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.

Video solution:

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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 shirts, 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 is working 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 whos 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:

Consider the following diagram where the lighter colored area makes up region S:

The circles can be in only two locations. We first place the largest circle and then the second largest circle in the opposite location. After this, the circle of radius $3$ must be placed on one of the two sides. To minimize the area lost with the last circle, we put it inside the third circle.

This yields an area of $7^2 \pi + 5^2 \pi -3^2\pi = 65\pi.$

Thus, the answer is D.

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Written solution:

The number on the top left corner is $211,$ so the number under it would be $210.$ This is the greatest number on the second row from the top since $211$ to $225$ are the largest numbers in the square and they are in the top row. now, we must find the least number in the row. Besides the left most number, the squares are increasing by $1$ from left to right.

Therefore, the second square is the least square. Note that the diagonal from the center to the upper right corner has all the odd squares, so the $14$th number in the row is $169.$ Thus, the second number in the row is $169-12 = 157,$ so that is the least number in the row. The sum is $157+210=367.$

Thus, the answer is A.

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Written solution:

To undo these steps, we first reflect $(-6,3)$ back across $y=-x.$ This would be $(-3,6).$

Next, we need to undo the rotation, so we rotate $(-3,6)$ around $(1,5)$ by $90^\circ$ clockwise.

We can do this by translating our point , rotating around the origin, and translate back. The first translation would be $(-4,1).$ Then rotating would be $(1,4).$ Then, translating back would be $2,9.$ This would make $b=9,a=2,$ so $b-a = 7.$

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 area 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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Written solution:

The number of interior pieces is $(l-2)(w-2).$ This would be half of the total number of pieces, which is $lw.$ This means $\frac{lw}2 = (l-2)(w-2).$

Multiplying out yields $lw -4l-4w+8=0,$ so $(l-4)(w-4)=8.$ This yields that $(l,w) = (5,12),$ $(6,8),$ or some permutation of it. Thus, $lw = 48$ or $60.$

Therefore, the largest number of brownies 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 of the sum of the odd divisors.

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:

Consider the following diagram:

Let the radius be $r.$ Then, we can make the distance between two chords be $2d.$ The two chords of length $38$ are equidistant from the center, so the distance from the center to the other chord is $3d.$ Thus, $r^2 +19^2 = (3r)^2 +17^2 \implies$ $19^2-17^2 = 8d^2,$ so $8d^2 = 72.$ Therefore, $2d=6.$

Thus, the answer is B.

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Written solution:

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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Written solution:

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:

First, we can find the length of $AC$ and $AD.$ Since the altitude of an equilateral triangle of side length $2$ is $\sqrt 3,$ we know $$AC =AD = 2\sqrt 3 = \sqrt{12} .$$ Note that the area of an equilateral triangle with side length $2$ has area $\sqrt 3.$ This means that $ABC$ and $AED$ combined have an area of $2\sqrt 3 = \sqrt{12}$ since they are $4$ halves of an equilateral triangle.

The rest of the area is $ACD.$ Since $AC=AD = \sqrt 2$ and $CD=2,$ the altitude from $A$ to $CD$ is $$\sqrt{(\sqrt{12})^2 - \left(\dfrac 22\right)^2} = \sqrt{11}$$ as it is an equilateral triangle. Since the height is $\sqrt{11}$ and the base is $2,$ the area is $$\frac{2 \cdot \sqrt{11}}2 = \sqrt{11}.$$ Thus, the total area is $\sqrt{12} + \sqrt{11}.$ Therefore, $m+n = 23.$

Thus, the answer is D.

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Written solution:

Consider the following diagram, which is identical to the one in the problem, only with all the points labelled.

Since $\angle FC'E = 90^\circ,$ we know $$\angle FC'D = 90^\circ - EC'A.$$

Since $\angle C'AE = 90^\circ,$ we know $$\angle C'EA = 90^\circ - EC'A = \angle FC'D.$$ Therefore, by angle angle similarity, we know $AC'E \sim DFC'.$ Therefore, the perimeter of $AEC'$ is equal to the area of $DFC'$ times $\dfrac{AC'}{DF}.$

Since $C'F +FD = 1,$ the perimeter of $DFC'$ is $$\dfrac 13 + 1 = \dfrac 43.$$ Also, $$AC' = 1- DC'$$$$= 1- \dfrac 13$$$$= \dfrac 23.$$ There, the perimeter of $AEC'$ can be simplified to $$\dfrac 43 \cdot \dfrac{\frac 23}{DF} = \dfrac 8{9\cdot DF}.$$ By the Pythagorean theorem on $DFC',$ we know $$(C'D)^2 + (DF)^2 = (C'F)^2$$ $$= (1-DF)^2.$$

This means $$\dfrac 13^2 + DF^2 = 1- 2DF +DF^2$$ so $$1- 2DF=\dfrac 19.$$ Therefore, $DF = \dfrac 49.$

This means the area is $$\dfrac{8}{9\cdot \frac 49} = \dfrac 84$$$$= 2.$$

Thus, the answer is A.

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Written solution:

First, by WLOG, we can index the boxes by how Ang put his blocks in the boxes. Then, we can count the total number of ways to place all of Ben and Jasmin's blocks. This has a total of $(5!)^2$ ways it can be done.

Now, we find the total number of configurations that have at least one box that have the same color. We can do this with the principle of inclusion exclusion.

First, we get that there are $\binom{5}{1} \cdot (4!)^2$ distributions as there are $\binom{5}{1}$ ways to choose the box that has $3$ of the same color, and there are $4!$ ways to choose the ordering for the other two boxes. However, this overcounts the number of configurations with 2 boxes that have 3 of the same block.

That would be counted as $\binom{5}{2} \cdot (3!)^2$ for similar reasons to above.

This also over counts the number of configurations with 3 boxes that are all the same color, so we add those back. We keep adding and removing configurations that are over and undercounted.

This leads to our number of configurations being $$\dfrac{1}{{(5!)^2}}\left(\binom{5}{1} \cdot (4!)^2+\binom{5}{2} \cdot (3!)^2\right.$$ $$+\binom{5}{3} \cdot (2!)^2+\binom{5}{4} \cdot (1!)^2$$ $$\left. +\binom{5}{5} \cdot (0!)^2\right),$$ which is equal to $$\frac{2256}{14400} = \frac{639}{3600}$$$$= \frac{71}{400} .$$ This makes $m+n=471.$

Thus, the answer is D.

Solution:

The radius of the circle is $\frac 12.$ This makes the total possible region for the circle's center to be in a square of length $8 - \frac 12 - \frac 12 = 7.$ Shown below is the region in which the circle can be.

Now, the circle's center must be inside the black region or within $\frac 12$ of the black region.

For each of the corner, we can take a right triangle for this region. Now, we can find the altitude to the hypotenuse. This would be useful to find the as the area of a right triangle is equal to the altitude times the hypotenuse divided by $2,$ and the hypotenuse is twice the altitude, making the area equal to the altitude squared.

The altitude of this region is equal to the altitude of the portion of the right triangle in the black region and the blue region plus $\frac 12.$ When finding the portion of the black right triangle in the blue region, we get a right triangle with length $2-\frac 12 - \frac 12 = 1.$ This has an altitude of $\frac{\sqrt 2}2 .$ The total altitude therefore is $\frac{\sqrt 2 + 1}2.$ Thus, the answer is $$(\dfrac{\sqrt 2 + 1}2)^2 = \dfrac{3+2\sqrt 2}4.$$ There are $4$ corners, so the combined area of the right triangles is $3+ 2\sqrt 2.$

For the center square, we find the area within it, the area that is within $\frac 12$ of the edge, and within $\frac 12$ of the corners. The area in the square is $(2\sqrt 2)^2 = 8.$ The area within the $\frac 12$ of the edges is $\frac 12 \cdot 2\sqrt 2 = \sqrt 2$ since its a rectangle of with lengths $\frac 12$ and $2\sqrt 2.$ The total area from this is $4\sqrt 2.$ The area within $\frac 12$ of the points are 4 quarter circles of radius $\frac 12,$ so their combined area is $\frac \pi 4 .$ The total area is $$3+ 2\sqrt 2 + 9 + 4\sqrt 2 + \frac \pi 4$$ $$=8 + 6\sqrt 2 + \frac \pi 4.$$

The probability is $$\dfrac{11 + 6\sqrt 2 + \frac \pi 4} {49}$$ $$=\dfrac{44 + 24\sqrt 2 + \pi } {196}.$$ This makes the answer $44+24=68.$

Thus, the answer is C.

Solution:

If Arjun can force the walls to have symmetry after his turn, in whichevery wall has matching wall of the same size, then he can always copy Beth's move on the matching wall. I claim that Arjun can do this with the setups from choices A,C,D, and E.

With A, Arjun can split it into $(2,2,1,1)$ which is symmetric.

With C, Arjun can split it into $(2,2,2,2)$ which is symmetric.

With D, Arjun can split it into $(3,2,3,2)$ which is symmetric.

With E, Arjun can split it into $(3,2,3,2)$ which is symmetric.

This leaves just B as a possible answer. Now we must prove it works. With $(6,2,1),$ Arjun can make $$(6,2),(6,1),(6,1,1),(5,2,1),$$ $$(4,2,1),(4,2,1,1),(3,2,1,1),$$ $$(2,2,2,1),(3,1,2,1).$$ If given the last seven configurations by Arjun, Beth can make it symmetric by turning it into $(2,2,1,1),$ which is a winning situtation for her since it is symmetric. Thus, we must now verify that $(6,1)$ and $(6,2)$ are winning for her. Both of these lead to $(3,2,1),$ so it suffices to prove that $(3,2,1)$ is winning for Beth when given to Arjun.

With $(3,2,1),$ Arjun can make $$(3,2),(3,1),(3,1,1),$$$$(2,2,1),(1,2,1).$$ Each can be made symmetric by Beth in the following ways:

$(3,2) \implies (2,2)$

$(3,1) \implies (1,1)$

$(3,1,1) \implies (1,1,1,1)$

$(2,2,1) \implies (2,2)$

$(1,2,1) \implies (1,1)$

This makes Beth have a winning strategy with $(6,2,1).$

Thus, the answer is B.

Solution:

First, lets create a way to find the number of lattice points under $(31,y),$ if $y$ is relatively prime with $31.$ With $(0,0)$ and $(31,y),$ we can make a rectangle with $32(y+1)$ points. There are $32(y+1)-2$ points not on the diagonal, and half of them are under the line. Also, there are $y$ points under the line that aren't in $S$ since their x-coordinate is $31$ and $30$ other points that aren't in $S$ since their y-coordinate is $0.$ Therefore, if the line goes through $(31,y),$ then the total number of points is $$\dfrac{32(y+1)-2}2 - y - 30$$$$= 15(y-1).$$ If $y=21,$ then there are $300$ points in $S,$ making $m = \frac{21}{31}$ in the interval.

Also, notice that the number of lattice points is equal to $$\sum_{i=1}^{30} \lfloor mx \rfloor.$$ As $m$ increases, each term stays constant or increases. Since $m = \frac{21}{31}$ is in the interval, the lower bound can be found by finding the largest $m \leq \frac{21}{31}$ such that at least one term increases as $m$ increases.

If a number was $\frac{a}{b}$ and $b \leq 30, \frac ab > 23,$ then $$\dfrac ab = \dfrac{2x+1}{3x+1}$$ if $b = 3x+1,$ $$\dfrac ab = \dfrac{2x+2}{3x+2}$$ if $b = 3x+2,$ or $$\dfrac ab = \dfrac{2x+3}{3x+3}$$ if $b = 3x+3.$ These would all be greater than $\frac {21}{31} .$

Therefore, $\frac 23$ must be the lower bound. Also, we now know that the fractions from above are the possible upper bounds. Furthermore, a greater $x$ would create a smaller interval, so we need to just look at the interval being $$\dfrac{19}{28}, \dfrac{20}{29}, \dfrac{21}{30} .$$ The least of these is $\frac{19}{28},$ so the interval is of length $$\dfrac {19}{28} - \dfrac 23 = \dfrac 1{84} .$$ This makes our answer $85.$

Thus, the correct answer is E.