Small Hint:

Turn subtraction of a negative into addition before combining terms

Big Hint:

Pair the positive and negative terms so the arithmetic stays small

Solution:

Subtracting a negative is the same as adding the corresponding positive number, so $$\begin{aligned} &1-(-2)-3-(-4) \\ &\quad {}-5-(-6) \\ &\quad = 1+2-3+4-5+6. \end{aligned}$$ Now combine the terms: $$\begin{aligned} &(1+2+4+6)-(3+5) \\ &\quad = 13-8=5. \end{aligned}$$

Thus, D is the correct answer.

Small Hint:

A cube with side length $s$ has volume $s^3$

Big Hint:

Find the volume from Carl’s cubes and Kate’s cubes separately

Solution:

A cube with side length $s$ has volume $s^3.$ Carl’s five cubes have total volume $5\cdot 1^3=5,$ and Kate’s five cubes have total volume $5\cdot 2^3=40.$

The total volume is $5+40=45.$

Thus, the correct answer is E .

Small Hint:

Rewrite each ratio as a fraction involving the same variables

Big Hint:

Express $\frac{w}{y}$ as $(\frac{w}{x})(\frac{x}{z})(\frac{z}{y})$

Solution:

The ratios give $$\begin{aligned} &\frac wx=\frac43,\qquad \frac yz=\frac32, \\ &\quad \frac zx=\frac16. \end{aligned}$$ Then $$\begin{aligned} &\frac wy=\frac wx\cdot\frac xz\cdot\frac zy \\ &\quad =\frac43\cdot 6\cdot\frac23=\frac{16}{3}. \end{aligned}$$ Therefore $w:y=16:3.$

Thus, E is the correct answer.

Small Hint:

The two acute angles in a right triangle add to $90^\circ$

Big Hint:

To make $b$ small, try the largest possible prime value of $a$

Solution:

The acute angles satisfy $a+b=90.$ Testing primes $b$ from least to greatest, $b=2,3,5$ would give $a=88,87,85,$ none of which is prime. For $b=7,$ however, $a=83$ is prime and is greater than $b.$

Therefore the least possible value of $b$ is $7.$ Thus, D is the correct answer.

Small Hint:

First arrange the seven tiles as if they were all different

Big Hint:

Divide by the orders of the identical green and yellow tiles

Solution:

There are $1+1+2+3=7$ total tiles. If all seven tiles were distinct, there would be $7!$ arrangements. The two green tiles are indistinguishable, and the three yellow tiles are indistinguishable, so we divide by $2!$ and $3!.$

Thus the number of arrangements is $$\frac{7!}{2!3!}=420.$$

Thus, B is the correct answer.

Small Hint:

Find the next palindrome after $15951$ before computing the speed

Big Hint:

Increasing the middle digit forces a carry to the next palindrome

Solution:

A five-digit palindrome is determined by its first three digits. The first half of $15951$ is $159$; increasing this to $160$ and reflecting the first two digits gives the next palindrome, $16061.$

Megan traveled $16061-15951=110$ miles in $2$ hours, so her average speed was $\frac{110}{2}=55$ miles per hour. Thus, B is the correct answer.

Small Hint:

A square that is an even multiple of $3$ has a square root divisible by $6$

Big Hint:

Count positive multiples of $6$ whose squares are less than $2020$

Solution:

A number that is both even and a multiple of $3$ is a multiple of $6.$ If such a number is also a perfect square, its square root must be divisible by both $2$ and $3,$ hence by $6.$ Therefore the numbers counted are exactly $$(6k)^2<2020$$ for positive integers $k.$

Since $42^2=1764<2020$ and $48^2=2304>2020,$ we have $k=1,2,\ldots,7,$ for a total of $7$ numbers.

Thus, A is the correct answer.

Small Hint:

The area condition fixes the distance from $R$ to line $PQ$

Big Hint:

Separate the cases where the right angle is at $P,$ $Q,$ or $R$

Solution:

Place $P=(-4,0)$ and $Q=(4,0).$ Since the area is $12$ and $PQ=8,$ the distance from $R$ to line $PQ$ is $3,$ so $R=(x,\pm 3).$

If the right angle is at $P,$ then $R=(-4,\pm 3),$ giving $2$ points. If it is at $Q,$ then $R=(4,\pm 3),$ giving $2$ more points.

If the right angle is at $R,$ then $PQ$ is the hypotenuse, so $$\begin{aligned} &64=PR^2+QR^2 \\ &\quad =(x+4)^2+9 \\ &\quad {}+(x-4)^2+9 \\ &\quad =2x^2+50. \end{aligned}$$ Thus $x^2=7,$ giving $R=(\pm\sqrt7,\pm 3),$ another $4$ points.

The total is $2+2+4=8.$

Thus, the correct answer is D .

Small Hint:

Complete the square in $y$

Big Hint:

The equation forces $x^{2020}\le 1,$ so only a few integer $x$-values are possible

Solution:

Move all terms to one side and complete the square: $$\begin{aligned} &x^{2020}+y^2=2y \\ &\quad \Longrightarrow x^{2020}+(y-1)^2=1. \end{aligned}$$ Because $(y-1)^2\ge 0,$ we must have $x^{2020}\le 1.$ Since $x$ is an integer, $x=-1,0,1.$

If $x=\pm1,$ then $(y-1)^2=0,$ so $y=1.$ If $x=0,$ then $(y-1)^2=1,$ so $y=0$ or $2.$ This gives $4$ ordered pairs.

Thus, D is the correct answer.

Small Hint:

The sector arc becomes the circumference of the cone base

Big Hint:

The sector radius becomes the slant height of the cone

Solution:

The sector’s arc becomes the base circumference, so if the base radius is $r,$ then $$2\pi r=\frac34(2\pi\cdot4)=6\pi,$$ giving $r=3.$ The sector radius becomes the cone’s slant height, so the cone’s height is $$h=\sqrt{4^2-3^2}=\sqrt7.$$

Hence the volume is $$\frac13\pi r^2h=\frac13\pi(3)^2\sqrt7=3\pi\sqrt7.$$ Thus, C is the correct answer.

Small Hint:

Fix Harold’s five books and count Betty’s choices

Big Hint:

Betty must choose exactly two from Harold’s books and three from the other five

Solution:

Assume that Harold has already picked his $5$ books. Of these five books, there are $\binom{5}{2}$ ways that Betty can have picked exactly two of the same books as Harold, and $\binom{5}{3}$ ways that Betty can choose her other three books from the $5$ books not on Harold’s list.

Thus there are $\binom{5}{2}\binom{5}{3}=100$ favorable choices for Betty out of $\binom{10}{5}=252$ equally likely choices. The probability is $$\frac{100}{252}=\frac{25}{63}.$$ Thus, D is the correct answer.

Small Hint:

Rewrite $\frac{1}{20^{20}}$ as $\frac{5^{20}}{10^{40}}$

Big Hint:

Find how many digits $5^{20}$ has to know how many leading zeros appear

Solution:

We can write $$\frac{1}{20^{20}}=\frac{1}{(4\cdot5)^{20}}=\frac{5^{20}}{10^{40}}.$$ Now $5^{20}=95{,}367{,}431{,}640{,}625,$ which has $14$ digits and begins with $9.$ Dividing by $10^{40}$ places this $14$-digit number after the decimal point with $40-14=26$ zeros before the first digit.

Thus, the correct answer is D .

Small Hint:

Track the net displacement over four moves

Big Hint:

There are exactly $505$ complete four-move cycles in $2020$ moves

Solution:

In the first four moves, Andy goes $1$ east, $2$ north, $3$ west, and $4$ south. The net change is $$(1-3,2-4)=(-2,-2),$$ and he is again facing east.

Since $2020=4\cdot 505,$ Andy completes $505$ such cycles. Starting from $(-20,20),$ his final position is $$\begin{aligned} &(-20,20)+505(-2,-2) \\ &\quad =(-1030,-990). \end{aligned}$$

Thus, B is the correct answer.

Small Hint:

Use the symmetry of the regular hexagon to split the shaded region into six equal pieces

Big Hint:

Each piece is two equilateral triangles minus a $60^\circ$ sector of radius $1$

Solution:

By symmetry, the shaded region is made of six congruent pieces. One such piece is the union of two equilateral triangles with side length $1,$ minus a $60^\circ$ sector of a circle of radius $1.$

The two equilateral triangles have total area $$2\cdot\frac{\sqrt3}{4}=\frac{\sqrt3}{2}.$$ The sector has area $$\frac{60^\circ}{360^\circ}\cdot\pi(1)^2=\frac{\pi}{6}.$$ Thus one shaded piece has area $\frac{\sqrt3}{2}-\frac{\pi}{6},$ and the total shaded area is $$6\left(\frac{\sqrt3}{2}-\frac{\pi}{6}\right)=3\sqrt3-\pi.$$

Thus, D is the correct answer.

Small Hint:

After each deletion step, find the new repeating block

Big Hint:

Use the final period to reduce positions $2019,$ $2020,$ and $2021$ modulo the period length

Solution:

Start with the repeating block $12345.$ Deleting every third digit repeats over $\operatorname{lcm}(3,5)=15$ original positions: $$\begin{aligned} &123451234512345 \\ &\quad \longrightarrow 1245235134, \end{aligned}$$ so the new period has length $10.$

Deleting every fourth digit from this period repeats over $\operatorname{lcm}(4,10)=20$ positions: $$\begin{aligned} &12452351341245235134 \\ &\quad \longrightarrow 124235341452513, \end{aligned}$$ so the new period has length $15.$

Deleting every fifth digit from this period gives $$\begin{aligned} &124235341452513 \\ &\quad \longrightarrow 124253415251, \end{aligned}$$ which has period $12.$ Since $2019\equiv 3\pmod{12},$ the digits in positions $2019,$ $2020,$ and $2021$ are the $3$rd, $4$th, and $5$th digits of this period: $4,2,5.$ Their sum is $11.$

Thus, the correct answer is D .

Small Hint:

Look for a first move that leaves the interval symmetric

Big Hint:

After Jenn moves to $x,$ Bela can respond at $n-x$

Solution:

Bela can first choose the midpoint $\frac{n}{2}.$ After that, whenever Jenn chooses a number $x,$ Bela chooses the reflected number $n-x.$

This reflected number is legal whenever Jenn’s move is legal: distances from previously chosen numbers are preserved by the reflection about $\frac{n}{2},$ and Jenn cannot choose $\frac{n}{2}$ because it was Bela’s first move. Therefore every Jenn move has a matching Bela response, so Jenn is the first player who can run out of legal moves.

Thus, A is the correct answer.

Small Hint:

Classify pairings by how many opposite pairs are used

Big Hint:

Once the opposite pairs are chosen, the remaining people must be matched along arcs of the circle

Solution:

Label the people around the circle. Count by the number of pairs of opposite people.

With no opposite pairs, everyone must be paired with a neighbor around the $10$-cycle. There are exactly $2$ alternating neighbor matchings.

With one opposite pair, choose that pair in $5$ ways. The remaining people form two paths of four vertices, and each path has only one perfect matching by neighbor pairs, so this gives $5$ matchings.

With two or four opposite pairs, the remaining neighbor-pairing paths have odd length somewhere, so no perfect matching is possible.

With three opposite pairs, the two opposite pairs not chosen must be adjacent around the five opposite-pair positions; otherwise the remaining people cannot be matched by neighbor pairs. There are $5$ adjacent choices for the two unchosen opposite pairs, so there are $5$ matchings.

With all five opposite pairs, there is $1$ matching. The total is $$2+5+5+1=13.$$

Thus, the correct answer is C .

Small Hint:

The final urn has three of each color exactly when the four draws contain two red and two blue

Big Hint:

Each fixed color order with two red draws and two blue draws has the same probability

Solution:

The urn ends with three red and three blue balls exactly when the four draws contain two red draws and two blue draws. There are $\binom42=6$ possible color orders of this type.

For any fixed order with two red draws and two blue draws, the probability is $$\frac{1\cdot 2\cdot 1\cdot 2}{2\cdot 3\cdot 4\cdot 5}=\frac{4}{120}=\frac{1}{30},$$ because the first and second draws of each color have numerators $1$ and $2,$ while the total number of balls before the four draws is $2,3,4,5.$

Thus the desired probability is $6\cdot\frac{1}{30}=\frac15.$

Thus, the correct answer is B .

Small Hint:

The number of hands is $\binom{52}{10}$

Big Hint:

After factoring out a final $10,$ only the units digit of the remaining product is needed

Video solution:

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Written solution:

After canceling factors in the binomial coefficient, $$\begin{aligned} \binom{52}{10} &=10\cdot17\cdot13\cdot7\\ &\quad\cdot47\cdot46\cdot11\cdot43. \end{aligned}$$ Dividing $158A00A4AA0$ by the final factor $10,$ the units digit of the remaining product is $A.$

Working modulo $10,$ $$\begin{aligned} A&\equiv7\cdot3\cdot7\cdot7\cdot6\cdot1\cdot3\\ &\equiv2\pmod{10}. \end{aligned}$$ Hence $A=2.$ Thus, A is the correct answer.