divisors.

A square divisor must use only even exponents, giving $6\cdot3\cdot2\cdot1\cdot1=36$ square divisors. The probability is $36/792=1/22$, so $m+n=1+22=23$. Thus, E is the correct answer.

Answer: B

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For $d<\dfrac12$, the points within $d$ of lattice points occupy, in each unit square, four quarter-circles whose total area is $\pi d^2$. The enormous square is tiled by unit squares, so the desired probability is $\pi d^2$.

Setting $\pi d^2=\dfrac12$ gives $d=\sqrt{\dfrac{1}{2\pi}}\approx0.399$, which rounds to $0.4$. Thus, B is the correct answer.

Answer: E

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The polynomial changes sign at each square $1^2,2^2,\ldots,100^2$, and its leading coefficient is positive. Thus $P(n)\le0$ for integers in the intervals $[1^2,2^2]$, $[3^2,4^2]$, $\ldots$, $[99^2,100^2]$.

For odd $k$, the interval $[k^2,(k+1)^2]$ contains $(k+1)^2-k^2+1=2k+2$ integers. Summing over odd $k=1,3,\ldots,99$ gives $2(1+3+\cdots+99)+2\cdot50=5000+100=5100$. Thus, E is the correct answer.

Answer: C

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Only parity matters. Modulo $2$, the condition is that $ad-bc$ is $1$, meaning the matrix $\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ is invertible over $\mathbb F_2$.

There are $(4-1)(4-2)=6$ invertible $2\times2$ matrices over $\mathbb F_2$. Each parity pattern lifts to $2^4=16$ choices from $\{0,1,2,3\}$, so there are $6\cdot16=96$ quadruples. Thus, C is the correct answer.

Answer: E

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After leaving the top face, choose one of the $5$ top-ring faces. Because moves from the bottom ring to the top ring are forbidden, every valid path has a top-ring phase, then one move down to the bottom ring, then a bottom-ring phase.

Fix the first top-ring face. On the top ring, the path can move around the 5-cycle without revisiting a face and then stop at any point: there are $1+2\cdot4=9$ possible top-ring paths. From the stopping face, there are $2$ possible downward moves to the bottom ring, so the top part has $18$ choices.

Once in the bottom ring, the path can move around the bottom 5-cycle without revisiting a face and then enter the bottom face; this gives $1+2\cdot4=9$ choices. The total is $5\cdot18\cdot9=810$. Thus, E is the correct answer.

Answer: D

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Place $A=(0,0)$ and $C=(20,0)$. Since $\angle ACD=90^\circ$ and $CD=30$, take $D=(20,30)$. The point $E$ is $(5,0)$, so line $BD$ has equation $y=2(x-5)$.

Because $\angle ABC=90^\circ$, point $B$ lies on the circle with diameter $AC$: $(x-10)^2+y^2=100$. Intersecting with $y=2(x-5)$ gives $x=2$ or $10$. The convex quadrilateral uses $B=(2,-6)$.

Then $[ACD]=\dfrac12\cdot20\cdot30=300$, and $[ABC]=\dfrac12\cdot20\cdot6=60$. The total area is $360$. Thus, D is the correct answer.

Answer: C

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Let $X=2^{17}$. Then $\dfrac{2^{289}+1}{2^{17}+1}=\dfrac{X^{17}+1}{X+1}=X^{16}-X^{15}+X^{14}-\cdots-X+1$.

Pair consecutive terms: $X^{16}-X^{15}$, $X^{14}-X^{13}$, $\ldots$, $X^2-X$, and then the final $+1$. Each pair is $2^{17m}(2^{17}-1)$, contributing $17$ ones in binary. There are $8$ such pairs plus the final $1$, so $k=8\cdot17+1=137$. Thus, C is the correct answer.

Answer: A

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Write $1000=qn+r$, where $0\le r<n$. Then $\left\lfloor\dfrac{1000}{n}\right\rfloor=q$. The other two floors are usually also $q$, except that subtracting $1$ or $2$ from $1000$ crosses a multiple of $n$ when $r$ is small.

For $n>1$, the sum is not divisible by $3$ exactly when $r=0$ or $r=1$. The case $r=0$ gives divisors of $1000$, excluding $1$, for $15$ values. The case $r=1$ gives divisors of $999$, excluding $1$, for $(3+1)(1+1)-1=7$ values. The total is $22$. Thus, A is the correct answer.

Answer: A

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Let $R$ be a $90^\circ$ rotation, so the allowed rotations are $R,R^2,R^3$. Let $X$ and $Y$ be the reflections across the coordinate axes. Once the first two transformations are chosen, the third is forced to be the inverse of their product.

The ordered first-two choices whose forced third transformation is still in the allowed set are $(R,R),(R,R^2),(R^2,R),(R^2,R^3),(R^3,R^2),(R^3,R^3)$, and $(R^2,X),(R^2,Y),(X,R^2),(Y,R^2),(X,Y),(Y,X)$. There are $12$ such sequences. Thus, A is the correct answer.

Answer: C

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The first gcd condition gives $n+120\equiv0\pmod{21}$, so $n\equiv6\pmod{21}$, but $n+120$ must not be divisible by $63$. The second gives $n+63\equiv0\pmod{60}$, so $n\equiv57\pmod{60}$, but $n+63$ must not be divisible by $120$.

Solving $n\equiv6\pmod{21}$ and $n\equiv57\pmod{60}$ gives $n\equiv237\pmod{420}$. The candidates above $1000$ are $1077,1497,1917,\ldots$. The first fails the first gcd condition, the second fails the second gcd condition, and $1917$ works. The digit sum is $18$. Thus, C is the correct answer.

Answer: A

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For any initial roll, Jason compares the best probabilities from rerolling $0,1,2,$ or $3$ dice. Rerolling all three dice has probability $15/216=5/72$. Rerolling one die has probability $1/6$ whenever some pair of kept dice has sum at most $6$.

If he rerolls exactly two dice, he keeps one die. Keeping a die showing $1,2,3,4,5,6$ gives probabilities $5,4,3,2,1,0$ out of $36$, respectively. This can be optimal only when the two smallest dice sum at least $7$ and the smallest die is $1,2,$ or $3$.

The sorted rolls satisfying this are $(1,6,6)$, $(2,5,5),(2,5,6),(2,6,6)$, and $(3,4,4),(3,4,5),(3,4,6),(3,5,5),(3,5,6),(3,6,6)$. Counting permutations gives $3+12+27=42$ rolls out of $216$, so the probability is $42/216=7/36$. Thus, A is the correct answer.