We can simplify the expression as follows: $$\begin{align*} \dfrac{(2112 - 2021)^2}{169} &= \dfrac{91^2}{169} \\ &= \dfrac{91^2}{13^2} \\ &= \left(\dfrac{91}{13}\right)^2 \\ &= 7^2 \\ &= 49. \end{align*}$$

Thus, C is the correct answer.

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.

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.

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.

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 $11.$

The only remaining choice is $A=9$, and $202109$ is prime: none of the primes at most $\sqrt{202109}<450$ divides it.

Thus, E is the correct answer.

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 strides is $$132 \div 12 = 11$$ feet. This makes Oscar's leap $11 - 3 = 8$ feet longer.

Thus, B is the correct answer.

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.

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

Then $$10a + b = a + b^2.$$ Rearranging, we get $$9a = b(b - 1).$$ This means that $9$ divides either $b$ or $b - 1$ ($3$ cannot divide both $b$ and $b - 1$).

Thus $b$ is either $0$ or $9$. If $b=0$, then $a=0$, which does not give 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.

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.

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.

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.

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.

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.

The second equation gives $|x|+|y|-4=\pm1$, so $|x|+|y|=3$ or $|x|+|y|=5$. Also the first equation gives $y=3-\frac{x^2}{3}$.

If $|x|+|y|=3$ and $y\ge0$, then $|x|+y=3$. For $x\ge0$, this gives $x+3-\frac{x^2}{3}=3$, so $x=0,3$. For $x\le0$, it gives $-x+3-\frac{x^2}{3}=3$, so $x=0,-3$. These give $3$ distinct points.

If $|x|+|y|=3$ and $y\lt0$, then $|x|-y=3$. For $x\ge0$, $x-3+\frac{x^2}{3}=3$, so $x=3$, which has $y=0$, not $y\lt0$. The case $x\le0$ similarly only gives $x=-3$, also already counted.

If $|x|+|y|=5$ and $y\ge0$, then $|x|+3-\frac{x^2}{3}=5$, which has no real solution in the required sign ranges. If $y\lt0$, then $|x|-3+\frac{x^2}{3}=5$. This gives $x=4$ for $x\ge0$ and $x=-4$ for $x\le0$, producing $2$ more points.

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

Thus, D is the correct answer.

Let $\odot C_1$ be the circle that is tangent to $AB$ and $AC.$ Then $$\angle ABO_1 = \angle ACO_1 = 90^{\circ},$$ making the two angles supplementary. This makes $ABO_1C$ cyclic.

Let $\odot O_2$ be the circumcircle of $ABO_1C.$ This makes $\odot O_2$ the circumcircle of $\triangle ABC$ as well.

We also know that $AO_1$ is the diameter of $O_2$ since $AO_1$ bisects $\angle BAC.$

By the Pythagorean theorem, we get that $$\begin{align*} AO_1 &= \sqrt{AB^2 + BO_1^2} \\ &= \sqrt{54 + 50} \\ &= 2\sqrt{26}. \end{align*}$$ This makes the area of $\odot O_2$ $26 \pi.$

Thus, C is the correct answer.

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.

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.