Solution:

The expression is equivalent to $$2a^{-2}+\frac{a^{-2}}{2} = 2.5 (a^{-1})^2.$$ Then $a^{-1}$ is equal to $\dfrac{1}{\frac 12} = 2,$ so our expression is equal to $2.5\cdot 2^2 = 10.$

Thus, the correct answer is D .

Solution:

For any nonzero $a,b,$ we have $$\dfrac{a\heartsuit b}{b\heartsuit a} = \dfrac{a^3b^2}{b^3 a^2} = \dfrac ab .$$ Using $a=2,b=4,$ we get that $\frac{ 2}{4} = \frac 12.$

Thus, the correct answer is B .

Solution:

Observe that: $$\begin{align*} &\Bigg\vert\Big\vert |x|-x\Big\vert-|x|\Bigg\vert-x\\ &=\Bigg\vert \Big\vert -x-x\Big\vert-|x|\Bigg\vert-x \\ &=\Bigg\vert\Big\vert -2x\Big\vert-|x|\Bigg\vert-x \end{align*}$$ since $x < 0$ implies that $-x = |x|.$ Substituting values, we can see that $$|4032-2016|-(-2016)$$$$= 2016 +2016$$$$= 4032.$$

Thus, the correct answer is D .

Solution:

The number of days it takes to read $15$ books is $\dfrac{ 15\cdot 16}{2} =120.$ Therefore, it is $120-1=119$ days after the first book is read. This is a multiple of $7,$ so the $15$th book was finished on the same day as the $1$st book. Therefore, it was finished on a Monday.

Thus, the correct answer is B .

Solution:

The total of the four cousin ages is $4\cdot8=32$. Since the median is $5$, the two middle ages have average $5$, so their sum is $10$.

Therefore the youngest and oldest ages sum to $32-10=22$.

Thus, the correct answer is D.

Solution:

The smallest possible sum of two three-digit numbers with six distinct digits is at least $104+235=339$. Any three-digit number at least $339$ with digit sum less than $4$ would have to be one of $100,101,110,200,201,210,300,301,310$, all of which are less than $339$. So the digit sum of $S$ is at least $4$.

The example $157+243=400$ uses six distinct digits and has digit sum $4$. Thus the smallest possible digit sum is $4$.

Thus, the correct answer is B.

Solution:

Let the smaller angle be $x.$ Then, the larger angle is $\frac 54 x.$ Note that their sum is $\frac 94 x.$

Their complements are then $90-x$ and $90 - \frac 54x.$ Since $90-x$ is larger than $90-\frac 54 x,$ we know $$\begin{align*}90 -x &= 2\left(90 - \frac 54 x\right) \\90-x&= 180-\frac 52 x\\x&=60.\end{align*}$$

Therefore, the sum is $$\dfrac 94 \cdot 60=135.$$

Thus, the correct answer is C .

Solution:

To find the tens digit, we first need to find the number modulo $100.$ First, we can find $2015^{2016} \mod 100.$ We can do this with the Chinese Remainder Theorem by first getting the number $\mod 4$ and then $\mod 25.$

The number $$2015^{2016} \equiv 0 \mod 25$$ since it is a multiple of $25.$

Then, observe that $$2015^{2016} \equiv (-1)^{2016}$$$$\equiv 1 \mod 4 .$$

By the Chinese remainder theorem, we can get that our number is congruent to $25 \mod 100.$

Therefore, $$2015^{2016}-2017$$$$\equiv 25 - 17 \mod 100$$$$\equiv 8 \mod 100 .$$ This means the tens digit is $0.$

Thus, the correct answer is A .

Solution:

Let the points be $$A=(0,0),$$$$B=(x_1,y_1),$$$$C=(x_2,y_2).$$

Then, since $BC$ is parallel with the $x$-axis, we know $y_1=y_2,$ which we will let be $y.$ Then, $x_1^2 = y=x_2^2,$ so $x_1^2=x_2^2 .$ This implies that either $x_1=-x_2$ or $x_1 = x_2.$ The second option cannot happen since that would set two points as the same, which would create an area of $0.$ As such, let $x= -x_1=x_2.$ Then $y=x^2$ also.

Then, the points are $$A=(0,0),$$$$B=(-x,x^2),$$$$C=(x,x^2).$$ With a base of $BC,$ the length is $2x$ and the height is $x^2.$ This would make the area $x^3 = 64.$ Therefore, $x=4,$ so $BC = 2\cdot 4 = 8.$

Thus, the correct answer is C .

Solution:

The surface area is increased by a factor of $\frac 53 ^2$ since the side lengths are increased by a factor of $\frac 53.$ Also since the thickness is constant, the volume is scaled up by that much. Then, the wood having the same density makes the weight increase by a factor of $\frac 53^2$ as well, so the new volume is $$12 \cdot \dfrac {25}{9} = \dfrac{100}3.$$ This is approximately $33.3.$

Thus, the correct answer is D .

Solution:

Let $l$ be the number of posts on a long side, including corners, and let $s$ be the number of posts on a short side, including corners. The total number of posts is $$2l+2s-4=20,$$ where the $-4$ avoids double-counting the corners.

Since a long side has twice as many posts as a short side, $l=2s$. Thus $2(2s)+2s-4=20$, so $s=4$ and $l=8$.

There are $3$ equal gaps on a short side and $7$ equal gaps on a long side. With $4$ yards between neighboring posts, the side lengths are $12$ and $28$, so the area is $12\cdot28=336$.

Thus, the correct answer is B.

Solution:

The product is odd if and only if both numbers are odd. There are $\binom 32=3$ ways to do this out of a possible $\binom 52=10$ ways. This makes the probability of it being odd equal to $\frac 3{10} = 0.3.$ This means the probability it is even is $1-0.3=0.7.$

Thus, the correct answer is D .

Solution:

Let $w,r,q$ be the numbers of sets of twins, triplets, and quadruplets. Then $$2w+3r+4q=1000.$$ The statement gives $r=4q$ and $w=3r=12q$.

Substituting gives $2(12q)+3(4q)+4q=40q=1000$, so $q=25$. The number of babies in sets of quadruplets is $4q=100$.

Thus, the correct answer is D.

Solution:

A square must lie above $y=-0.1$, to the left of $x=5.1$, and below $y=\pi x$. Since $\pi$ is a little more than $3$, the lattice heights available at $x=1,2,3,4$ are $3,6,9,12$, and side lengths larger than $3$ cannot fit.

Count by side length using the top-left lattice point. For side length $1$, there are $3+6+9+12=30$ choices. For side length $2$, there are $2+5+8=15$ choices. For side length $3$, there are $1+4=5$ choices.

The total is $30+15+5=50$.

Thus, the correct answer is D.

Solution:

We firstly claim that everything that is either in the center or a corner is odd. This is because every number is next to a consecutive number and therefore has the opposite parity as it.

Therefore, all of the points that are the center or a corner are the same parity. The are $5$ such points, but only $4$ even numbers, so their parity isn't even. Thus, they are all odd. This means the sum of the corners plus the center is $$1+3+5+7+9=25.$$

Since the corners have a sum of $18,$ the center has a value of $25-18=7.$

Thus, the correct answer is C .

Solution:

Let the first value of the series be $a,$ and let the ratio be $r.$ Thus, $$\begin{align*}S&=\dfrac{a}{1-r}\\ &= \dfrac{ar}{r(1-r)} \\&= \dfrac{1}{r(1-r)}.\end{align*}$$ This means we have to find $r$ that maximizes $$r(1-r)= 0.25-(r-0.5)^2.$$ This maximization will happen with $r=0.5.$

Therefore, $S = \dfrac{1}{0.5(0.5)}= 4.$

Thus, the correct answer is E .

Solution:

Pair opposite faces as $(a_1,a_2)$, $(b_1,b_2)$, and $(c_1,c_2)$. Each vertex product uses one number from each pair, so the sum of all eight vertex products is $$(a_1+a_2)(b_1+b_2)(c_1+c_2).$$

The six face labels sum to $2+3+4+5+6+7=27$, so the three opposite-pair sums have total $27$. Their product is maximized when the sums are as equal as possible, namely $9,9,9$, giving at most $9^3=729$.

This maximum is attainable by pairing $2$ with $7$, $3$ with $6$, and $4$ with $5$. Hence the greatest possible sum is $729$.

Thus, the correct answer is D.

Solution:

Suppose the sequence has length $s$ and first term $x$. Then $$345=s\left(x+\frac{s-1}{2}\right),$$ or $$s(2x+s-1)=690.$$

Thus $s$ must be a divisor of $690$, with $x=\frac{690/s-s+1}{2}$ a positive integer. Checking the possible divisor lengths gives $$s=2,3,5,6,10,15,23.$$ These are the only lengths that keep $x$ positive and integral.

Therefore there are $7$ representations.

Thus, the correct answer is E.

Solution:

Observe that the value $$AE = AB-EB$$$$= 5-1$$$$=4.$$ Also, the value $$FC = DC-DF$$$$= 5-2$$$$=3.$$ Then, since $AE \mid \mid FC,$ we know $\triangle AEP \sim \triangle CFP$ by angle angle symmetry. Thus, $$\dfrac{PF}{PE} = \dfrac{FC}{EA} = \dfrac 34 .$$ This makes $$\dfrac{PF}{EF} = \dfrac{3}{3+4} = \dfrac 37.$$

Then, let $X$ be the extension of $AG$ to meet $CD.$ Since two sides are parallel, we have $$\dfrac{AD}{DX} = \dfrac{BG}{AB}.$$ Thus, $$\dfrac{4}{DX} = \dfrac{3}{5}$$ $$DX = \dfrac{20}3.$$ This makes $$FX = \dfrac{20}3 - 2 = \dfrac{14}3 .$$ Then, since $AE \mid \mid FX,$ we know $\triangle AEQ \sim \triangle XFQ$ by angle angle symmetry. Therefore, $$\dfrac{QF}{QE} = \dfrac{FX}{EA} = \dfrac {\frac{14}{3}}4 = \dfrac 76.$$ This makes $$\dfrac{QF}{EF} = \dfrac{7}{7+6} = \dfrac 7{13}.$$

As such, $$\dfrac{QP}{EF} = \dfrac{QF}{EF} - \dfrac{PF}{EF}$$$$= \dfrac{7}{13} - \frac 37$$$$= \dfrac{10}{91}.$$

Thus, the correct answer is D .

Solution:

The dilation scale factor is $\frac32$, since the radius changes from $2$ to $3$. The center of dilation $C$ lies on the line through $A(2,2)$ and $A'(5,6)$.

The vector from $A$ to $A'$ is $(3,4)$. Since $CA:CA'=1:\frac32$, the vector from $C$ to $A$ is twice the vector from $A'$ back to $A$, so $$C=(2,2)-2(3,4)=(-4,-6).$$

Under a scale factor $\frac32$ dilation about $C$, a point moves by half its distance from $C$. Since $CO=\sqrt{(-4)^2+(-6)^2}=\sqrt{52}$, the origin moves $\frac12\sqrt{52}=\sqrt{13}$.

Thus, the correct answer is C.

Solution:

The equation is symmetric in all four quadrants. In the first quadrant it becomes $$x^2+y^2=x+y,$$ or $$(x-\tfrac12)^2+(y-\tfrac12)^2=\tfrac12.$$

In the first quadrant, the enclosed region is the triangle under $x+y=1$, with area $\frac12$, plus a semicircle of radius $\sqrt{1/2}$, with area $\frac\pi4$.

Multiplying by $4$, the total area is $$4\left(\frac12+\frac\pi4\right)=2+\pi.$$

Thus, the correct answer is B.

Solution:

The total number of teams is $10+10+1=21.$ The total number of sets is therefore $\binom{21}{3} = 1330.$

Now, we must subtract the total number of sets such that there is no cycle. This only happens if one team beats the other two teams. There are $21$ choices for the team that beat the other two and $\binom{10}{2} = 45$ ways to choose the teams they beat. Thus, the total of non-cycles is $21\cdot 45=945.$ This means the total number of cycles is $1330-945=385.$

Thus, the correct answer is A .

Solution:

Consider the following diagram:

The shape is symmetric, so we can find the ratio of the areas of $EFCD$ to the area of $ZFCW.$ For this, we can extend $FE$ and $CD$ until they meet each other, and let this point be $P.$

Also, the distance from $ED$ to $ZW$ is the same as the distance from $ZW$ to $YX,$ and the distance from $ZW$ to $YX$ is twice the distance as the distance from $ZW$ to $FC.$

Thus, the distance from $ED$ to $ZW$ is twice the distance as the distance from $ZW$ to $FC.$ Let the altitude from $P$ to $ED$ be $s.$ Then, if the altitude from $P$ to $ZW$ is $s+d,$ the altitude from $P$ to $CF$ is $s+1.5d$ because of the ratio. Then, since $$\triangle FPC \sim \triangle PED,$$$$\dfrac{FP}{EP} = \dfrac{s+1.5d}{s}.$$ Also, since $FE= ED = PE$ as $PED$ is equilateral, we have $2s = s+1.5d,$ and so $s=1.5d.$

Thus, the ratio between side lengths of $\triangle PED$ and $\triangle PZW$ is $\frac 53$ which makes the ratio between the area $\frac{25}9.$ Also the ratio between side lengths of $\triangle PED$ and $\triangle PFC$ is $2,$ which makes the ratio between the area $4.$

This makes the area of $EDCF$ $3$ times the area of $\triangle PED$ and the area of $EDWZ$ $\frac{16}{9}$ times the area of $\triangle PED$ Therefore, the area of $ZWCF$ is $\frac{11}{9}$ times the area of $\triangle PED.$

Thus, the ratio between $ZWCF$ and $EDCF$ is $$\dfrac{\frac{11}9}{3} = \dfrac{11}{27} .$$

Thus, the correct answer is C .

Solution:

We know $a \leq b \leq c$ by analyzing $ab < bc < cd.$ Also, $bc$ is the average of $ab$ and $cd$ so $$\dfrac {10a+b+10c+d}2 = 10b+c.$$ This means that $$10(c-2b+a)=-b+2c-d,$$ making the right hand side a multiple of $10.$ Thus, it must be $0$ or $10$ since it is digits that satisfy $a \leq b \leq c .$ Thus, we can case on that value.

Case 1: $$-b+2c-d=10$$$$2b-a-c=1$$

We can look at the possible values of $c.$

$c=6:$ $b+d=2, 2b-5=a.$ Thus, $b\leq 2$ from the first equation, but can't work for the second equation.

$c=7:$ $b+d=4, 2b-6=a.$ Thus, $b\leq 4$ from the first equation, and $b > 3$ from the second equation. This makes one case for $b=4.$

$c=8:$ $b+d=6, 2b-7=a.$ Thus, $b\leq 6$ from the first equation, and $b > 3$ from the second equation. This makes three cases for $b=4,5,6.$

$c=9:$ $b+d=8, 2b-8=a.$ Thus, $b\leq 8$ from the first equation, and $b > 4$ from the second equation. This makes four cases for $b=5,6,7,8.$ This case has $8$ solutions.

Case 2: $$-b+2c-d=0,$$$$2b-a-c=0,$$ which means the digits are an arithmetic sequence.

If the difference is $1,$ then $1 \leq a \leq 6$ makes $6$ solutions.

If the difference is $2,$ then $1 \leq a \leq 3$ makes $3$ solutions. This case makes $9$ solutions.

In total, the number of solutions is $8+9 = 17.$

Thus, the correct answer is D .

Solution:

Write $x=\lfloor x\rfloor+t$, where $0\le t\lt1$. Then $$\lfloor kx\rfloor-k\lfloor x\rfloor=\lfloor kt\rfloor,$$ so $f(x)$ depends only on the fractional part $t$.

The value of $f$ changes only when $t$ crosses a fraction $i/k$, where $2\le k\le10$ and $1\le i\lt k$. The number of distinct such fractions in $(0,1)$ is $$\varphi(2)+\varphi(3)+\cdots+\varphi(10)=1+2+2+4+2+6+4+6+4=31.$$

Including the initial value before the first breakpoint, $f$ assumes $31+1=32$ distinct values.

Thus, the correct answer is A.