Answer: D

Solution:

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

Thus, D is the correct answer.

Answer: E

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 .

Answer: E

Solution:

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

Thus, E is the correct answer.

Answer: D

Solution:

We know that the interior angles of a triangle add up to $180^{\circ},$ and since the triangle in question is a right triangle, by definition one of the interior angles must measure $90^{\circ}.$ The remaining two acute angles, $a^{\circ}$ and $b^{\circ},$ must therefore have a sum of $180^{\circ}-90^{\circ}=90^{\circ}.$

Let's begin by exploring the largest values of $a^{\circ}$ and going from there, as those will naturally yield the smallest values of $b^{\circ}.$

The greatest possible value of $a^{\circ}$ is $89^{\circ}.$ This makes $b^{\circ}=90^{\circ}-89^{\circ}=1^{\circ},$ which is not prime, so this is not a valid possible case.

Moving on, the next largest possible value of $a^{\circ}$ is $83^{\circ}.$ This makes $b^{\circ}=90^{\circ}-83^{\circ}=7^{\circ},$ which is prime! Therefore, this is the smallest possible value of $b.$

Thus, D is the correct answer.

Answer: B

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.

Answer: B

Solution:

We want to find the smallest palindrome larger than $15951,$ and to do so, we want to increase the value of the middle digit. This is because if we increased the value of any other digit with less place value (say, the ones place), we would also need to increase the value of the digit with greater place value (i.e. ten-thousands place).

Therefore, we must replace the number $9$ to the next greatest integer: $10.$ As all digits must be between $0$ and $9$ (inclusive), we write a zero in place of the nine, and increase the thousands digit (and the tens — to preserve the palindrome).

As such, the next smallest palindrome larger than $15951$ is $16061.$

As Megan drives $16061-15951$$=110$ miles in $2$ hours, she drives at an average of $55$ mph.

Thus, B is the correct answer.

Answer: A

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.

Answer: D

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 $$64=PR^2+QR^2=(x+4)^2+9+(x-4)^2+9=2x^2+50.$$ 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 .

Answer: D

Solution:

Move all terms to one side and complete the square: $$x^{2020}+y^2=2y\quad\Longrightarrow\quad x^{2020}+(y-1)^2=1.$$ 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.

Answer: C

Solution:

Remember that the volume of a cone is equal to $\dfrac 13 \pi r^2 h.$ Further notice that the circumference of the base of the cone is equal to the remaining circumference of the circle: $\dfrac 34$ the circumference of the complete circle.

Therefore, the circumference of the base of the cone is equal to: $$\begin{align*} C &= \dfrac 34 \cdot 2(\pi) (4)\\ &= 6\pi \end{align*}$$ This suggests that the radius of the cone's base ($r'$) is equal to: $$\begin{align*} C &= 2\pi r' \\ 6\pi &= 2\pi r' \\ r' &= 3. \end{align*}$$ Also, when we tape together the marked radii to form the cone, the radii in question become the slanted height of the cone. This can be used to find the actual height of the cone, which by the Pythagorean Theorem, is equal to: $$\begin{align*}SL^2 &= h^2 + (r')^2\\ 4^2 &= h^2 +3^2 \\ 16 &= h^2 + 9\\ 7 &= h^2 \\ h &= \sqrt{7}. \end{align*}$$ Thus, the volume of the cone is equal to: $$\begin{align*}V &= \dfrac 13 \pi r^2 h \\ &= \dfrac 13 \pi (r')^2 h\\ &= \dfrac 13 \pi (3)^2 \cdot \sqrt{7} \\ &= \dfrac 13 \pi 9\sqrt{7} \\ &= 3\pi\sqrt{7} \end{align*}$$

Thus, the correct answer is C .

Answer: D

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.

As such, there are $\binom{5}{2}\binom{5}{3}=100$ ways for Betty to choose her books such that she chooses exactly two books on Harold's list and three books not on Harold's list.

Therefore, as there are $\binom{10}{5}=252$ ways that Betty can choose her books arbitrarily, and $100$ of those choices satisfy the above conditions, the probability that they have exactly two books in common is: $$\dfrac{100}{252} = \dfrac{25}{63}$$

Thus, D is the correct answer.

Answer: D

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 .

Answer: B

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 $$(-20,20)+505(-2,-2)=(-1030,-990).$$

Thus, B is the correct answer.

Answer: D

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.

Answer: D

Solution:

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

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

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

Thus, the correct answer is D .

Answer: A

Solution:

Bela can first choose the midpoint $n/2$. 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 $n/2$, and Jenn cannot choose $n/2$ 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.

Answer: C

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 .

Answer: B

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 .

Answer: A

Video solution:

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Written solution:

The number of distinct hands that can be dealt to the player is equal to: $$\begin{align*} &\binom{52}{10} \\ &= \dfrac{52!}{42!\cdot 10!}\\ &=10\cdot 17\cdot 13\cdot 7\cdot 47\cdot 46\cdot 11\cdot 43\\ &= 158A00A4AA0 \end{align*}$$

Therefore: $$\begin{align*} &17\cdot 13\cdot 7\cdot 47\cdot 46\cdot 11\cdot 43\\ &= 158A00A4AA \end{align*}$$

To find the units digit, we can find the value of this expression mod $10$: $$\begin{align*} A &\equiv 17\cdot 13\cdot 7\cdot 47\cdot \\&46\cdot 11\cdot 43 \bmod{10}\\ &\equiv 7\cdot 3\cdot 7\cdot 7\cdot 6\cdot 1\cdot 3 \bmod{10}\\ &\equiv 1\cdot 9\cdot 6\cdot 3 \bmod{10}\\ &\equiv 9\cdot 8 \bmod{10}\\ &\equiv 2 \bmod{10}\\ \end{align*}$$

Therefore, as $0\le A \le 9,$ and $A\equiv 2\bmod{10},$ it follows that $A=2.$

Thus, A is the correct answer.

Answer: B

Video solution:

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Written solution:

Decompose $S(r)$ by where the added volume lies relative to the box.

The original box has volume $d=1\cdot3\cdot4=12$. The face slabs contribute surface area times $r$, so $$c=2(1\cdot3+1\cdot4+3\cdot4)=38.$$

Along each edge is a quarter-cylinder of radius $r$. The sum of all edge lengths is $4(1+3+4)=32$, so $$b=\frac14\pi\cdot 32=8\pi.$$ At the eight corners, the eighth-spheres combine to one full sphere, so $$a=\frac43\pi.$$

Therefore $$\frac{bc}{ad}=\frac{(8\pi)(38)}{(\frac43\pi)(12)}=19.$$

Thus, the correct answer is B .

Answer: B

Video solution:

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Written solution:

The four named regions fill the square and each has area $1$, so the square has area $4$ and side length $2$. Since triangle $AEH$ is right isosceles with area $1$, we have $AE=AH=\sqrt2$.

Extend $FI$ to meet $AB$ at $K$. Then $BE=2-\sqrt2$, and triangle $BEK$ is right isosceles, so its area is $$\frac12(2-\sqrt2)^2=3-2\sqrt2.$$ The region $BFIE$ has area $1$, so triangle $FIK$ has area $$1+(3-2\sqrt2)=4-2\sqrt2.$$ Triangle $FIK$ is also right isosceles, so its area is $\frac12 FI^2$. Hence $$\frac12 FI^2=4-2\sqrt2,\qquad FI^2=8-4\sqrt2.$$

Thus, the correct answer is B .

Answer: D

Video solution:

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Written solution:

Let $m=2^{101}+2^{51}+1$. We factor the numerator around this divisor: $$2^{202}+202=(2^{101}+1)^2-(2^{51})^2+201.$$ By the difference of squares, $$(2^{101}+1)^2-(2^{51})^2=(2^{101}+2^{51}+1)(2^{101}-2^{51}+1),$$ which is a multiple of $m$. Therefore $$2^{202}+202\equiv 201\pmod m.$$

Thus, the correct answer is D .

Answer: C

Video solution:

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Written solution:

Each of $L,R,H,V$ moves every vertex to an adjacent corner of the square. Therefore after an odd number of transformations the labeling is in one of the four odd-parity states, and after an even number it is in one of the four even-parity states.

After any first $19$ transformations, the square is in an odd-parity state. From each odd-parity state, exactly one of $L,R,H,V$ sends the labeled vertices back to their original positions. Thus every sequence of the first $19$ transformations has exactly one valid final transformation.

There are $4^{19}=2^{38}$ choices for the first $19$ transformations, so there are $2^{38}$ valid sequences.

Thus, C is the correct answer.

Answer: C

Video solution:

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Written solution:

Let $$k=\left\lfloor\sqrt n\right\rfloor.$$ The equation gives $n=70k-1000$. Also, by the definition of the floor function, $$k^2\le n<(k+1)^2.$$ Substituting $n=70k-1000$, we get $$k^2\le 70k-1000<(k+1)^2.$$

The left inequality is $$k^2-70k+1000\le0\quad\Longrightarrow\quad (k-20)(k-50)\le0,$$ so $20\le k\le50$. The right inequality is $$70k-1000<k^2+2k+1\quad\Longrightarrow\quad k^2-68k+1001>0.$$ The roots of $k^2-68k+1001$ are $34\pm\sqrt{155}$, which are approximately $21.55$ and $46.45$. Thus, together with $20\le k\le50$, the possible integer values are $$k=20,21,47,48,49,50.$$ There are $6$ such values.

Thus, C is the correct answer.

Answer: A

Video solution:

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Written solution:

Write $96=2^5\cdot3$. Suppose an ordered factorization has $k$ factors. Exactly one factor contains the single prime factor $3$; choose its position in $k$ ways.

The other $k-1$ factors must each contain at least one factor of $2$, while the factor containing $3$ may contain any number of factors of $2$. Distributing the five factors of $2$ under these conditions can be done in $$\binom{5}{k-1}$$ ways. Therefore the number of ordered factorizations with $k$ factors is $k\binom{5}{k-1}$, where $1\le k\le6$.

Thus $$D(96)=\sum_{k=1}^6 k\binom{5}{k-1}.$$ Letting $j=k-1$, this becomes $$\sum_{j=0}^5 (j+1)\binom5j=\sum_{j=0}^5 j\binom5j+\sum_{j=0}^5\binom5j=5\cdot2^4+2^5=80+32=112.$$

Thus, A is the correct answer.