Solution:

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.

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 $202109.$ 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 $11.$

This means that $A$ must be $9.$

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 strides 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 $$9a = b(b - 1).$$ This means that $9$ divides either $b$ or $b - 1$ ($3$ cannot divide both $b$ and $b - 1$).

The only way this is possible is if $b = 9$ ($b$ is one digit, so it can't be anything else). 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 $$\text{Expected value} =$$$$\Sigma (\text{Outcome} \cdot \text{Probability}).$$

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:

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.

Solution:

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.

Solution:

For every real $x$, $$f(1-x)=|\lfloor 1-x\rfloor|-|\lfloor x\rfloor|=-f(x).$$

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

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 which crop is in the top-left section. Then there are $3$ options for the top-right and $2$ for the bottom-left. This leaves $2$ options for the bottom-right section. 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 $$b^2-4c\le0\quad\text{and}\quad c^2-4b\le0,$$ 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 $b\le4$.

Now check $b=1,2,3,4$. The inequalities give respectively $$(c=1,2),\quad (c=1,2),\quad (c=3),\quad (c=4).$$

There are $2+2+1+1=6$ ordered pairs.

Thus, B is the correct answer.

Solution:

For the sake of simplicity, we can assume the balls and bins are both distinguishable.

Since each case includes having $4$ balls in $3$ bins, we can leave those out during our calculation.

For $p,$ there are $5$ choices for the bin with $3$ balls and then $4$ choices for the bin with $5$ balls. Finally, there are $\binom{8}{3} = 56$ ways to choose which balls go in the bins.

For $q,$ after cancelling out $3$ of the $4$ s, there are $\binom{8}{4} = 70$ ways to ensure $4$ balls go in each of the remaining bins.

Since the total number of distributions is the same for both $p$ and $q,$ we can let $\dfrac{p}{q}$ be the ratio of the numerators. Therefore, $$\dfrac{p}{q} = \dfrac{20 \cdot 56}{70} = 16.$$

Thus, E is the correct answer.

Solution:

Let the cone have base in the plane $z=0$, center at the origin, and vertex on the $z$-axis. The centers of the three spheres form an equilateral triangle of side $2r$, so one sphere center may be taken at horizontal distance $\frac{2r}{\sqrt3}$ from the cone axis and height $r$ above the base.

In the axial cross-section through that center and the cone axis, the side of the cone is the line $12\rho+5z=60$, where $\rho$ is horizontal distance from the axis. The distance from $\left(\frac{2r}{\sqrt3},r\right)$ to this line must be $r$: $$r=\frac{60-12\cdot\frac{2r}{\sqrt3}-5r}{13}.$$

Thus $(18+8\sqrt3)r=60$, so $$r=\frac{60}{18+8\sqrt3}=\frac{90-40\sqrt3}{11}.$$

Thus, B is the correct answer.

Solution:

The value $12$ is fixed by the function, since $12$ has $6$ positive divisors and therefore $f_1(12)=12$.

First find all $n\le50$ with $f_1(n)=12$, meaning $n$ has $6$ divisors. These are $$12,18,20,28,32,44,45,50.$$

Now check whether $f_1(n)$ can be one of these values before reaching $12$. Since $f_1(n)$ is twice a divisor count, the only useful possibilities in that list are $18$ and $20$, meaning $n$ has $9$ or $10$ divisors.

For $n\le50$, the additional possibilities are $36$, which has $9$ divisors, and $48$, which has $10$ divisors. Therefore there are $8+2=10$ values of $n$.

Thus, D is the correct answer.

Solution:

Label one face $ABCD$. Each face must contain two $0$s and two $1$s, so first split by the pattern on the four edges of $ABCD$.

Case 1: opposite edges of $ABCD$ have the same label. There are $2$ such patterns on $ABCD$. For either pattern, once the label of one vertical edge, say $AE$, is chosen, the face-sum conditions force all remaining labels. This gives $2\cdot2=4$ labelings.

Case 2: opposite edges of $ABCD$ have different labels. There are $4$ such patterns on $ABCD$. For each, the labels of two adjacent vertical edges may be chosen in $4$ ways, and then the remaining labels are forced by the face-sum conditions. This gives $4\cdot4=16$ labelings.

The total is $4+16=20$.

Thus, E is the correct answer.

Solution:

Let the roots of $p(x)$ be $r$ and $s$, so $$p(x)=(x-r)(x-s)=x^2-(r+s)x+rs.$$ The equation $p(p(x))=0$ is equivalent to $p(x)=r$ or $p(x)=s$.

For exactly three real solutions, one of these two quadratic equations must have a double root and the other must have two distinct real roots. Suppose $p(x)=r$ has the double root. Its discriminant is $$(r+s)^2-4(rs-r)=(r-s)^2+4r,$$ so $(r-s)^2=-4r$, forcing $r\le0$.

The other equation, $p(x)=s$, has discriminant $(r-s)^2+4s=-4r+4s=4(s-r)$, which must be positive. Hence $s\gt r$, so $r-s=-2\sqrt{-r}$ and $s=r+2\sqrt{-r}$.

The sum of the roots is $r+s=2r+2\sqrt{-r}$. Let $u=\sqrt{-r}$, so this is $-2u^2+2u$, maximized at $u=\frac{1}{2}$. Thus $r=-\frac{1}{4}$ and $s=\frac{3}{4}$.

Therefore $p(x)=x^2-\frac{1}{2}x-\frac{3}{16}$, and $$p(1)=1-\frac{1}{2}-\frac{3}{16}=\frac{5}{16}.$$

Thus, A is the correct answer.