Solution:

We can add each individual digit, yielding $11,110.$

We can also get the sum by noticing that each digit has a sum of $10,$ so the sum is equal to $10\cdot 1111 = 11,110.$

Thus, the answer is E .

Solution:

The area is a triangle of area $$\dfrac {4\cdot 5}2 - \dfrac{2\cdot 4}2 = 10-4$$$$= 6$$ since we subtract the area of a smaller triangle from a larger triangle.

Thus, the answer is B .

Solution:

We can rewrite this as $$\dfrac{2021\cdot 2021}{2020\cdot 2021} -\dfrac{2020\cdot 2020}{2020\cdot 2021}$$ $$= \dfrac{2021^2-2020^2}{2020\cdot 2021}.$$

This can be simplified to $$\dfrac{(2021-2020)(2021+2020)}{2020\cdot 2021}$$ $$=\dfrac{4041}{2020\cdot 2021}.$$

Since $4041$ is coprime with both $2020$ and $2021,$ we know $4041$ is the numerator.

Thus, the answer is E .

Solution:

Let $N=m-s$ be the noon temperature difference. At $4{:}00$, the new difference is $(m-5)-(s+3)=N-8$.

The temperatures then differ by $2$ degrees, so $|N-8|=2$. Hence $N=6$ or $N=10$, and the product of all possible values is $6\cdot10=60$.

Thus, the answer is C .

Solution:

Since $8=2^3$, we have $$n=8^{2022}=2^{6066}=4^{3033}.$$

Therefore $$\frac n4=\frac{4^{3033}}4=4^{3032}.$$

Thus, the answer is E .

Solution:

Before starting, note that if we can represent the prime factorization of an integer $z$ as $$z=p_1^{e_1} p_2^{e_2} \cdots,$$ then there are $(e_1+1)(e_2+1) \cdots$ distinct positive factors.

If the number in question has $2021$ factors, by the previous logic, $$2021=(e_1+1)(e_2+1) \cdots,$$ and as the prime factorization of $2021 = 43\cdot 47,$ then our number must be $p_1^{46}p_2^{42}$ or $p^{2020}.$

The smallest number we can make in either of these is making $p_1 = 2,p_2=3$ in the first configuration, yielding $$2^{46}3^{42} = 16\cdot 6^{42} .$$

Therefore, $$m=16$$$$k=42,$$ so $$m+k = 42+16 = 58.$$

Thus, the answer is B .

Solution:

A special fraction with denominator $b$ equals $\frac{15-b}{b}=\frac{15}{b}-1$, where $1\le b\le14$. We need integer values of $\frac{15}{x}+\frac{15}{y}-2$.

Checking the possible denominators by fractional part gives the distinct integer sums $$1,2,3,4,6,7,8,13,16,18,28.$$

There are $11$ such integers.

Thus, the answer is C .

Solution:

We know $$\begin{align*}16,383 &= 16384-1 \\&= 2^{14}-1 \\&= (2^7-1)(2^7+1) \\&= 127\cdot 129.\end{align*}$$ Since $129=3\cdot 43,$ we get $$16383 = 3\cdot 43\cdot 127.$$ Therefore, $127$ is the largest prime factor, and the sum of its digits is $10.$

Thus, the answer is C .

Solution:

Let $x$ be the fraction of blue knights who are magical. Then the fraction of red knights who are magical is $2x$.

The total magical fraction is a weighted average over the red and blue groups: $$\frac27(2x)+\frac57(x)=\frac16.$$

Thus $\frac97x=\frac16$, so $x=\frac7{54}$. The red magical fraction is $2x=\frac7{27}$.

Thus, the answer is C .

Solution:

If Alice had drawn $1$ or $40$, she would know who had the larger number. Her first statement tells Bob that Alice has neither $1$ nor $40$.

Bob can then know who has the larger number only if his number is $1,2,39,$ or $40$. Since Bob says his number is prime, his number must be $2$.

Now $100\cdot2$ plus Alice's number is a square between $201$ and $240$. The only square in that interval is $225$, so Alice's number is $25$. The sum is $2+25=27$.

Thus, the answer is A .

Solution:

The original circle is made from the regular hexagon plus $6$ equal circular segments. Reflecting each minor arc over its side puts those same $6$ segments inside the hexagon instead.

Therefore the average of the circle's area and the reflected-arc region's area is the area of the regular hexagon. The hexagon has area $6\cdot\frac{\sqrt3}{4}=\frac{3\sqrt3}{2}$, and the circle has radius $1$, so its area is $\pi$.

If the desired area is $A$, then $\frac{A+\pi}{2}=\frac{3\sqrt3}{2}$, so $A=3\sqrt3-\pi$.

Thus, the answer is B .

Solution:

Expand and rewrite: $$\begin{gathered} x(x-y)+y(y-z)+z(z-x)\\ \small =\frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}. \end{gathered}$$

For the value to be $1$, the three nonnegative square terms must sum to $2$. Since $x,y,z$ are integers, this means the squared differences are $1,1,0$.

Thus two of the variables must be equal, and the third must differ from them by $1$. The condition $x=z$ and $y-1=x$ guarantees exactly that.

Thus, the answer is D .

Solution:

Let the isosceles triangle have height $H$ and base $B$. By similarity, horizontal widths in the triangle are proportional to distance from the top vertex.

The top side of the larger square has width $3$ and is $H-3$ units from the top. The top side of the smaller square has width $2$ and is $H-5$ units from the top. Hence $$\frac{3}{H-3}=\frac{2}{H-5}.$$

Solving gives $H=9$. Also $\frac{B}{H}=\frac{3}{H-3}$, so $B=\frac{9}{2}$. The area is $\frac12\cdot\frac92\cdot9=\frac{81}{4}=20\frac14$.

Thus, the answer is B .

Solution:

Count the complement, where the product is not divisible by $4$. This happens if the product has no factor of $2$, or exactly one factor of $2$.

All dice odd has probability $(\frac12)^6=\frac1{64}$. Exactly one factor of $2$ means exactly one die is $2$ or $6$, and the other five dice are odd. This has probability $$6\cdot\frac26\cdot\left(\frac12\right)^5=\frac4{64}.$$

The complement has probability $\frac5{64}$, so the desired probability is $1-\frac5{64}=\frac{59}{64}$.

Thus, the answer is C .

Solution:

Since $BR=6$ and $PR=7$, we have $BP=13$. The right-angle and square-angle chasing gives $\triangle PAB\cong\triangle QBC$, so $CQ=BP=13$.

Because $BP\perp CQ$, point $R$ is the foot of the altitude from $B$ to the hypotenuse $CQ$ of right triangle $BQC$. Thus $$QR\cdot RC=BR^2=36$$ and $$QR+RC=CQ=13.$$

So $QR$ and $RC$ are $4$ and $9$. From the diagram $RC=9$, and therefore $$\begin{aligned} BC^2&=BR^2+RC^2\\ &=6^2+9^2\\ &=117. \end{aligned}$$

The area of the square is $117$.

Thus, the answer is D .

Solution:

After Chris chooses an adjacent pair, Silva has $5$ equally likely adjacent pairs to choose.

If Silva chooses the same pair, all $5$ balls return to their original positions. This has probability $\frac15$.

If Silva chooses a pair sharing exactly one ball with Chris's pair, then $2$ balls are in their original positions. There are $2$ such pairs, so this has probability $\frac25$.

If Silva chooses a disjoint adjacent pair, then $1$ ball is in its original position. This also has probability $\frac25$.

The expected number is $$\begin{aligned} 5\cdot\frac15+2\cdot\frac25+1\cdot\frac25&=\frac{11}{5}\\ &=2.2. \end{aligned}$$

Thus, the answer is D .

Solution:

Two reflections across lines through the origin are equivalent to a rotation by twice the angle from the first reflecting line to the second.

The point $(-1,4)$ is sent to $(4,1)$, which is a $90^\circ$ clockwise rotation. Therefore line $m$ is $45^\circ$ clockwise from line $\ell$.

The slope of $\ell$ is $5$. If $\theta_m=\theta_\ell-45^\circ$, then $$\tan\theta_m=\frac{5-1}{1+5}=\frac23.$$

Thus line $m$ is $y=\frac23x$, or $2x-3y=0$.

Thus, the answer is D .

Solution:

The boundary can be split into $24$ congruent triangles. Each has angles $15^\circ$, $45^\circ$, and $120^\circ$.

For one such triangle, draw the altitude from the center-side direction. The altitude is $3$, half the side length of a square. The adjacent right triangle has a $30^\circ$ angle, so the part cut off from a length $3$ base is $3\tan30^\circ=\sqrt3$.

Thus each small triangle has base $3-\sqrt3$ and height $3$, giving area $\frac{3(3-\sqrt3)}{2}=\frac{9-3\sqrt3}{2}$.

The total area is $$24\cdot\frac{9-3\sqrt3}{2}=108-36\sqrt3.$$ Hence $a+b+c=108+36+3=147$.

Thus, the answer is E .

Solution:

The number $N$ satisfies $$7\cdot10^{312}\lt N\lt 8\cdot10^{312}.$$ Multiplying or dividing by a power of $10$ only shifts the decimal point, so we only need the leading factor left after taking out the largest convenient power of $10$.

For $r=2$, $\sqrt N$ has leading factor between $\sqrt7$ and $\sqrt8$, so $f(2)=2$.

For $r=3$, the leading factor is between $\sqrt[3]{7}$ and $\sqrt[3]{8}$, so $f(3)=1$. For $r=4$, the leading factor is between $\sqrt[4]{7}$ and $\sqrt[4]{8}$, so $f(4)=1$.

For $r=5$, since $312=5\cdot62+2$, the leading factor is between $\sqrt[5]{700}$ and $\sqrt[5]{800}$. Since $3^5\lt700\lt800\lt4^5$, $f(5)=3$.

For $r=6$, the leading factor is between $\sqrt[6]{7}$ and $\sqrt[6]{8}$, so $f(6)=1$. The sum is $2+1+1+3+1=8$.

Thus, the answer is A .

Solution:

By symmetry, $P(\text{Hugo wins})=\frac14$. So the desired probability is $4\cdot P(\text{Hugo first rolls }5\text{ and wins})$.

If Hugo rolls $5$, then no other player can roll $6$. Case on how many of the other three players also roll $5$. If $t$ other players tie Hugo, then Hugo wins the eventual tiebreaker with probability $\frac1{t+1}$.

Thus $$\begin{gathered} P(\text{Hugo rolls }5\text{ and wins})\\ {}=\frac{64+24+4+\frac14}{6^4}\\ {}=\frac{369}{4\cdot1296}. \end{gathered}$$

Multiplying by $4$ gives $\frac{369}{1296}=\frac{41}{144}$.

Thus, the answer is C .

Solution:

For a regular $k$-gon and a regular $n$-gon inscribed in the same circle with $k\lt n$ and no shared vertices, their boundaries intersect in $2k$ points. Each side of the smaller polygon is crossed twice by the boundary of the larger polygon.

Therefore, sum over all pairs of polygons. The $5$-gon contributes $2\cdot5$ intersections with each of the $6$-, $7$-, and $8$-gons. The $6$-gon contributes $2\cdot6$ intersections with each of the $7$- and $8$-gons. The $7$-gon contributes $2\cdot7$ with the $8$-gon.

The total is $$3(10)+2(12)+14=68.$$

Thus, the answer is E .

Solution:

When passing from $S_{n-1}$ to $S_n$, the new terms are $jn$ for $1\le j\lt n$. Their sum is $$\begin{gathered} n(1+2+\cdots+(n-1))\\ =\frac{n^2(n-1)}2. \end{gathered}$$

Modulo $3$, this increment is $0$ when $n\equiv0$ or $1\pmod3$, and is $2$ when $n\equiv2\pmod3$.

Since $S_2=2$, the sequence becomes divisible by $3$ after the third occurrence of a number congruent to $2\pmod3$, namely at $n=8$. Then $S_n$ stays divisible by $3$ for $n=8,9,10$, and the same pattern repeats every $9$ in $n$.

The ten least values are $8,9,10,17,18,19,26,27,28,35$. Their sum is $197$.

Thus, the answer is B .

Solution:

Count the complement: colorings of the $10$ edges of $K_5$ with no monochromatic triangle.

At any vertex, if $3$ incident edges had the same color, then the edges among their other endpoints would all have to be the other color, making a monochromatic triangle. Thus each vertex has exactly $2$ red and $2$ blue incident edges.

So the red edges form a $2$-regular graph on $5$ vertices, which must be a $5$-cycle. The number of labeled $5$-cycles is $\frac{(5-1)!}{2}=12$.

There are $2^{10}=1024$ total colorings, so the desired probability is $$1-\frac{12}{1024}=\frac{253}{256}.$$

Thus, the answer is D .

Solution:

This is the number of ways to choose which $4$ of the $8$ cube vertices are blue, up to cube rotation. Use Burnside's lemma on the $24$ rotations of the cube.

The identity fixes $\binom84=70$ colorings. The $6$ quarter-turn face rotations each fix $2$. The $3$ half-turn face rotations each fix $6$. The $8$ rotations about opposite vertices each fix $4$. The $6$ half-turn rotations about opposite edges each fix $6$.

Thus the number of inequivalent constructions is $$\begin{gathered} \frac{70+6\cdot2+3\cdot6+8\cdot4+6\cdot6}{24}\\ =7. \end{gathered}$$

Thus, the answer is A .

Solution:

Use similar triangles as shown in the diagram. The left side of the large square has length $4x+2y$, and the bottom side has length $3y+x$. Since these are equal, $3y+x=4x+2y$, so $y=3x$. The side length of the large square is therefore $10x$.

In the upper part of the figure, let the marked horizontal segment be $m$. The two right triangles formed by the sides of rectangle $R$ have parallel corresponding sides and equal hypotenuses, so they are congruent. This gives the lengths shown below.

Similar triangles give $$\frac{m}{3x}=\frac{4x-\frac m3}{6x-m}.$$ Hence $6xm-m^2=12x^2-xm$, so $$\begin{gathered} m^2-7xm+12x^2=0\\ =(m-3x)(m-4x). \end{gathered}$$

If $m=3x$, rectangle $R$ has side length $3\sqrt2x$ in both directions, so its area is $18x^2$. If $m=4x$, its side lengths are $5x$ and $\frac{10}{3}x$, so its area is $\frac{50}{3}x^2$.

The two possible areas sum to $\frac{104}{3}x^2$. Since the $1\times3$ rectangle gives $x^2+(3x)^2=1$, we have $x^2=\frac1{10}$. The sum of the possible areas is $$\frac{104}{3}\cdot\frac1{10}=\frac{52}{15}.$$

Thus $m+n=52+15=67$, and the answer is E .