Solution:

We can factor a $9!$ out the numerator and simplify. $$\begin{align*} \dfrac{9!(11 \cdot 10 - 10)}{9!} &=11 \cdot 10 - 10 \\&= 110 - 10 \\&= 100. \end{align*}$$ Thus, the correct answer is B .

Solution:

Rewrite all bases as powers of $10$: $$10^x\cdot100^{2x}=10^x\cdot(10^2)^{2x}=10^{5x},$$ and $$1000^5=(10^3)^5=10^{15}.$$

Therefore $5x=15$, so $x=3$.

Thus, the correct answer is C.

Solution:

For each dollar Ben spent, David spent $75$ cents, so the difference was $25$ cents. The total difference $\$12.50$ is therefore one quarter of Ben's spending.

Ben spent $12.50\div0.25=50$ dollars, and David spent $37.50$ dollars. Together they spent $87.50$ dollars.

Thus, the correct answer is C.

Solution:

Using the formula, we get $$\begin{align*} \text{rem} \left(\dfrac{3}{8}, -\dfrac{2}{5}\right) &= \dfrac{3}{8} + \dfrac{2}{5} \left \lfloor \dfrac{\frac{3}{8}}{-\frac{2}{5}}\right \rfloor \\ &= \dfrac{3}{8} + \dfrac{2}{5} \left \lfloor -\dfrac{15}{16} \right \rfloor \\ &= \dfrac{3}{8} + \dfrac{2}{5} \cdot -1 \\ &= \dfrac{3}{8} - \dfrac{2}{5} \\ &= - \dfrac{1}{40} \end{align*}$$ Thus, the correct answer is B .

Solution:

Let $s$ be the side length of the smallest side. Then the other two sides are $3s$ and $4s.$

The volume is therefore $$s \cdot 3s \cdot 4s = 12s^3.$$ Testing out values of $s,$ we see that if $s = 2,$ then $12s^3 = 96,$ which is an answer choice.

Thus, the correct answer is D .

Solution:

When Emilio changes a units digit $2$ to $1$, his copied number is smaller by $1$. This happens in $2,12,22$, for a loss of $3$.

When he changes a tens digit $2$ to $1$, his copied number is smaller by $10$. This happens for $20,21,\ldots,29$, for a loss of $100$.

Thus Emilio's sum is $103$ less than Ximena's sum, so Ximena's sum is $103$ larger.

Thus, the correct answer is D.

Solution:

The sum of the seven data values is $540+x$. If the mean equals $x$, then $$\frac{540+x}{7}=x,$$ so $540=6x$ and $x=90$.

With $x=90$, the ordered data set is $40,50,60,90,90,100,200$. Its median is $90$, and its mode is also $90$, so the value works.

Thus, the correct answer is D.

Solution:

We know that Fox has $0$ coins at the end. Then before paying the final toll, Fox had $40$ coins.

Then he had $40 \div 2 = 20$ coins before the doubling. Then before paying the toll for the second crossing, he had $20 + 40 = 60$ coins.

Before the doubling on the second crossing, he had $60 \div 2 = 30$ coins. On the first crossing before the toll, Fox had $30 + 40 = 70$ coins.

Finally, before the first doubling, Fox had $70 \div 2 = 35$ coins.

Thus, the correct answer is C .

Solution:

Recall that the sum of the first $N$ number is $\dfrac{N(N + 1)}{2}.$

We want to find $N$ such that $$\dfrac{N(N + 1)}{2} = 2016.$$ Cross-multiplying and simplifying gives us $$N^2 + N - 4032 = 0.$$ Factoring gives us $$(N - 63)(N + 64) = 0$$ We want the positive value so $N = 63.$ Adding together the digits gives us $9.$

Thus, the correct answer is D .

Solution:

Let $l$ be the length of the inner rectangle. Then the area of the inner rectangle is $l.$

The area of the middle region is going to be $$(l + 2) \cdot 3 - l = 2l + 6.$$ The area of the outer region is $$(l + 4) \cdot 5 - (l + 2) \cdot 3 = 2l + 14.$$

We know that these $3$ values form an arithmetic sequence. That means that $$\begin{align*} l + 2l + 14 &= 2(2l + 6) \\ 3l + 14 &= 4l + 12 \\ l &= 2\end{align*}$$

Thus, the correct answer is B .

Solution:

We can split the region into $4$ triangles with bases of $1.$

Two of the triangles have bases $8 \div 2 = 4$ and the other two have bases $5 \div 2 = \dfrac{5}{2}$

The sum of the areas of the triangles is $$2 \cdot \dfrac{1}{2} \left(1 \cdot \dfrac{5}{2}\right) + 2 \cdot \dfrac{1}{2} \left(1 \cdot 4\right) = 6\dfrac{1}{2}$$

Thus, the correct answer is D .

Solution:

The product is odd exactly when all three selected integers are odd. There are $1008$ odd and $1008$ even integers from $1$ to $2016$.

Because the integers are selected without replacement, $$p=\frac{1008}{2016}\cdot\frac{1007}{2015}\cdot\frac{1006}{2014}.$$ The first factor is $\frac12$, and each of the next two factors is slightly less than $\frac12$. Therefore $p<\frac18$.

Thus, the correct answer is A.

Solution:

Note that Dee and Edie do not change the answers since their seats remain occupied throughout.

Since Bea moves to the right, this forces Ada and Ceci to move to offset this disruption.

Ceci only moves one seat, so Ada must also move one to cancel out the two seat shift that Bea did.

Bea moved to the right and Ceci moved to the left, so Ada must also move to the left to get a total displacement of $0.$

Therefore, Ada must have started off in seat $2$ to move one to the left to end up in seat $1.$

Thus, the correct answer is B .

Solution:

The problem can be rewritten as an equation $2x + 3y = 2016,$ where $x$ is the number of twos and $y$ is the number of threes.

The goal is to find the number of multiples of $3$ that can be subtracted from 2016 to result in an even number.

This can be achieved by the pairs of $(1008, 0)$ up to $(0, 672)$ with $y$ being incremented by $2.$

This gives us $$\dfrac{672}{2} + 1 = 337$$ solutions for $y$ and $x.$

Thus, the correct answer is C .

Solution:

The circle of cookie dough has a radius of $3$ inches since it is the same as the diameter plus the radius of a cookie.

The area of the cookie dough is $3^2\pi = 9\pi,$ and the cookies have an area of $7 \cdot 1^2\pi = 7\pi.$

The area of the leftover scrap is therefore $9\pi - 7\pi = 2\pi.$ This means that its radius is $\sqrt{2}.$

Thus, the correct answer is A .

Solution:

To figure out how to reverse the transformations, we can analyze a single point and see what happens to it.

Let $(x, y)$ be the point. After being reflected about the $x$-axis, the point would go to $(x, -y).$

Rotating this counterclockwise would put it at $(y, x).$ The only transformation that puts this back at $(x, y)$ is reflection about $y = x.$

Thus, the correct answer is D .

Solution:

Think of first arranging the $N$ green balls, then placing the red ball in one of the $N+1$ gaps. If $k$ green balls are to the left of the red ball, then $N-k$ are to its right.

At least $\frac35N$ green balls are on one side exactly when $k\le\frac25N$ or $k\ge\frac35N$. Thus the bad gaps are $$\frac25N+1,\frac25N+2,\ldots,\frac35N-1,$$ a total of $\frac15N-1$ gaps.

Therefore $$P(N)=1-\frac{N/5-1}{N+1}=\frac{4N+10}{5N+5}.$$ Solving $\frac{4N+10}{5N+5}<\frac{321}{400}$ gives $N>479$. The least positive multiple of $5$ is $480$, whose digit sum is $12$.

Thus, the correct answer is A.

Solution:

Opposite faces together use all eight labels, whose sum is $36$, so every face must have sum $18$.

Put label $1$ at one vertex, and let the three adjacent labels be $a,b,c$. The three vertices adjacent to those across faces are then forced to be $$17-a-b,\quad 17-a-c,\quad 17-b-c,$$ and the opposite vertex is $a+b+c-16$.

Assume the three neighbor labels are listed in increasing order. Checking possible triples from $2,3,\ldots,8$, the valid triples are $(4,6,8)$, $(4,7,8)$, and $(6,7,8)$. Each gives two non-rotationally equivalent arrangements, so there are $3\cdot2=6$ arrangements.

Thus, the correct answer is C.

Solution:

Since $BC=3$ and $BE=EF=FC$, we have $BE=1$ and $BF=2$. Also $AD=3$.

From $\triangle APD\sim\triangle EPB$, $$\frac{BP}{BD}=\frac{BE}{AD+BE}=\frac14.$$ From $\triangle AQD\sim\triangle FQB$, $$\frac{BQ}{BD}=\frac{BF}{AD+BF}=\frac25.$$

Therefore $BP=\frac14BD$, $PQ=\left(\frac25-\frac14\right)BD=\frac3{20}BD$, and $QD=\frac35BD$. Thus $$BP:PQ:QD=\frac14:\frac3{20}:\frac35=5:3:12.$$ The sum is $5+3+12=20$.

Thus, the correct answer is E.

Solution:

A term that includes all four variables has form $a^vb^wc^xd^y$, with $v,w,x,y\ge1$, multiplied by some power of $1$. If that power is $z$, then $$v+w+x+y+z=N.$$

After subtracting $1$ from each of $v,w,x,y$, this becomes a nonnegative stars-and-bars count with $5$ variables summing to $N-4$. The number of such terms is $$\binom{N}{4}.$$

Since $\binom{14}{4}=1001$, we get $N=14$.

Thus, the correct answer is B.

Solution:

Using the Pythagorean theorem, we get that $$P'Q' = \sqrt{3^2 - 1^2} = 2\sqrt{2}$$ and $$Q'R' = \sqrt{5^2 - 1^2} = 2\sqrt{6}.$$

This follows from $PQ = 1 + 2 = 3$ and $QR = 2 + 3 = 5.$ The heights of the triangles are also just $1.$

Then, we get that $$[Q'QPP'] = \dfrac{1}{2}(1 + 2)2\sqrt{2}$$$$= 3\sqrt{2}.$$

We also get that $$[R'RQQ'] = \dfrac{1}{2}(2 + 3)2\sqrt{6}$$$$= 5\sqrt{6}.$$

Finally, we have that $$[R'RPP'] =$$$$\dfrac{1}{2}(1 + 3)(2\sqrt{2} + 2\sqrt{6})$$$$= 4\sqrt{2} + 4\sqrt{6}.$$

Now, we can express $[PQR]$ as $$[Q'QPP'] + [R'RQQ']$$$$- [R'RPP'].$$

This evaluates to $$3\sqrt{2} + 5\sqrt{6} - 4\sqrt{2} - 4\sqrt{6}$$ $$= \sqrt{6} - \sqrt{2}.$$

Thus, the correct answer is D .

Solution:

The prime factorization is $110=2\cdot5\cdot11$. If $110n^3$ has $110=2\cdot5\cdot11$ divisors, then it has exactly three prime factors, with exponents $1,4,10$ in some order.

For each of the primes $2,5,11$, its exponent in $110n^3$ is $1$ more than a multiple of $3$. The exponents $1,4,10$ all have this form, so the corresponding exponents in $n$ are $0,1,3$ in some order.

In $81n^4=3^4n^4$, the exponents from $n^4$ are therefore $0,4,12$, in some order, along with the exponent $4$ on prime $3$. Hence the divisor count is $$(0+1)(4+1)(12+1)(4+1)=325.$$

Thus, the correct answer is D.

Solution:

Since $a\diamondsuit a=1$, substituting $b=c$ in $$a\diamondsuit(b\diamondsuit c)=(a\diamondsuit b)c$$ gives $a\diamondsuit1=(a\diamondsuit b)b$. Also, using $a\diamondsuit(a\diamondsuit a)=(a\diamondsuit a)a$ gives $a\diamondsuit1=a$. Therefore $a\diamondsuit b=\frac ab$.

The equation becomes $$2016\diamondsuit(6\diamondsuit x)=2016\diamondsuit\frac6x=\frac{2016}{6/x}=336x=100.$$ Thus $x=\frac{25}{84}$, so $p+q=25+84=109$.

Thus, the correct answer is A.

Solution:

Let $O$ be the circle's center, and let $AD$ meet $OB$ and $OC$ at $E$ and $F$. Equal chords $AB,BC,CD$ give equal central angles.

From the equal-angle relationships, $\triangle OAB\sim\triangle ABE$. Since $OA=OB=200\sqrt{2}$ and $AB=200$, the similarity gives $AE=AB=200$ and $BE=100\sqrt{2}$. Similarly, $FD=CD=200$.

In triangle $OBC$, the same scaling gives $EF=\frac12BC=100$. Therefore $$AD=AE+EF+FD=200+100+200=500.$$

Thus, the correct answer is E.

Solution:

Factor the given least common multiples: $$72=2^3\cdot3^2,\quad 600=2^3\cdot3\cdot5^2,\quad 900=2^2\cdot3^2\cdot5^2.$$ Since $\operatorname{lcm}(x,y)$ has no factor of $5$, neither $x$ nor $y$ has a factor of $5$, and $z$ must contain $5^2$.

Write $x=2^a3^b$, $y=2^c3^d$, and $z=2^e3^f5^2$. The power of $3$ in $\operatorname{lcm}(y,z)$ forces $d=2$, and the power of $2$ in $\operatorname{lcm}(x,z)$ forces $a=3$.

The remaining independent conditions are $\max(b,f)=1$ and $\max(c,e)=2$. These have $3$ and $5$ ordered choices, respectively, so there are $3\cdot5=15$ triples.

Thus, the correct answer is A.