Solution:

We can evaluate this as follows.$$\begin{align*} &2^{\left(0^{\left(1^9\right)}\right)}+\left(\left(2^0\right)^1\right)^9\\ &= 2^{\left(0^1\right)} + \left(1^1\right)^9 \\&= 2^0 + 1^9 \\ &= 1 + 1 \\&= 2 \end{align*}$$

Thus, C is the correct answer.

Solution:

Note that $20!$ and $15!$ both have factors of $5^3$ in them.

This means that they are both divisible by a $1000,$ making their difference also a multiple of $1000.$

Being a multiple of $1000$ makes the last three digits $0,$ which shows that the hundreds digit is also $0.$

Thus, A is the correct answer.

Solution:

Let $a$ be Ana's age and $b$ be Bonita's age. The statement then gives us that $$\begin{gather*} a - 1 = 5(b - 1), \\ a = b^2. \end{gather*}$$

We can substitute the second equation into the first to get $$\begin{gather*} b^2 - 1 = 5b - 5 \\ b^2 - 5b + 4 = 0 \\ (b - 4)(b - 1) = 0. \end{gather*}$$

We can see that $b \neq 1$ since that would make Ana and Bonita the same age, so we know that $b = 4.$

This gives us that $a = 4^2 = 16$ and $n = 16 - 4 = 12.$

Thus, D is the correct answer.

Solution:

Note that we can pull as many as $14$ balls of each color without ensuring that $15$ balls of one color are drawn.

This means that we can draw all of the black, white and blue balls, along with $14$ red, green, and yellow balls.

This gives us a total of $$9 + 11 + 13 + 3 \cdot 14$$$$= 33 + 42$$$$= 75.$$ We need to add one at the end, however, to ensure that we get that $15$th ball of some color, $75 + 1 = 76.$

Thus, B is the correct answer.

Solution:

Note that negative integers are allowed to be in the sequence.

This means that we could form the sequence $$-44, -43, \cdots, 44, 45,$$ which clearly adds up to $45.$ There are $90$ terms in this sequence.

Adding another negative term wouldn't work, since that would require having to go up to $46$ in the positive numbers, which puts the sum over $45.$

Thus, D is the correct answer.

Solution:

Note that if a point is equidistant from all the vertices, than that point is the center of the shape's circumcircle.

The question then becomes which of these shapes is cyclic (has a circumcircle). One condition that we can use is that opposite angles are supplementary.

Clearly, a square and rectangle that is not a square work (opposite angles are right, adding up to $180^{\circ}$).

A rhombus that is not a square does not work, since opposite angles are equal, but they are not $90^{\circ}.$

A parallelogram that is not a rectangle or a rhombus faces the same problem as above, making it not cyclic as well.

An isosceles trapezoid that is not a parallelogram by definition has supplementary opposite angles, making it cyclic.

Thus, C is the correct answer.

Solution:

Let us first find the equations of the two lines. Using slope-intercept form, we know they are the form $y = ax + b.$

For the first line, we know that $a = \dfrac{1}{2},$ so we get $$2 = \dfrac{1}{2} \cdot 2 + b$$$$b = 1.$$

Similarly, for the second line, we get that $a = 2,$ which gives us $$2 = 2 \cdot 2 + b$$$$b = -2.$$

Our two lines are now $y = \dfrac{1}{2}x + 1$ and $y = 2x - 2.$

We can rewrite $x + y = 10$ as $y = 10 - x.$ Substituting this into the first line yields $$10 - x = \dfrac{1}{2}x + 1$$$$x = 6, y = 4.$$

Similarly, for the second line, $$10 - x = 2x - 2$$$$x = 4, y = 6.$$

The three vertices of the triangle are therefore $(2, 2), (6, 4),$ and $(4, 6).$

Note that these vertices form an isosceles triangle (distance formula yields the three sides as $2\sqrt{5}, 2\sqrt{5},$ and $2\sqrt{2}.$

The midpoint of the base is $(5, 5),$ and applying the distance formula again tells us that the height is $3\sqrt{2}.$

The area is therefore $$\dfrac{1}{2} \cdot 2\sqrt{2} \cdot 3 \sqrt{2} = 6.$$ Thus, C is the correct answer.

Solution:

The first transformation works, as we can rotate $\ell$ $180^{\circ}$ around the midpoint between an upward-facing and downward-facing square.

The second also works, as we can just move $ell$ to the right until the squares line up with each other again.

The third fails, as a reflection would cause the line segments to face the opposite direction.

The fourth transformation also doesn't work since the diagonal lines would again be facing in the wrong direction.

Thus, C is the correct answer.

Solution:

The sum of the first $n$ numbers is $$\dfrac{n(n + 1)}{2}.$$ We need this to not divide $n!.$

Note that if $n + 1$ is composite, then it can be broken down into factors that divide $n!.$

This means that we need $n + 1$ to be prime. The largest three-digit prime is $997,$ so the largest $n$ value is $$997 - 1 = 996.$$

Thus, B is the correct answer.

Solution:

Note that every time the bug crosses a vertical or horizontal line, the bug visits one new tile.

This means that the number of tiles the bug visits is $1$ (the first tile) plus the number of lines it crosses.

The bug never walks over a corner since $10$ and $17$ are relatively prime, so we don't have to worry about that.

The bug crosses $16$ horizontal lines and $9$ vertical lines for a total of $$1 + 16 + 9 = 26$$ tiles.

Thus, C see the correct answer.

Solution:

Taking the prime factorization of $201^9,$ we get $3^9 \cdot 67^9.$

Note that a perfect square has even exponents for its prime factors, and a cube's exponents are divisible by $3.$

There are $5$ options for an even exponent $(0$ through $8,)$ and $4$ options for multiples of $3$ $(0$ through $9).$

This gives us $5^2$ options for the squares and $4^2$ options for the cubes. We have to subtract out the powers of $6,$ however.

Using the same logic, sixth powers have to have exponents of prime factors be divisible by $6.$ There are $2$ options $(0$ and $6).$

This means that there are $2^2 = 4$ sixth powers. This gives us a total of $$25 + 16 - 4 = 37$$ perfect squares or perfect cubes.

Thus, C is the correct answer.

Solution:

$d$ must have to be less than $M$ because $M$ accounts for the larger numbers that are not modes ($29, 30$ and $31$).

There are $365$ entries, so $m$ is the $183$ rd number. The first $15$ numbers take up $15 \cdot 12 = 180$ spots, so $m$ is $16.$

$\mu$ is less than $16$ since there are fewer occurrences of the larger numbers, making the distribution left skewed.

$d$ is also less than $\mu$ since $\mu$ accounts for $29, 30$ and $31.$ Therefore, we get that $$d \lt \mu \lt M.$$ Thus, E is the correct answer.

Solution:

Since $\overline{BC}$ is the diameter of the circle, we get that $\angle BDC$ and $\angle BEC$ are right angles.

We know that $\angle ABC = 70^{\circ}$ from the fact that $\triangle ABC$ is isosceles.

Using the fact that the angles of a triangle add up to $180^{\circ},$ we get that $$\begin{align*} \angle ECB &= 180^{\circ} - 70^{\circ} - 90^{\circ}\\&= 20^{\circ} \end{align*}$$ and $$\begin{align*} \angle DBC &= 180^{\circ} - 40^{\circ} - 90^{\circ}\\&= 50^{\circ}.\end{align*}$$

Now, from $\triangle BFC,$ we get that $$\begin{align*} \angle BFC &= 180^{\circ} - 50^{\circ} - 20^{\circ} \\&= 110^{\circ}.\end{align*}$$ Thus, D is the correct answer.

Solution:

It is shown below that $$0, 1, 3, 4, \text{ and } 5$$ are attainable.

We need to show that it is impossible to attain only $2$ intersection points. Let $X$ and $Y$ be the intersection points.

Case $1:$ no lines go through both $X$ and $Y$

This means that there are $2$ unique pairs of lines that go through each of $X$ and $Y.$

Take one line that goes through $X$ and call it $\ell.$ For $\ell$ to not intersect the lines that go through $Y,$ it must be parallel to them.

This means that all $3$ lines must be parallel, but this is not possible since the other two lines intersect at $Y.$

Case $2:$ one line goes through both $X$ and $Y$

Let this common line be $\ell.$ Then the other two lines that go through $X$ and $Y$ must be parallel.

For there to be no other intersections, every other line must also be parallel to this two lines.

This, however, ensures that all the other lines are not parallel with $\ell,$ which results in more intersections.

In both cases, $2$ intersections is not possible. Therefore, the sum of all values of $N$ is $$1 + 3 + 4 + 5 + 6 = 19.$$ Thus, D is the correct answer.

Solution:

We can rewrite the recursive formula as $$\begin{align*} \dfrac{1}{a_n} &= \dfrac{a_{n - 2} \cdot a_{n - 1}}{2a_{n - 2} - a_{n - 1}} \\&= \dfrac{2}{a_{n - 1}} - \dfrac{1}{a_{n - 2}}. \end{align*}$$

This means that $$\dfrac{1}{a_n} - \dfrac{1}{a_{n - 1}} = \dfrac{1}{a_{n - 1}} - \dfrac{1}{a_{n - 2}},$$ which tells us that $\left\{\dfrac{1}{a_n}\right\}$ is an arithmetic sequence.

Using $a_1$ and $a_2,$ we get that the common difference is $$\dfrac{1}{\frac{3}{7}} - \dfrac{1}{1} = \dfrac{7}{3} - 1 = \dfrac{4}{3}.$$

From this, we get that $$\begin{align*} \dfrac{1}{a_{2019}} &= 1 + 2918 \cdot \dfrac{4}{3} \\&= \dfrac{8075}{3}. \end{align*}$$

$p + q$ is therefore $$8075 + 3 = 8078.$$

Thus, E is the correct answer.

Solution:

We know $\triangle ABC$ and $\triangle ABO$ are equilateral triangles.

We get that $OC = 2\sqrt{3}$ using special right triangles to find the altitudes of the triangles.

The radius of the larger circle is therefore $2\sqrt{3} + 1,$ since there is the extra unit radius after $\overline{OC}.$

The area of the larger circle is $$(2\sqrt{3} + 1)^2\pi = (13 + 4\sqrt{3})\pi.$$

The area of all the inner circles is $13\pi.$

The area of the shaded region is $$(13 + 4\sqrt{3})\pi - 13\pi = 4\pi\sqrt{3}.$$

Thus, A is the correct answer.

Solution:

Every tower of height $8$ could have been formed by creating a tower of height $9$ and removing the top cube.

This shows that there is a one-to-one correspondence between towers of height $8$ and $9.$

There are $9!$ ways to make a tower of height $9,$ but we are overcounting since there are multiple cubes of the same color.

We have to divide through by $2!$ ways to arrange the red cubes, $3!$ for the blue cubes, and $4!$ for the green cubes.

Therefore, the number of valid arrangements is $$\dfrac{9!}{2! \cdot 3! \cdot 4!} = 1,260.$$

Thus, D is the correct answer.

Solution:

We can expand the repeating fraction as $$0.\overline{23}_k = 2 \cdot k^{-1} + 3 \cdot k^{-2} + \cdots.$$

We can rearrange this to get the sum of two infinite sequences: $$2(k^{-1} + k^{-3} + k^{-5} + k^{-7} \cdots)$$ and $$3(k^{-2} + k^{-4} + k^{-6} + k^{-8} \cdots).$$

These sums evaluate to $$2 \cdot \dfrac{k^{-1}}{1 - k^{-2}} = \dfrac{2k}{k^2 + 1}$$ and $$3 \cdot \dfrac{k^{-2}}{1 - k^{-2}} = \dfrac{3}{k^2 + 1}.$$

Adding these together yields $$\dfrac{2k}{k^2 + 1} + \dfrac{3}{k^2 + 1} = \dfrac{2k + 3}{k^2 + 1}.$$

We know that these equals $\dfrac{7}{51},$ which gives us the quadratic $$\begin{gather*} 51(2k + 3) = 7(k^2 - 1) \\ 7k^2 - 102k - 160 = 0 \\ (k - 16)(7k + 10) = 0. \end{gather*}$$ $k$ can't be negative, so we get that $k = 16.$

Thus, D is the correct answer.

Solution:

Multiplying the first two terms and the last terms yields $$(x^2 + 5x + 4)(x^2 + 5x + 6).$$

Note that these two terms differ by $2.$ We can try to express this as a difference of squares, which is $$(x^2 + 5x + 5)^2 - 1.$$

Adding $2019$ to this gets us $$(x^2 + 5x + 5)^2 + 2018.$$

Squares are non-negative, so as long as we find a way to make the inner expression $0,$ we can make the square $0.$

The discriminant is $5^2 - 4 \cdot 5 = 5,$ which is positive meaning that there is a value that makes the square $0.$

This means that the minimum value would be $$0^2 + 2018 = 2018.$$ Thus, B is the correct answer.

Solution:

Note that the only way to get an odd sum is if there are either $0$ or $2$ even numbers in the row or column.

The only way for this to happen is if the $4$ even numbers form a rectangle with sides parallel to the large square.

The way to see this is we choose a spot for the first even number. Then we need to choose another square in the same row, $x,$ and column, $y,$ to be even.

The final even has to be in same column as $x$ and the same row as $y.$ This forms the aforementioned rectangle.

There are four $2 \times 2$ rectangles, two $3 \times 2$ rectangles, two $2 \times 3$ rectangles, and one $3 \times 3$ rectangle.

This gives us a total of $$4 + 2 + 2 + 1 = 9$$ rectangles, are arrangements for the even numbers.

There are $4!$ ways to arrange the even numbers and $5!$ ways to arrange the odd numbers.

This means that there are a total of $$9 \cdot 4! \cdot 5!$$ configurations of squares that satisfy the condition.

There are a total of $9!$ arrangements with no restrictions. The probability is therefore $$\dfrac{9 \cdot 4! \cdot 5!}{9!} = \dfrac{1}{14}.$$

Thus, B is the correct answer.

Solution:

We get the following diagrams by taking the cross-section of the plane of the triangle.

Note that $AC = 9$ by the Pythagorean theorem. We also get that $\triangle ADP \sim \triangle ACB.$

We can see that $PCBD$ is a kite, so we know that $$DB = BC = 12,$$ which makes $AD = 15 - 12 = 3.$ Using the similar triangles, above, we get that $$\dfrac{r}{3} = \dfrac{12}{9}$$$$r = 4.$$

Let $d$ be the distance from the sphere to this plane. $d$ is also the distance from $O$ to $P.$

Once again using the Pythagorean theorem, we get that $$d = \sqrt{6^2 - 4^2} = 2\sqrt{5}.$$

Thus, D is the correct answer.

Solution:

We can case on whether $x$ and $y$ are chosen from the interval or from $0$ and $1.$ Each case has a $\frac{1}{4}$ chance of happening, since they depend on two coin flips.

Case $1: x$ and $y$ are either $0$ or $1$

$x$ and $y$ need to be different, which happens with a $\frac{1}{2}$ probability.

Case $2: x$ is either $0$ or $1,$ and $y$ is chosen from $[0, 1]$

If $x = 0,$ then $y$ has to be chosen from $\left(\dfrac{1}{2}, 1\right],$ and if $x = 1,$ then $y$ has to be chosen from $\left[0, \dfrac{1}{2}\right).$

This means that $y$ always has a $\frac{1}{2}$ probability of being chosen from the correct interval.

Case $3: x$ is chosen from $[0, 1],$ and $y$ is either $0$ or $1$

This has the same probability as case $2$ due to symmetry.

$4: x$ and $y$ are chosen from $[0, 1]$

We can use geometric probability since we are working with an infinite number of $(x, y)$ pairs. We graph $|x - y| \gt \frac{1}{2}.$

The shaded area covers $\frac{1}{4}$ of the graph, showing that there is a $\frac{1}{4}$ probability of this case working.

Adding up all the probabilities, we get $$\begin{align*} \dfrac{1}{4}\left(3 \cdot \dfrac{1}{2} + \dfrac{1}{4}\right) &= \dfrac{1}{4} \cdot \dfrac{7}{4}\\&= \dfrac{7}{16}. \end{align*}$$

Thus, B is the correct answer.

Solution:

We can find how many numbers each triplet says in one round.

$$\begin{align*} &\text{Tadd: } 1, 4, 7, 10, 13 \cdots \\ &\text{Todd: } 2, 5, 8, 11, 14 \cdots \\ &\text{Tucker: } 3, 6, 9, 12, 15 \cdots \end{align*}$$

Now we can find a general formula for the number of numbers Tadd says after the $n$ th round.

$$\begin{align*} \sum_{i = 1}^n 3n - 2 &= -2n + 3 \sum_{i = 1}^n \\ &= -2n + \dfrac{3n(n + 1)}{2} \\&= \dfrac{3n^2 - n}{2} \end{align*}$$

Now we to find the largest $n$ such that $$\dfrac{3n^2 - n}{2} \leq 2019.$$

We can guess and check to find that $36$ is the largest such value.

Note that Todd and Tucker also go through $36$ turns each before Tadd says the $2019$ th number.

The number of numbers that Todd and Tucker go through, plus the $2019$ numbers that Tadd says, is $$\begin{align*} &\sum_{i = 1}^{36} 3n + \sum_{i = 1}^{36} (3n - 1) + 2019 \\ &= \sum_{i = 1}^{36} (6n - 1) + 2019 \\&= (5 + 11 + \cdots + 215) + 2019 \\&= \dfrac{36 \cdot 220}{2} + 2019 \\&= 5979. \end{align*}$$

Thus, C is the correct answer.

Solution:

We can multiply each side by $$(s - p)(s - q)(s - r)$$ to get $$A(s - q)(s - r) + B(s - p)(s - r)$$$$+ C(s - p)(s - q) = 1.$$

We can expand to get $$s^2(A + B + C) - s \cdot$$ $$(Aq + Ar + Bp + Br + Cp + Cq)$$ $$+ (Aqr + Bpr + Cpq - 1) = 0.$$

Note that the coefficients of $s$ and $s^2$ must both be $0.$

From $$A + B + C = 0, \tag*{(1)}$$ we get $$A = -(B + C),$$$$B = -(A + C),$$ and $$C = -(A + B).$$

Plugging this into the coefficient of $s,$ we get $$Ap + Bq + Cr = 0.$$

Subtracting $(1) \cdot r$ form this above equation, we get $$A(p - r) + B(q - r) = 0 (2)$$ We also know that $$Aqr + Bpr + Cpq = 1.$$ Subtracting $(1) \cdot pq$ from this equation, we get $$Aq(r - p) + Bp(r - q) = 1.$$

Adding $(2) \cdot p$ to this equation, we get $$A(r - p)(q - p) = 1,$$ which gets us $$A = \dfrac{1}{(r - p)(q - p)}.$$

Similarly, we get $$B = \dfrac{1}{(r - q)(p - q)}$$ and $$C = \dfrac{1}{(q - r)(p - r)}.$$

This gives us $$\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C} =$$ $$p^2 + q^2 + r^2 - pq - qr - pr.$$

Using Vieta's formulas, we get $$p^2 + q^2 + r^2 = (p + q + r)^2 -$$ $$2(pq + qr + pr) = 324.$$ Finally, since $$pq + qr + pr = 80,$$ we get our desired value of $324 - 80 = 244.$

Thus, B is the correct answer.

Solution:

One fact that greatly helps with this problem is realizing that $$\dfrac{(n^2)!}{(n!)^{n + 1}}$$ is always an integer.

This is because it is the number of ways to split up $n^2$ objects into $n$ unordered groups of size $n.$

Now, we get that $$\dfrac{(n^2 - 1)!}{(n!)^n} = \dfrac{(n^2)!}{(n!)^{n + 1}} \cdot \dfrac{n!}{n^2}.$$

Therefore, we need to find when $n^2 \div n!,$ or when $n \div (n - 1)!.$

This condition is false if $n = 4,$ or if $n$ is prime. $n = 4$ is too large for it to divide $3!.$

$n$ cannot be prime because $(n - 1)!$ does not contain any numbers where $n$ could be a factor.

If $n$ is not $n$ and not prime, this works since $n$ can be decomposed into $2$ numbers both less than $n$ that are found in $(n - 1)!.$

There are $15$ primes less than $50,$ and adding on $4,$ we get that there are $16$ values for $n$ that do not work.

Therefore, the desired answer is $50 - 16 = 34.$

Thus, D is the correct answer.