Small Hint:

Evaluate exponent towers from the top down

Big Hint:

Remember that $0^1=0$ and $2^0=1$

Solution:

We can evaluate from the top of each exponent tower downward: $$\begin{aligned} &2^{\left(0^{\left(1^9\right)}\right)}\\ &\quad+\left(\left(2^0\right)^1\right)^9\\ &=2^{0^1}+(1^1)^9\\ &=2^0+1^9\\ &=2. \end{aligned}$$

Thus, C is the correct answer.

Small Hint:

Look only at divisibility by $1000$

Big Hint:

Both factorials contain at least three factors of $5$ and three factors of $2$

Solution:

Both $20!$ and $15!$ contain at least three factors of $5$ and three factors of $2$, so both are divisible by $2^3\cdot5^3=1000.$

Their difference is therefore also divisible by $1000.$

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.

Small Hint:

Let Bonita’s current age be $b$

Big Hint:

Use last year’s relation and this year’s square relation together

Solution:

Let $a$ be Ana’s current age and $b$ be Bonita’s current age. Then $$\begin{aligned} a-1&=5(b-1),\\ a&=b^2. \end{aligned}$$

Substitution gives $$\begin{aligned} b^2-1&=5b-5,\\ b^2-5b+4&=0,\\ (b-4)(b-1)&=0. \end{aligned}$$

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.

Small Hint:

To avoid getting $15$ of a color, draw at most $14$ from each large color

Big Hint:

For colors with fewer than $15$, all of that color can be drawn

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.

Small Hint:

A long consecutive block can be centered around zero

Big Hint:

Use divisibility of $90$ by the number of terms

Solution:

Suppose there are $k$ consecutive integers with first term $a.$ Their sum is $\dfrac{k(2a+k-1)}{2}=45,$ so $k$ must divide $90.$ Therefore the number of terms cannot exceed $90.$

This bound is attained by $$-44, -43, \cdots, 44, 45$$ which has $90$ terms and sum $45.$

Thus the greatest possible number of consecutive integers is $90.$

Thus, D is the correct answer.

Small Hint:

Equidistant from all four vertices means cyclic

Big Hint:

For quadrilaterals, check whether opposite angles can be supplementary

Solution:

Note that if a point is equidistant from all the vertices, then 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.

Small Hint:

Find all three pairwise intersections of the lines

Big Hint:

Use the distance or shoelace formula for the resulting triangle

Solution:

The two lines through $(2,2)$ have equations $$\begin{aligned} y&=\frac12x+1,\\ y&=2x-2. \end{aligned}$$

Their intersections with $x+y=10$ are $(6,4)$ and $(4,6)$, respectively.

Thus the vertices are $(2,2),(6,4),$ and $(4,6)$. The segment joining the last two points has length $2\sqrt2$, its midpoint is $(5,5)$, and the distance from $(2,2)$ to that midpoint is $3\sqrt2$.

The area is therefore $$\frac12(2\sqrt2)(3\sqrt2)=6.$$ Thus, C is the correct answer.

Small Hint:

Translations along $\ell$ preserve the repeating pattern

Big Hint:

Check whether reflections reverse the slanted segments

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.

Small Hint:

The sum $1+2+\cdots+n$ equals $\frac{n(n+1)}{2}$

Big Hint:

The obstruction comes from when $n+1$ contributes a new prime factor

Solution:

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

Put $m=n+1$. If $m$ is composite, write $m=ab$ with $2\le a\le b$. When $a<b$, the distinct factors $a$ and $b$ both occur in $(m-2)!=(n-1)!$. When $a=b$, we have $a\ge3$, and the two multiples $a$ and $2a$ both occur in $(m-2)!$, so $a^2=m$ divides that factorial as well. Thus $(n-1)!$ is divisible by $m$, and consequently $$n! \text{ is divisible by } \frac{n(n+1)}2.$$

Conversely, if $n+1$ is prime, that prime factor does not occur in $n!$, so the divisibility fails. Since $997$ is prime while $998,999,$ and $1000$ are composite, the greatest three-digit value is $$997-1=996.$$

Thus, B is the correct answer.

Small Hint:

Count every grid line the diagonal crosses

Big Hint:

Because $10$ and $17$ are relatively prime, it never passes through an interior grid corner

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 is the correct answer.

Small Hint:

Factor $201$ first

Big Hint:

Use inclusion-exclusion on divisor exponents that are even or multiples of $3$

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, from $0$ through $8,$ and $4$ options for multiples of $3,$ from $0$ through $9.$

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

Small Hint:

Understand which dates appear $12$ times and which appear fewer times

Big Hint:

Compare the mean, median, and modes from the calendar counts

Solution:

The modes are all the integers from $1$ through $28,$ so their median is $d=\dfrac{14+15}{2}=14.5.$

There are $365$ entries, so $M$ is the $183$rd number. The dates from $1$ through $15$ occupy $15 \cdot 12 = 180$ positions, so $M=16.$

The sum of all dates is $$\begin{aligned}12(1+\cdots+28)&+11(29+30)\\&+7\cdot31=5738.\end{aligned}$$ Hence $\mu=\dfrac{5738}{365}\approx15.72.$

Therefore $$d \lt \mu \lt M.$$ Thus, E is the correct answer.

Small Hint:

Angles subtended by a diameter are right angles

Big Hint:

Use the isosceles angles of $\triangle ABC$ to chase the remaining angles

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.

In $\triangle BCE$ and $\triangle BCD$, respectively, $$\begin{aligned} \angle ECB&=180^{\circ}-70^{\circ}-90^{\circ}\\ &=20^{\circ} \end{aligned}$$ and $$\begin{aligned} \angle DBC&=180^{\circ}-40^{\circ}-90^{\circ}\\ &=50^{\circ}. \end{aligned}$$

Because $F$ lies on $BD$ and $CE$, the other two angles of $\triangle BFC$ are $50^{\circ}$ and $20^{\circ}$. Hence $$\begin{aligned} \angle BFC&=180^{\circ}-50^{\circ}-20^{\circ}\\ &=110^{\circ}. \end{aligned}$$ Thus, D is the correct answer.

Small Hint:

List attainable intersection counts for four lines

Big Hint:

Then separately rule out exactly two intersection points

Solution:

The values $0,1,3,4,5,$ and $6$ are attainable. Four parallel lines give $0$, four concurrent lines give $1$, three parallel lines cut by a fourth give $3$, three concurrent lines plus a fourth not through that point give $4$, three lines forming a triangle plus a fourth parallel to one side give $5$, and four lines in general position give $\binom42=6$.

It remains to rule out $2$. Choose two nonparallel lines, meeting at $X$. If a third line also passes through $X$, then a fourth line not through $X$ intersects at least two of those three concurrent lines at two different new points; otherwise all four lines pass through $X$, giving only one point. If the third line does not pass through $X$, then to create only one new point it must be parallel to one of the first two lines. A fourth distinct line cannot pass through either existing intersection without meeting the parallel line at a new point, and if it passes through neither, it creates a new intersection immediately. Thus exactly two intersection points are impossible.

Thus the possible values are $0,1,3,4,5,6$, whose sum is $19$. Thus, D is the correct answer.

Small Hint:

Take reciprocals of the recurrence

Big Hint:

The reciprocal sequence becomes arithmetic

Solution:

Taking reciprocals in the recursive formula gives $$\begin{aligned} \frac1{a_n} &=\frac{2a_{n-2}-a_{n-1}}{a_{n-2}a_{n-1}}\\ &=\frac2{a_{n-1}}-\frac1{a_{n-2}}. \end{aligned}$$

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 $$\dfrac{1}{\frac{3}{7}} - \dfrac{1}{1} = \dfrac{7}{3} - 1 = \dfrac{4}{3}.$$

Therefore $$\frac1{a_{2019}}=1+2018\cdot\frac43=\frac{8075}{3},$$ so $a_{2019}=\frac3{8075}.$

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

Thus, E is the correct answer.

Small Hint:

Connect centers of tangent circles

Big Hint:

The shaded area is the large circle area minus the areas of the small circles

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.

Small Hint:

Count all arrangements of eight cubes after choosing the omitted color

Big Hint:

Correct for repeated colors with factorial denominators

Solution:

Given a valid height-$8$ tower, its color counts determine the one unused cube; place that cube on top. Conversely, removing the top cube from any arrangement of all $9$ cubes gives a valid height-$8$ tower. These operations are inverses, so the desired towers are in one-to-one correspondence with arrangements of all $9$ cubes.

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.

Small Hint:

Expand $0.\overline{23}_k$ as a geometric series

Big Hint:

The common ratio after two digits is $k^{-2}$

Solution:

The repeating base-$k$ fraction is $2k^{-1}+3k^{-2}$ $+2k^{-3}+3k^{-4}+\cdots$. Grouping odd and even powers gives $$\begin{aligned} &2(k^{-1}+k^{-3}+\cdots) \\ &\quad {}+3(k^{-2}+k^{-4}+\cdots). \end{aligned}$$

Using geometric series, these sums are $\dfrac{2k}{k^2-1}$ and $\dfrac{3}{k^2-1}$, so $0.\overline{23}_k=\dfrac{2k+3}{k^2-1}$.

Setting $\dfrac{2k+3}{k^2-1}=\dfrac{7}{51}$ gives $51(2k+3)=7(k^2-1)$, so $7k^2-102k-160=0$. Hence $k=16$. Thus, D is the correct answer.