Answer: E

Small Hint:

First simplify the right side using denominator $12$

Big Hint:

After finding the right side, add $\frac34$ to both sides

Video solution:

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Written solution:

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

Small Hint:

Convert the average of five numbers into a total sum

Big Hint:

Subtract the known three numbers before averaging $a$ and $b$

Video solution:

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Written solution:

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

Small Hint:

Each denominator is the negative of a matching numerator factor

Big Hint:

Count the number of negative signs after cancellation

Video solution:

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Written solution:

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

Small Hint:

Find the total miles driven in two hours

Big Hint:

Subtract gasoline cost before dividing by the number of hours

Video solution:

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Written solution:

The driver travels $2\cdot60=120$ miles, so she is paid $120\cdot \$0.50=\$60$. The trip uses $\frac{120}{30}=4$ 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

Small Hint:

Split the absolute-value equation into two quadratic equations

Big Hint:

Remember to count the double root only once in the sum of real numbers

Video solution:

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Written solution:

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

Small Hint:

The last digit must be compatible with both evenness and divisibility by $5$

Big Hint:

Choose the thousands, hundreds, and tens digits independently after fixing the last digit

Video solution:

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Written solution:

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

Small Hint:

The five row sums together equal the sum of all $25$ integers

Big Hint:

Compute the arithmetic-sequence sum from $-10$ through $14$

Video solution:

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Written solution:

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

Small Hint:

Group the expression into blocks of four terms

Big Hint:

Find a formula for the $j$th block and sum over $50$ blocks

Video solution:

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Written solution:

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

Small Hint:

Let the equal number of adults and children be $P$

Big Hint:

The number of bench sections is $\frac{P}{7}+\frac{P}{11}$

Video solution:

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Written solution:

If the equal number of adults and children is $P$, then the adults use $\frac{P}{7}$ bench sections and the children use $\frac{P}{11}$ 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

Small Hint:

Start with the surface areas of all seven separate cubes

Big Hint:

Each contact hides two faces with the side length of the smaller cube

Video solution:

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Written solution:

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

Small Hint:

Near the median, count ordinary integers and square numbers separately

Big Hint:

Use $44^2<2020<45^2$

Video solution:

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Written solution:

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

Small Hint:

Put the centroid at the origin and align the perpendicular medians with the axes

Big Hint:

The centroid divides each median in a $2:1$ ratio

Video solution:

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Written solution:

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

Small Hint:

Let each interior lattice point store the probability of hitting a vertical side first

Big Hint:

Use symmetry to reduce the random-walk equations to four unknowns

Video solution:

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Written solution:

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

Small Hint:

Use power sums $S_k=x^k+y^k$

Big Hint:

The relation $t^2-4t-2=0$ gives a recurrence for $S_k$

Video solution:

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Written solution:

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

Small Hint:

Factor $12!$ into primes

Big Hint:

A square divisor must choose only even prime exponents

Video solution:

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Written solution:

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

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

Answer: B

Small Hint:

Look at one unit square of the lattice

Big Hint:

The four quarter-circles around its corners have total area $\pi d^2$

Video solution:

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Written solution:

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

Small Hint:

Make a sign chart across the square roots $1^2,2^2,\ldots,100^2$

Big Hint:

Count integer points in every other interval, including endpoints

Video solution:

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Written solution:

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

Small Hint:

Only the parities of $a,b,c,d$ matter

Big Hint:

Count invertible $2\times2$ matrices over the two-element field

Video solution:

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Written solution:

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

Small Hint:

Because upward moves are forbidden, split every path into a top-ring phase and a bottom-ring phase

Big Hint:

On a five-cycle, a self-avoiding walk has $1+2\cdot4$ possible stopping paths