Solution:

Since $2112-2021=91=7\cdot13,$ $$\begin{aligned} \frac{(2112-2021)^2}{169} &=\frac{(7\cdot13)^2}{13^2}\\ &=7^2=49. \end{aligned}$$

Thus, C is the correct answer.

Solution:

If she shortens the $4$-inch side by $1,$ she has a $3\times 6$ card, whose area is $18$ square inches.

Therefore the given shortening was on the $4$-inch side. Shortening the other side instead gives a $4\times 5$ card, with area $20$ square inches.

Thus, E is the correct answer.

Solution:

The cube has volume $6^3=216.$ One ball of clay has volume $\frac{4}{3}\pi\cdot 2^3=\frac{32\pi}{3}.$

Because the clay may be reshaped but not compressed, the maximum number of balls is $$\left\lfloor \frac{216}{32\pi/3}\right\rfloor=\left\lfloor\frac{81}{4\pi}\right\rfloor.$$

Since $12\lt 4\pi\lt 13,$ we have $6\lt \frac{81}{4\pi}\lt \frac{81}{12}\lt 7.$ Therefore the floor is $6.$

Thus, D is the correct answer.

Solution:

Mr. Lopez would take $$\dfrac{6}{30} \cdot 60 = 12$$ minutes to travel on Route A.

On Route B, he would take $$\left(\dfrac{5 - .5}{40} + \dfrac{.5}{20}\right) \cdot 60 = 8.25$$ minutes.

The difference in times along these routes is $12 - 8.25 = 3.75$ minutes.

Thus, B is the correct answer.

Solution:

Note that $A$ cannot be even, as then the number would be divisible by $2.$

$A$ also cannot be $5,$ as that would make the number divisible by $5.$

If $A$ equaled $1$ or $7,$ then the sum of the digits of the number would be $6$ and $12$ respectively.

This would make the number divisible by $3,$ so that rules out $A$ equaling either of these numbers.

Finally, if $A$ equals $3,$ then the whole number becomes $202103.$ If we look at the difference of the sums of alternating digits, we get $$2 + 2 - 1 - 3 = 0,$$ which means the number is divisible by $11.$

Every choice except $9$ makes the number composite. Because the problem states that exactly one digit works, that digit must be $A=9.$ (Indeed, trial division by the primes at most $\sqrt{202109}<450$ confirms that $202109$ is prime.)

Thus, E is the correct answer.

Solution:

There are $40$ gaps between the $1$st and $41$st pole, which means that the distance between consecutive poles is $$5280 \div 40 = 132$$ feet.

This means that each of Elmer's strides is $$132 \div 44 = 3$$ feet. Similarly, each of Oscar's leaps is $$132 \div 12 = 11$$ feet. This makes Oscar's leap $11 - 3 = 8$ feet longer.

Thus, B is the correct answer.

Solution:

Since $\angle ADC = 90^{\circ},$ we get that $$\angle FDE = 360^{\circ} - 90^{\circ} - 110^{\circ}$$$$= 160^{\circ}.$$ Also since $\triangle FDE$ is isosceles, we get that $$\angle EFD = \dfrac{180^{\circ} - 160^{\circ}}{2} = 10^{\circ}.$$ Finally, we get that $$\angle AFE = 180^{\circ} - 10^{\circ} = 170^{\circ}.$$

Thus, D is the correct answer.

Solution:

Let $\underline{a} \ \underline{b}$ be a $2$-digit cuddly number.

Then $$10a + b = a + b^2.$$ Rearranging, we get $$9a = b(b - 1).$$ Because consecutive integers cannot both be divisible by $3,$ one of $b$ and $b-1$ must contain both factors of $3.$

For a digit $b,$ this leaves $b=0,1,$ or $9.$ The first two choices give $a=0,$ which does not make a two-digit number. If $b=9,$ then $a=8.$ Checking, we get that $89$ is a cuddly number. This shows that there is only $1$ two-digit cuddly number.

Thus, B is the correct answer.

Solution:

Let $p$ be the probability that an odd number is rolled. Then $3p$ is the probability an even number is rolled. We know that $$p + 3p = 1 \Rightarrow p = \dfrac{1}{4}.$$

The only way for the sum to be even is if both rolls have the same parity. This happens with a probability of $$\dfrac{1}{4}^2 + \dfrac{3}{4}^2 = \dfrac{10}{16} = \dfrac{5}{8}.$$

Thus, E is the correct answer.

Solution:

Recall the expected-value formula $$\mathbb E[X]=\sum_x x\,\Pr(X=x).$$

Therefore, $$t = \dfrac{1}{5} (50 + 20 + 20 + 5 + 5)$$ $$= \dfrac{1}{5} \cdot 100 = 20$$ and $$s = 50 \cdot \dfrac{50}{100} + 20 \cdot \dfrac{20}{100}$$$$+ 20 \cdot \dfrac{20}{100} + 5 \cdot \dfrac{5}{100} + 5 \cdot \dfrac{5}{100}$$ $$= 25 + 4 + 4 + .25 + .25 = 33.5$$

$t - s = -13.5.$

Thus, B is the correct answer.

Solution:

Let $x$ be the length of the ship. Then in the time that Emily moves $210$ steps, the ship moves $210 - x$ steps.

In the time that Emily moves $42$ steps, the ship moves $x - 42$ steps. Since the ship and Emily move at a constant rate $$\dfrac{210}{210 - x} = \dfrac{42}{x - 42}.$$ Cross-multiplying yields $$210x - 210 \cdot 42 = 210 \cdot 42 - 42x$$ $$42 \cdot 6 x = 2 \cdot 42 \cdot 210$$ $$x = 70.$$

Thus, A is the correct answer.

Solution:

Note that $$9 \equiv -1 \pmod{5}.$$ Then if expand $N$ using the definition of bases, we get $$N = 2 \cdot 9^{10} + 7 \cdot 9^9 + 6 \cdot 9^6 +$$$$5 \cdot 9 + 2$$ $$\equiv 2 (-1)^{10} + 7 (-1)^9 + 6 (-1)^6 +$$$$5 (-1) + 2 \pmod{5}$$ $$\equiv 2 - 7 + 6 - 5 + 2 \pmod{5}$$ $$\equiv 3 \pmod{5}.$$

Thus, D is the correct answer.

Solution:

Note that for this restriction to hold, there must be $3$ balls of each color.

There are $2^6 = 64$ ways to color the balls and $\binom{6}{3} = 20$ to choose which balls are white.

The desired probability is therefore $\dfrac{20}{64} = \dfrac{5}{16}.$

Thus, D is the correct answer.

Solution:

Put $t=|x|.$ The second equation gives $t+|y|=3$ or $5,$ while the first gives $y=3-\frac{t^2}{3}.$

If $y\ge0,$ then $0\le t\le3.$ For $t+y=3,$ substitution gives $t-\frac{t^2}{3}=0,$ so $t=0$ or $3.$ These yield $x=0,3,-3,$ for three points. For $t+y=5,$ substitution gives $t^2-3t+6=0,$ which has no real root.

If $y<0,$ then $t>3.$ The equation $t-y=3$ gives $t^2+3t-18=0,$ whose only nonnegative root is the excluded boundary value $t=3.$ The equation $t-y=5$ gives $t^2+3t-24=0,$ with exactly one positive root $t=\frac{-3+\sqrt{105}}2>3.$ It yields two points, one for each sign of $x.$

The total number of ordered pairs is $3+2=5.$

Thus, D is the correct answer.

Solution:

Let $O_1$ be the center of the circle tangent to $AB$ and $AC.$ Then $\angle ABO_1=\angle ACO_1=90^\circ,$ so $A,B,O_1,C$ are concyclic.

Because the right angles at $B$ and $C$ subtend $AO_1,$ the segment $AO_1$ is a diameter of this circle. Let $O_2$ be its center. The same circle passes through $A,B,$ and $C,$ so it is the desired circumcircle.

By the Pythagorean Theorem in $\triangle ABO_1,$ $$\begin{aligned} AO_1&=\sqrt{AB^2+BO_1^2}\\ &=\sqrt{54+50}=2\sqrt{26}. \end{aligned}$$ Therefore, the circumradius is $\sqrt{26},$ and the requested area is $26\pi.$

Thus, C is the correct answer.

Solution:

For every real $x,$ $$\begin{aligned} &f(1-x)=|\lfloor 1-x\rfloor| \\ &\quad {}-|\lfloor x\rfloor|=-f(x). \end{aligned}$$

Equivalently, replacing $x$ by $\frac{1}{2}+t$ gives $f\!\left(\frac{1}{2}-t\right)=-f\!\left(\frac{1}{2}+t\right).$ This is point symmetry about $\left(\frac{1}{2},0\right).$

Thus, D is the correct answer.

Solution:

Put a regular hexagon in coordinates with $A=(-1,0),$ $B=(-\frac{1}{2},\frac{\sqrt3}{2}),$ $C=(\frac{1}{2},\frac{\sqrt3}{2}),$ and $E=(\frac{1}{2},-\frac{\sqrt3}{2}).$ Because the solar panel is flat, the height is an affine function $h(x,y)=ux+vy+w.$

From $h(A)=12,$ $h(B)=9,$ and $h(C)=10,$ subtracting the last two equations gives $u=1.$ Then $-u+w=12,$ so $w=13.$ Using $h(B)=9$ gives $-\frac{1}{2}+\frac{\sqrt3}{2}v+13=9,$ so $\sqrt3v=-7.$

Therefore $$\begin{aligned} h(E) &=\frac{1}{2}-\frac{\sqrt3}{2}v+13 \\ &=\frac{1}{2}+\frac{7}{2}+13 \\ &=17. \end{aligned}$$

Thus, D is the correct answer.

Solution:

There are $2$ cases.

Case $1:$ the top-right and bottom-left sections have the same crop

There are $4$ options for which crop is in those sections. The other two sections have $3$ options for the crop since each crop has one restriction for crops next to it. This gives us $$4 \cdot 3 \cdot 3 = 36$$ combinations.

Case $2:$ the top-right and bottom-left sections have different crops

There are $4$ options for the top-right crop and $3$ different options for the bottom-left crop. Each of the other two sections borders both of these crops, leaving $2$ choices for each. This gives us another $$4 \cdot 3 \cdot 2 \cdot 2 = 48$$ configurations.

In total, there are $36 + 48 = 84$ combinations.

Thus, C is the correct answer.

Solution:

The side length of the inner square traced out by the inner circle is $s - 4.$

There are also the small pieces remaining in the corner. These form a total area of $$(1 + 1)^2 - \pi 1^2 = 4 - \pi.$$

Therefore, $$A = s^2 - (s - 4)^2 - (4 - \pi)$$$$= 8s - 20 + \pi.$$

The outer disk traces out an area that is comprised of $4$ rectangles and $4$ quarter-circles. The rectangles have area $s \cdot 2 = 2s$ and the quarter-circles form a circle with radius $2$ and area $4 \pi.$

This gives us $$2A = 8s + 4 \pi.$$

Equating the two equations we get $$8s + 4 \pi = 2(8s - 20 + \pi).$$ Solving yields $$8s = 40 + 2 \pi$$$$s = 5 + \dfrac{\pi}{4}.$$

Thus, A is the correct answer.

Solution:

A quadratic fails to have two distinct real solutions exactly when its discriminant is nonpositive. Thus we need $$b^2-4c\le0\qquad c^2-4b\le0,$$ or $b^2\le4c$ and $c^2\le4b.$

From $b^2\le4c,$ we get $b^4\le16c^2.$ Combining this with $c^2\le4b$ gives $b^4\le64b,$ so $b\le4.$