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:

We can express the $100$ as $10^2$ and $1000$ as $10^3$ to get $$\begin{align*} 10^x \cdot (10^2)^{2x} &= (10^3)^5 10^x \cdot 10^{4x} \\&= 10^{15}10^{5x} \\&= 10^{15} \end{align*}$$ Since the bases are the same, the exponents must also be the same. Therefore, $$5x = 15$$$$x = 3$$ Thus, the correct answer is C.

Solution:

This means that for every dollar Ben spent, he spent $25$ cents more than David.

This means that Ben spent $12.50 \div 0.25 = 50$ dollars. Therefore, David spent $50 - $12.5 = $37.5

The total amount of money they spent is $50 + $37.5 = $87.5.

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:

Whenever Ximena replaces a units digit of $2$ with a $1,$ her total sum decreases by $1.$

Similarly, whenever a tens digit of $2$ is replaced with a $1,$ her total sum decreases by $10.$

$2$ appears as a unit digit $3$ times ($2, 12,$ and $22$) and it appears in the tens digit $10$ times ($20-29$).

Her total sum, therefore, will be $$3 \cdot 1 + 10 \cdot 10 = 103$$ less than Emilio's sum.

Thus, the correct answer is D.

Solution:

The sum of the elements in this set is $540 + x,$ making the mean $\dfrac17(540 + x).$ Therefore, $$\dfrac17(540 + x) = x$$$$540 + x = 7x$$$$6x = 540$$$$x = 90.$$ 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 only way for the product of the three integers to be odd is if all the numbers themselves are odd.

There are an even number of consecutive integers, which means that there is a $\frac{1}{2}$ chance that a randomly chosen number is odd.

For all $3$ numbers to be add, the probability is $$\left(\dfrac{1}{2}\right)^3 = \dfrac{1}{8}.$$ But note that all integers must be distinct. This lowers the probability since we have added an extra restriction.

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:

For the condition to be satisfied, the red ball cannot be placed in the middle fifth of the green balls.

This means that there are $\frac{N}{5} - 1$ spots where the red ball cannot be placed.

Placing the red ball anywhere else works, which means that $$\begin{align*}P(N) &= 1 - \dfrac{\frac{N}{5} - 1}{N + 1} \\&=\dfrac{4N + 10}{5N + 5} \\&\lt \dfrac{321}{400} \end{align*}$$

Cross-multiplying and simplifying gives us $$2395 \lt 5N$$$$479 \lt N.$$

The smallest value of $N$ is therefore $480.$ The sum of the digits is $4 + 8 = 12.$

Thus, the correct answer is A.

Solution:

We have that the sum of the vertices on each face is $$\dfrac{1 + 2 + \cdots + 8}{2} = 18.$$ This is because two opposite faces use all $8$ vertices, and their vertices have the sum.

Let $a, b,$ and $c$ be the vertices next to $1.$ Then the remaining vertices are $$17 - a - b,$$$$17 - a - c,$$$$17 - b - c,$$ and $$a + b + c - 16.$$

We can do casework on $a, b, c.$ The only restrictions we have are that $a + b + c \gt 17$ and all the vertices are distinct.

WLOG let $a \lt b \lt c:$

$3, 7, 8:$ $17 - 3 - 7 = 7,$ which is not allowed

$4, 6, 8:$ all the vertices work

$4, 7, 8:$ all the vertices work

$5, 6, 7:$ $17 - 5 - 6 = 6,$ which is not allowed

$5, 6, 8:$ $17 - 5 - 6 = 6,$ which is not allowed

$5, 7, 8:$ $17 - 5 - 7 = 5,$ which is not allowed

$6, 7, 8:$ all vertices work

For each triple, the only way we can arrange them to create unique configurations is $(x, y, z)$ and $(z, y, x.)$

These cannot be created with rotations by each other. Therefore, there are $3 \cdot 2 = 6$ arrangements.

Thus, the correct answer is C.

Solution:

Note that $\triangle APD \sim \triangle EPB$ by angle-angle using alternate interior angles.

This gives us $$\dfrac{DP}{PB} = \dfrac{AD}{BE}$$$$PB = \dfrac{1}{4} BD.$$

Similarly, we have that $\triangle AQD \sim \triangle FQB,$ which gives us $$\dfrac{DQ}{BQ} = \dfrac{AD}{BF}$$$$BQ = \dfrac{2}{5} BD.$$

Finally, we get that $$DQ = \left(1 - \dfrac{2}{5}\right) BD = \dfrac{3}{5} BD$$ Then the desired ratio is $$\left(\dfrac{1}{4} : \dfrac{2}{5}\right) - \left(\dfrac{1}{4} : \dfrac{3}{5}\right)$$$$= 5 : 3 : 12$$

The sum of these is $5 + 3 + 12 = 20.$

Thus, E is the correct answer.

Solution:

We want to find all the terms that are of the form $$a^vb^wc^xd^y1^z.$$ Note that $$v, w, x, y \gt 0.$$

These variables must satisfy $$v + w + x + y + z = N.$$

Since $v, w, x,$ and $y$ must be positive, we can define $x' = x - 1$ and similarly for all the other variables.

Then, we get that $$v', w', x', y' \geq 0.$$

Using these new variables, we get the new equation $$v' + w' + x' + y' + z = N - 4.$$ Now we can use stars and bars since all the values are non-negative. There are $$\binom{N - 4 + 4}{4} = \binom{N}{4}$$ solutions to the equation.

We need to find $N$ such that $\binom{N}{4} = 1001.$ Checking all the answer choices yields $14$ as the right answer.

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:

Note that the prime factorization of $110$ is $2 \cdot 5 \cdot 11.$

Recall that if a number is expressed as $$p_1^{e_1}p_2{e_2}\cdots p_n{e_n},$$ then it has $$(e_1 + 1)(e_2 + 1)\cdots(e_n + 1)$$ factors.

We know that $110n^3$ has at least $3$ factors, namely $2, 5,$ and $11.$

The only way to express $110$ as the product of at least $3$ numbers is $2 \cdot 5 \cdot 11.$

This means that $110n^3$ has no other prime factors. Then the exponents must be $1, 4,$ and $10$ from the above formula.

Let $2^3$ be a factor of $n.$ Then $n^3$ has a factor of $2^9,$ which makes $110n^3$ have a factor of $2^{10}.$

Then let $5$ be a factor of $n.$ Then $n^3$ has a factor of $5^3,$ which means that $110n^3$ has a factor of $5^4.$

We do not have to alter the number of $11$s since it already has $1$ as its exponent.

This means that we can let $n = 2^3 \cdot 5.$ Then $$81n^4 = 3^4 \cdot 2^12 \cdot 5^4.$$ This number has $$5 \cdot 13 \cdot 5 = 325$$ factors.

Thus, the correct answer is D.

Solution:

Let us see what properties we can gather from the given conditions. First, we see that $$a \diamondsuit (a \diamondsuit a) = (a \diamondsuit a) \cdot a,$$ which tells us that $$a \diamondsuit 1 = a.$$

Now, let's see what happens if we set $b = c$ and apply the second property. We get $$a \diamondsuit (b \diamondsuit b) = (a \diamondsuit b) \cdot b,$$ which simplifies to $$a = (a \diamondsuit b) \cdot b$$$$\dfrac{a}{b} = a \diamondsuit b.$$ We can divide by $b$ since $b$ is nonzero.

We can now calculate the value of $x$ directly.

$$\begin{gather*} \dfrac{2016}{\frac{6}{x}} = 100 \\ 2016 = \dfrac{600}{x} \\ x = \dfrac{600}{2016} = \dfrac{25}{84} \end{gather*}$$ Then we have that $p + q = 109.$

Thus, the correct answer is A.

Solution:

Let $\overline{AD}$ intersect $\overline{BO}$ and $\overline{CO}$ at $E$ and $F$ respectively.

Since $AB = BC = CD,$ we get that $$\overset{\Large\frown}{AB} = \overset{\Large\frown}{BC} = \overset{\Large\frown}{CD} = \theta.$$

Then we get that $\angle BAD$ is an interior angle with value $$\dfrac{1}{2} \overset{\huge\frown}{BCD} = \dfrac{1}{2} \cdot 2 \theta = \theta = \angle AOB.$$

Then, by angle-angle, we get that $\triangle OAB \sim \triangle ABE.$ This gives us $$\dfrac{OA}{AB} = \dfrac{AB}{BE} = \dfrac{OB}{AE}.$$ We have that $OA = OB$ as they are both radii, which means that $AB = AE.$ Similarly, we have that $CD = DF.$

Now, we get that $BE = 100\sqrt{2}$ from the above ratios, which means that $\dfrac{BE}{BO} = \dfrac{1}{2}.$

This tells us that $$EF = \dfrac{BC}{2} = \dfrac{200}{2} = 100.$$

Then we get that $$AD = AE + EF + FD$$ $$= AB + EF + CD = 500.$$ Thus, the correct answer is E.

Solution:

We can prime factorize $72$ into $2^3 \cdot 3^2,$ $600$ into $2^3 \cdot 3 \cdot 5^2,$ and $900$ into $2^2 \cdot 3^2 \cdot 5^2.$

Note that the lcm of $x$ and $y$ does not have a factor of $5,$ so will neither $x$ nor $y.$ This means $z$ must have a factor of $5^2.$

Then we can express $x = 2^a \cdot 3^b,$ $y = 2^c \cdot 3^d,$ and $z = 2^e \cdot 3^f \cdot 5^2.$

By definition of lcm, we get that $$\begin{gather*} \max{(a, c)} = 3, \\ \max{(b, d)} = 2, \\ \max{(a, e)} = 3, \\ \max{(b, f)} = 1, \\ \max{(c, e)} = 2, \\ \max{(d, f)} = 2. \end{gather*}$$

Since the max of $b$ and $f$ is $1,$ we have that $d = 2.$ Similarly, since the max of $c$ and $e$ is $2,$ we have that $a = 3.$

We are left with a few redundant equations, so we can trim them down to $$\max{(b, f)} = 1$$ and $$\max{(c, e)} = 2.$$ For the first equation, at least one of them must be $1,$ giving us $3$ options: $(0, 1), (1, 0), (1, 1).$

For the second equation, at least one of them must be $2,$ giving us $5$ options: $$(2, 0), (2, 1), (2, 2), (0, 2), (1, 2).$$

There are then $3 \cdot 5 = 15$ possible combinations for all the variables.

Thus, the correct answer is A.