Solution:

Every integer from $-9$ to $9,$ inclusive, works. This yields $9-(-9)+1 = 19$ solutions.

Thus, the correct answer is D .

Solution:

We know $$\sqrt{\left(3-2\sqrt{3}\right)^2}+\sqrt{\left(3+2\sqrt{3}\right)^2 }$$ $$= |3-2\sqrt{3}| + |3+2\sqrt{3}|.$$ Since $3 < 2 \sqrt 3,$ we know that $3-2\sqrt{3} < 0.$

Therefore, our desired equation expression is equal to $$-3+2\sqrt{3} + 3+2\sqrt{3}$$$$= 4 \sqrt 3.$$

Thus, the correct answer is D .

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Written solution:

Let the number of juniors be $j$ and the number of seniors be $s.$ Then, $j+s = 28$ and $0.25j = 0.1s.$ This means $2.5j = s,$ so $3.5j = 28.$ This makes $j=8.$

Thus, the correct answer is C .

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Written solution:

There are $2\cdot 23 = 46$ students with blue shirts that are in a pair with just blue shirts. This means there are $57-46=11$ students in blue shirts who are paired with someone wearing a yellow shirt, meaning exactly $11$ people wearing yellow shirts are paired with someone wearing a blue shirt.

This leaves just $64$ students wearing a yellow shirt who are paired with someone else wearing a yellow shirt. This yields $\dfrac{64}2 = 32$ pairs.

Thus, the answer is B .

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Written solution:

Since the last two are multiplied to $30$ and both are single-digit numbers, one of them must be $5,$ making the other person $6.$ The first two are of ages that multiply to $24.$ The only pair of single-digit numbers whose product is $24$ and none of them are $4$ or $6$ is the pair $3,8.$ Thus, the ages are $3,5,6,8,$ making their sum $22.$

Thus, the answer is B .

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Written solution:

Let the number of people in the first class be $3x.$ This means the number of people in the second class is $4x.$

Thus, the sum of the scores of the first class is $84\cdot 3x = 252x$ and the sum of the scores for the people in the second class is $70\cdot 4x = 280x.$ This means the total sum is $532x,$ with $7x$ people.

Therefore, the average of all the students is $\dfrac{532x}{7x} = 76.$

Thus, the correct answer is C .

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Written solution:

On one side of $\ell$, circles tangent at $A$ are nested. For nested circles with radii $r_1>r_2>\cdots$, the points inside exactly one of those circles have area $\pi(r_1^2-r_2^2)$ if there are at least two circles; a third smaller nested circle does not count because its points are inside three circles, not exactly one.

To maximize the area, put the circle of radius $7$ alone on one side, and put the circles of radii $5,3,1$ on the other side. This gives

$$\begin{aligned} &7^2\pi+(5^2-3^2)\pi \\ &=49\pi+16\pi=65\pi. \end{aligned}$$

Thus, the answer is D .

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Written solution:

In the outer $15\times15$ ring, the top row contains $211,212,\ldots,225$, so the number just below $211$ in the second row is $210$. This is the greatest number in the second row.

The inner $13\times13$ spiral has $13^2=169$ in its upper-right corner. In the second row of the full grid, the inner-ring entries run from $157$ to $169$. Thus the least entry in that row is $157$.

The required sum is $210+157=367$.

Thus, the answer is A .

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Written solution:

Work backward. Reflecting $(-6,3)$ across $y=-x$ gives $(-3,6)$.

Now undo the $90^\circ$ counterclockwise rotation by rotating $(-3,6)$ clockwise about $(1,5)$. Relative to $(1,5)$, the point is $(-4,1)$. A clockwise quarter-turn sends this to $(1,4)$, and translating back gives $(2,9)$.

Thus $a=2$, $b=9$, and $b-a=7$.

Thus, the answer is D .

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Written solution:

The volumes must be the same since the water is poured from one to another. The volume of the cone is $\frac{r^2h \pi}3 = \frac{12^2\cdot 18 \pi}3 = 864 \pi.$

The volume of the cylinder is $r^2h \pi =24^2h \pi = 576h \pi.$ This makes $576 h \pi = 864 \pi,$ so $h = 1.5.$

Thus, the answer is A .

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Suppose the cuts make an $l\times w$ grid of pieces. The number of interior pieces is $(l-2)(w-2)$, and the total number of pieces is $lw$. Since the number of interior pieces equals the number of perimeter pieces, the interior pieces make up half the total:

$$(l-2)(w-2)=\frac{lw}{2}.$$

Multiplying out gives $lw-4l-4w+8=0$, or

$$(l-4)(w-4)=8.$$

The positive factor pairs of $8$ give $(l,w)=(5,12)$ or $(6,8)$, up to order. These produce $60$ or $48$ pieces, respectively, so the greatest possible number is $60$.

Thus, the answer is D .

Solution:

Using prime factorization, we get $$N =2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2.$$

If we have an odd divisor $x$ of $N,$ then $2x,4x,8x$ are divisors of $N,$ which has a combined sum of $14x.$ If we take the sum of every odd divisor, then the even divisors must have a sum which is $14$ times the sum of the odd divisors. Therefore, the requested ratio is $1:14$.

Thus, the answer is C .

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Written solution:

The first statement means $$3n^2 + 2n+d = 263.$$

Similarly, the second statement means $$3n^2 +2n +4$$$$= 6^3 +6^2 +6d+1$$$$= 253 + 6d.$$

Subtracting these shows us that $$4-d = 6d-10$$ $$7d = 14$$ $$d=2.$$ Therefore, $3n^2 + 2n + 2 = 263,$ so $n(3n+2) = 261.$

This implies $n=9.$ Therefore, $n+d = 11.$

Thus, the answer is B .

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Written solution:

The two chords of length $38$ are equally far from the center of the circle. Because the three parallel lines are equally spaced, those two equal chords must lie on adjacent lines, with the center halfway between them. Let that half-distance be $d$. Then each $38$-chord is distance $d$ from the center, and the $34$-chord is distance $3d$ from the center.

If the circle has radius $r$, then

$$r^2=19^2+d^2=17^2+(3d)^2.$$

Thus $19^2-17^2=8d^2$, so $72=8d^2$, and $d=3$. The distance between adjacent parallel lines is $2d=6$.

Thus, the answer is B .

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Since $x+\frac{1}{x} = \sqrt{5},$ squaring yields $$x^2 + 2 + \frac 1{x^2} = 5$$ $$x^2 + \frac 1{x^2} = 3.$$ Squaring again yields $$x^4 + 2 + \frac 1{x^4} = 9$$$$x^4 -7 + \frac 1{x^4} = 0.$$ Multiplying by $x^7$ yields $$x^{11}-7x^{7}+x^3=0.$$

Thus, the answer is B .

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If a number is divisible by $15,$ it has a units digit of $0$ or $5.$ If the units digit is $0$ and the digits are strictly increasing, then the number is $0,$ which isn't positive. Therefore, we can just look at numbers with a units digit of $5.$

Next, we need to find uphill integers that are a multiple of $3.$ This means the other digits are a subset of $\{1,2,3,4\}.$ Taking the sum of the set must have a remainder of $1$ when divided by $3.$ Also, having or taking out $3$ wouldn't affect the remainder, so we can take the number of subsets without a $3$ and multiply it by $2.$ There are only $3$ such subsets, namely $\{1\}, \{4\},$ and $\{1,2,4\}.$ Thus, there are $6$ total subsets.

Thus, the correct answer is C .

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Written solution:

If there are $2$ cards for Oscar that add up to $4,$ he must have both $1$ and $3.$ This eliminates choices A and B.

If there are $2$ cards for Aditi that add up to $7,$ he must have both $2$ and $5$ so she doesn't have $1$ and $3.$

If someone has $4,$ their sum must be equal to or under $14$ since the other number must be under or equal to $10.$ Thus, Ravon must have the $4$ and $7,$ making C true and D and E false.

Thus, the answer is C .

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Written solution:

The probability that the first number is even is $\frac 36.$

The probability that the second distinct number is even is $\frac 25.$

The probability that the third distinct number is even is $\frac 14.$

The combined probability is $$\dfrac{3\cdot 2\cdot 1}{6\cdot 5\cdot 4} = \dfrac 1{20}.$$

Thus, the answer is C .

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Written solution:

Let the sum of all the integers be $s,$ the greatest number be $g,$ the least number be $l,$ and the size of $S$ be $n.$

From the info given, we know $$\dfrac{s-l}{n-1} = 40$$$$\dfrac{s-g}{n-1} = 32.$$ Subtracting these yields $$\dfrac{g-l}{n-1} = 8.$$ Since we know $g-l=72,$ we know $\frac{72}{n-1} =8,$ so $n=10.$

We also know $$\dfrac{s-g-l}{n-2} = 35,$$ so $$s-g-l = 8\cdot 35$$$$=280.$$ Since $$\dfrac{s-l}{n-1} = 40,$$ we know$s-l = 360.$ This makes $g=80.$ Using $g=l+72,$ we get $l=8.$ Thus, $s=368.$

The average is $$\dfrac sn = \dfrac{368}{10}$$$$= 36.8.$$

Thus, the answer is D .

Solution:

The equal length segments show that the side pieces near $B$ and $E$ are made from halves of equilateral triangles of side length $2$. An equilateral triangle of side length $2$ has altitude $\sqrt3$ and area $\sqrt3$, so the two side pieces together contribute area $2\sqrt3=\sqrt{12}$.

The remaining central triangle is $\triangle ACD$. From the same equilateral-triangle altitudes, $AC=AD=2\sqrt3=\sqrt{12}$, and $CD=2$. Its altitude to $CD$ is

$$\sqrt{(\sqrt{12})^2-1^2}=\sqrt{11}.$$

Therefore the area of $\triangle ACD$ is $\frac12\cdot2\cdot\sqrt{11}=\sqrt{11}$. The pentagon's total area is

$$\sqrt{12}+\sqrt{11},$$

so $m+n=12+11=23$.

Thus, the answer is D .

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Written solution:

Use coordinates with $A=(0,1)$, $B=(0,0)$, $C=(1,0)$, and $D=(1,1)$. Since $C'D=\frac13$, we have $C'=(\frac23,1)$, so $AC'=\frac23$.

The fold reflects $C$ to $C'$, so the image of side $BC$ is the line through $C'$ and $E$. Reflecting $B=(0,0)$ across the perpendicular bisector of $CC'$ gives $(-\frac{2}{15},\frac25)$. The line through this point and $C'$ meets $AB$ at $E=(0,\frac12)$.

Thus $AE=\frac12$, and

$$EC'=\sqrt{\left(\frac23\right)^2+\left(\frac12\right)^2}=\frac56.$$

The perimeter of $\triangle AEC'$ is

$$\frac12+\frac23+\frac56=2.$$

Thus, the answer is A .

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Written solution:

Fix Ang's placement and label each box by the color Ang put in it. Ben and Jasmin each choose a permutation of the five colors, so there are $(5!)^2$ equally likely pairs of placements.

For a specified set of $k$ boxes to receive three blocks of the same color, both Ben and Jasmin must match Ang in those $k$ boxes. This can happen in $((5-k)!)^2$ ways. By inclusion-exclusion, the number of successful placement pairs is

$$\begin{aligned} &\binom51(4!)^2-\binom52(3!)^2 \\ &\quad {}+\binom53(2!)^2-\binom54(1!)^2 \\ &\quad {}+\binom55(0!)^2. \end{aligned}$$

This equals

$$2880-360+40-5+1=2556.$$

Therefore the probability is

$$\frac{2556}{(5!)^2}=\frac{2556}{14400}=\frac{71}{400}.$$

Thus $m+n=71+400=471$.

Thus, the answer is D .

Solution:

The coin has radius $\frac12$, so its center is uniformly distributed over a $7\times7$ square of area $49$.

A shaded corner triangle contributes the set of center positions within distance $\frac12$ of that triangle, inside the allowed center square. For each corner this is a right isosceles triangle whose altitude is $\frac{1+\sqrt2}{2}$, so its area is

$$\left(\frac{1+\sqrt2}{2}\right)^2=\frac{3+2\sqrt2}{4}.$$

All four corners contribute $3+2\sqrt2$.

The center shaded diamond is a square of side $2\sqrt2$. Expanding it by distance $\frac12$ adds four rectangles of total area $4\sqrt2$ and four quarter-circles of total area $\frac\pi4$, in addition to the diamond's area $8$. Thus the center contribution is

$$8+4\sqrt2+\frac\pi4.$$

The favorable area is

$$\begin{aligned} &3+2\sqrt2+8+4\sqrt2+\frac\pi4 \\ &=11+6\sqrt2+\frac\pi4. \end{aligned}$$

The probability is

$$\begin{aligned} &\frac{11+6\sqrt2+\frac\pi4}{49} \\ &=\frac{44+24\sqrt2+\pi}{196}. \end{aligned}$$

So $a+b=44+24=68$.

Thus, the answer is C .

Solution:

For a single wall of length $n$, compute its Sprague-Grundy value from the possible moves. For the wall lengths needed here, the values are

$$\begin{aligned} &g(1)=1,\quad g(2)=2, \\ &g(3)=3,\quad g(4)=1, \\ &g(5)=4,\quad g(6)=3. \end{aligned}$$

For several walls, the position is losing for the player to move exactly when the xor of the wall values is $0$. Evaluating the choices gives

$$(6,1,1): 3\oplus1\oplus1=3,$$

$$(6,2,1): 3\oplus2\oplus1=0,$$

$$(6,2,2): 3\oplus2\oplus2=3,$$

$$(6,3,1): 3\oplus3\oplus1=1,$$

$$(6,3,2): 3\oplus3\oplus2=2.$$

Only $(6,2,1)$ is losing for the player to move, so Beth has a guaranteed win exactly for that starting configuration.

Thus, the answer is B .

Solution:

For a fixed slope $m$, the number of points in $S$ on or below $y=mx$ is

$$\sum_{x=1}^{30}\lfloor mx\rfloor,$$

for the slopes near the answer.

At $m=\frac23$, grouping $x=3k+1,3k+2,3k+3$ for $k=0,1,\ldots,9$ gives

$$\lfloor 2x/3\rfloor=2k,\ 2k+1, \ 2k+2,$$

whose sum over each block is $6k+3$. Thus the total is

$$\sum_{k=0}^9(6k+3)=270+30=300.$$

If $m<\frac23$, the ten points with ratios $y/x=2/3$ are no longer counted, so the count is less than $300$. Therefore the lower end is $\frac23$.

The next possible ratio $y/x$ greater than $\frac23$, with $1\le x,y\le30$, is minimized by checking $x$ modulo $3$. The best candidates are

$$\frac{19}{28},\qquad \frac{20}{29},\qquad \frac{21}{30}=\frac{7}{10},$$

and the smallest is $\frac{19}{28}$. Hence the interval length is

$$\frac{19}{28}-\frac23=\frac1{84}.$$

Thus $a+b=1+84=85$.

Thus, the answer is E .