Solution:

Factoring the numerator gives $$\dfrac{11!-10!}{9!}=\dfrac{10!(11-1)}{9!}=\dfrac{10\cdot 9!\cdot 10}{9!}=100.$$

Thus, the correct answer is B.

Solution:

Since $100=10^2$ and $1000=10^3,$ the equation becomes $$10^x\cdot 10^{4x}=10^{15},$$ so $10^{5x}=10^{15}.$ Then $5x=15,$ giving $x=3.$

Thus, the correct answer is C.

Solution:

First, $$\dfrac{x}{y}=\dfrac{3/8}{-2/5}=\dfrac{3}{8}\cdot\left(-\dfrac{5}{2}\right)=-\dfrac{15}{16},$$ and $\left\lfloor -\dfrac{15}{16}\right\rfloor=-1.$

Therefore $$\text{rem}\left(\dfrac{3}{8},-\dfrac{2}{5}\right)=\dfrac{3}{8}-\left(-\dfrac{2}{5}\right)(-1)=\dfrac{3}{8}-\dfrac{2}{5}=\dfrac{15-16}{40}=-\dfrac{1}{40}.$$

Thus, the correct answer is B.

Solution:

The mean condition gives $$\dfrac{60+100+x+40+50+200+90}{7}=\dfrac{540+x}{7}=x,$$ so $540+x=7x$ and $x=90.$

In nondecreasing order the data are $40, 50, 60, 90, 90, 100, 200,$ so the median is $90$ and the mode is $90,$ as required.

Thus, the correct answer is D.

Solution:

A counterexample must satisfy the hypothesis of being an even integer greater than $2$ while failing the conclusion that it can be written as the sum of two prime numbers.

Thus, the correct answer is E.

Solution:

The total number of coins is $$1+2+\cdots+N=\dfrac{N(N+1)}{2}=2016,$$ so $N(N+1)=4032.$ Since $63\cdot 64=4032,$ we have $N=63,$ and the sum of its digits is $6+3=9.$

Thus, the correct answer is D.

Solution:

Moving all terms to one side gives $$(x^2-y^2)(x+y+1)=0,$$ which factors as $$(x-y)(x+y)(x+y+1)=0.$$ The graph is therefore the union of the lines $x=y,$ $x=-y,$ and $x+y+1=0.$

The first two lines intersect at the origin, but the third line $x+y=-1$ is parallel to $x+y=0$ and does not pass through the origin. So the graph consists of three lines that do not all pass through a common point.

Thus, the correct answer is D.

Solution:

The diagonal of the rectangle from the upper-left corner to the lower-right corner divides the shaded region into four triangles, all meeting at the center of the rectangle.

Two of these triangles have a horizontal base of length $1$ and altitude $\frac12\cdot 5=\frac52,$ and the other two have a vertical base of length $1$ and altitude $\frac12\cdot 8=4.$ The total area is $$2\cdot\dfrac12\cdot 1\cdot\dfrac52+2\cdot\dfrac12\cdot 1\cdot 4=\dfrac52+4=\dfrac{13}{2}.$$

Thus, the correct answer is D.

Solution:

Let $x$ be the common side length. The diagonal of the unit square has length $\sqrt{2}$ and consists of two small-square diagonals (each $x\sqrt2$) plus one small-square side length $x,$ so $$2x\sqrt2+x=\sqrt2.$$

Solving, $$x=\dfrac{\sqrt2}{2\sqrt2+1}=\dfrac{\sqrt2\,(2\sqrt2-1)}{(2\sqrt2+1)(2\sqrt2-1)}=\dfrac{4-\sqrt2}{7}.$$ Thus $a=4,$ $b=7,$ and $a+b=11.$

Thus, the correct answer is E.

Solution:

The net displacement of all five friends is zero. Dee and Edie swapped seats, so their movements cancel. Bea moved $+2$ and Ceci moved $-1,$ a net of $+1,$ so Ada must move $-1$ to balance.

Ada returns to an end seat; since she moved one seat to the left, that seat must be seat $1,$ so she had been sitting in seat $2.$

Thus, the correct answer is B.

Solution:

The numbers who can sing, dance, and act are $100-42=58,$ $100-65=35,$ and $100-29=71,$ respectively, for a total of $58+35+71=164.$

Since no student has all three talents, each student has one or two talents, so single-talent students are counted once and two-talent students are counted twice. The number counted twice is $164-100=64.$

Thus, the correct answer is E.

Solution:

Applying the Angle Bisector Theorem to $\triangle ABC$ gives $BD:DC=AB:AC=6:8,$ so $BD=\dfrac{6}{6+8}\cdot 7=3.$

Now $BF$ lies along the bisector of $\angle ABD$ in $\triangle ABD,$ so by the Angle Bisector Theorem again, $$AF:FD=AB:BD=6:3=2:1.$$

Thus, the correct answer is C.

Solution:

Write $N=5k.$ Number the positions of the red ball $0,1,\ldots,5k$ from one end; there are $5k+1$ equally likely positions.

Fewer than $\dfrac{3}{5}$ of the green balls lie on each side exactly when the red ball is in one of the positions $2k+1,2k+2,\ldots,3k-1,$ which is $k-1$ positions. Hence $$P(N)=1-\dfrac{k-1}{5k+1}=\dfrac{4k+2}{5k+1}.$$

Solving $\dfrac{4k+2}{5k+1}\lt\dfrac{321}{400}$ gives $400(4k+2)\lt 321(5k+1),$ so $1600k+800\lt 1605k+321$ and $5k\gt 479,$ meaning $k\gt 95.8.$ Thus $k=96$ and $N=480,$ whose digit sum is $4+8+0=12.$

Thus, the correct answer is A.

Solution:

Each vertex belongs to $3$ faces, so $6S=3(1+2+\cdots+8)=108,$ giving each face-sum $S=18.$

The four-element subsets containing $1$ with sum $18$ are $\{1,2,7,8\},\{1,3,6,8\},\{1,4,5,8\},$ and $\{1,4,6,7\}.$ Three of these contain both $1$ and $8,$ so $1$ and $8$ must lie on two adjacent vertices.

Rotate the cube so that $1$ is at the lower-left-front vertex and $8$ at the lower-right-front vertex. The numbers $4,6,7$ must label the remaining vertices of the face containing $1,$ which can be done in $3!=6$ ways; then $5,3,2$ are forced onto the opposite vertices. Hence there are $6$ arrangements.

Thus, the correct answer is C.

Solution:

The centers lie at heights $1,$ $2,$ and $3$ above line $l.$ Since circle $Q$ is externally tangent to circle $P,$ we have $PQ=3,$ so the horizontal distance is $P'Q'=\sqrt{3^2-1^2}=\sqrt{8}.$ Since circle $Q$ is tangent to circle $R,$ we have $QR=5,$ so $Q'R'=\sqrt{5^2-1^2}=\sqrt{24}.$

Place $P=(0,1),$ $Q=(\sqrt8,2),$ and $R=(\sqrt8+\sqrt{24},3).$ By the shoelace formula, the area is $$\dfrac12\left|\sqrt8(3-1)+(\sqrt8+\sqrt{24})(1-2)\right|=\dfrac12\left(\sqrt{24}-\sqrt8\right)=\sqrt6-\sqrt2.$$

Thus, the correct answer is D.

Solution:

Let $u=\log_3 x.$ Then $\log_x 3=\dfrac1u,$ $\log_{1/3}x=-u,$ and $\log_x\dfrac13=-\dfrac1u.$ Two graphs meet where two of $u,\dfrac1u,-u,-\dfrac1u$ are equal for some valid $x\gt 0.$

Setting $u=\dfrac1u$ gives $u=\pm1,$ so $x=3$ or $x=\dfrac13;$ setting $-u=-\dfrac1u$ gives the same values. Setting $u=-u$ gives $u=0,$ i.e. $x=1,$ where $\log_3 x$ and $\log_{1/3}x$ are both $0.$ The remaining pairings have no real solution.

The distinct intersection points are $(1,0),$ $(3,1),$ $\left(\dfrac13,-1\right),$ $(3,-1),$ and $\left(\dfrac13,1\right),$ so there are $5.$

Thus, the correct answer is D.

Solution:

Let square $ABCD$ have side length $6.$ Each equilateral triangle has height $3\sqrt3,$ and its center lies $\frac13$ of that height, namely $\sqrt3,$ from the square's side.

Square $ABCD$ has diagonal $6\sqrt2.$ Square $EFGH$ has diagonal equal to the side of $ABCD$ plus twice $\sqrt3,$ namely $6+2\sqrt3.$ The area ratio is the square of the ratio of diagonals: $$\left(\dfrac{6+2\sqrt3}{6\sqrt2}\right)^2=\left(\dfrac{3+\sqrt3}{3\sqrt2}\right)^2=\dfrac{12+6\sqrt3}{18}=\dfrac{2+\sqrt3}{3}.$$

Thus, the correct answer is B.

Solution:

Write $110n^3=p_1^{r_1}p_2^{r_2}\cdots$ so that the number of divisors is $(r_1+1)(r_2+1)\cdots=110.$ Since $110=2\cdot5\cdot11,$ there are exactly three distinct primes, which must be $2,5,11,$ with exponents $1,4,10$ in some order.

Taking $r_1=1,$ $r_2=4,$ $r_3=10$ for the primes $2,5,11$ gives $$n^3=\dfrac{2^1\cdot 5^4\cdot 11^{10}}{2\cdot5\cdot11}=5^3\cdot 11^9,\quad\text{so}\quad n=5\cdot 11^3.$$

Then $81n^4=3^4\cdot 5^4\cdot 11^{12},$ and since $3,5,11$ are distinct primes, the number of divisors is $$(4+1)(4+1)(12+1)=5\cdot5\cdot13=325.$$

Thus, the correct answer is D.

Solution:

Count the sequences of $8$ tosses whose running total reaches $4.$ With at most $2$ tails he certainly reaches $4,$ contributing $$\binom80+\binom81+\binom82=1+8+28=37$$ sequences.

With exactly $3$ tails he reaches $4$ only if he does so before the second tail, which allows at most one tail in the first $5$ tosses; this gives $4+4=8$ sequences. With exactly $4$ tails, only HHHHTTTT works, giving $1.$ He cannot reach $4$ with fewer than $4$ heads.

So there are $37+8+1=46$ favorable sequences out of $2^8=256,$ a probability of $\dfrac{46}{256}=\dfrac{23}{128}.$ Then $a+b=23+128=151.$

Thus, the correct answer is B.

Solution:

Setting $b=c=a$ gives $a\diamond 1=a\diamond(a\diamond a)=(a\diamond a)\cdot a=a.$ Then setting $c=b$ gives $a=a\diamond 1=a\diamond(b\diamond b)=(a\diamond b)\cdot b,$ so $a\diamond b=\dfrac{a}{b}.$

Therefore $$2016\diamond(6\diamond x)=2016\diamond\dfrac{6}{x}=\dfrac{2016}{6/x}=336x=100,$$ so $x=\dfrac{100}{336}=\dfrac{25}{84}$ and $p+q=25+84=109.$

Thus, the correct answer is A.

Solution:

Let $\theta$ be the central angle subtending a side of length $200,$ with radius $R=200\sqrt2.$ By the law of cosines on the isosceles triangle from the center, $$200^2=2R^2(1-\cos\theta)=160000(1-\cos\theta),$$ so $\cos\theta=\dfrac34.$

The fourth side subtends the central angle $3\theta,$ and $$\cos 3\theta=4\cos^3\theta-3\cos\theta=4\cdot\dfrac{27}{64}-\dfrac94=-\dfrac{9}{16}.$$ Its length squared is $$2R^2(1-\cos 3\theta)=160000\left(1+\dfrac{9}{16}\right)=160000\cdot\dfrac{25}{16}=250000,$$ so the fourth side is $500.$

Thus, the correct answer is E.

Solution:

Because $\text{lcm}(x,y)=2^3\cdot3^2$ and $\text{lcm}(x,z)=2^3\cdot3\cdot5^2,$ the factor $5^2$ divides $z$ while neither $x$ nor $y$ is divisible by $5.$ Also $3^2$ divides $y,$ while neither $x$ nor $z$ is divisible by $3^2,$ and $x$ must have the factor $2^3.$

Writing $x=2^3\cdot3^{\,j},$ $y=2^{\,k}\cdot3^2,$ and $z=2^{\,m}\cdot3^{\,n}\cdot5^2,$ the lcm conditions require $\max(j,n)=1$ and $\max(k,m)=2.$ There are $3$ choices for $(j,n)$ and $5$ choices for $(k,m),$ giving $3\cdot5=15$ ordered triples.

Thus, the correct answer is A.

Solution:

The ordered triples $(x,y,z)$ fill the unit cube of volume $1.$ They fail to form a triangle exactly when one value is at least the sum of the other two.

The region $z\ge x+y$ is a tetrahedron with vertices $(0,0,0),(0,0,1),(0,1,1),(1,0,1)$ of volume $\frac16.$ The analogous regions $x\ge y+z$ and $y\ge x+z$ also have volume $\frac16$ and have disjoint interiors. So the failure probability is $3\cdot\frac16=\frac12,$ and the triangle probability is $1-\frac12=\frac12.$

Thus, the correct answer is C.

Solution:

Since $a$ and $b$ are positive, all roots $r,s,t$ must be positive. By Vieta's formulas, $r+s+t=a,$ $rs+st+tr=b,$ and $rst=a,$ so $r+s+t=rst.$

By the AM-GM inequality, $27rst\le(r+s+t)^3=(rst)^3,$ so $a=rst\ge 3\sqrt3,$ with equality if and only if $r=s=t=\sqrt3.$ At this smallest $a,$ $$b=rs+st+tr=3r^2=3\cdot 3=9.$$

Thus, the correct answer is B.

Solution:

Take $k=2j.$ The smallest perfect square with $k+1$ digits is $10^{k}=(10^{j})^2,$ and after Silvia erases, the numbers shown are $\left\lfloor n^2/10^{k}\right\rfloor$ for $n=10^{j}, 10^{j}+1,\ldots$ Consecutive terms increase by $0$ or $1$ until the first jump of at least $2.$

That first jump occurs at $n=\dfrac{10^{k}}{2}+m$ with $m=10^{j}-1,$ and one computes that the last number written before the gap gives $$f(2j)=\dfrac{10^{2j}}{4}+10^{j}.$$

Summing over $j=1,\ldots,1008,$ $$\sum_{j=1}^{1008}f(2j)=25\sum_{j=0}^{1007}10^{2j}+10\sum_{j=0}^{1007}10^{j}=\underbrace{2525\cdots25}_{2016\text{ digits}}+\underbrace{111\cdots10}_{1009\text{ digits}}.$$ There are no carries, so the digit sum is $1008\cdot(2+5)+1008\cdot 1=1008\cdot 8=8064.$

Thus, the correct answer is E.