Solution:

Distributing the negative signs gives $$(20-2010+201)+(2010-201+20)=40.$$

Thus, C is the correct answer.

Solution:

The ferry makes $6$ trips: at $10,$ $11,$ $12,$ $1,$ $2,$ and $3.$

The numbers of tourists are $100,99,98,97,96,95,$ so the total is $$6\cdot100-(1+2+3+4+5)=600-15=585.$$

Thus, A is the correct answer.

Solution:

Let $s$ be the side length of square $EFGH.$ The shaded overlap has width $s$ and height $AD,$ so its area is $s\cdot AD.$

Because the overlap is $50\%$ of the rectangle, $s\cdot AD=\tfrac12\,AB\cdot AD,$ so $AB=2s.$ Because it is $20\%$ of the square, $s\cdot AD=\tfrac15 s^2,$ so $AD=\tfrac{s}{5}.$

Therefore $$\dfrac{AB}{AD}=\dfrac{2s}{s/5}=10.$$

Thus, E is the correct answer.

Solution:

Choice (D) is $-x^{-1}=-\dfrac1x.$ When $x\lt0,$ $\dfrac1x\lt0,$ so $-\dfrac1x\gt0.$

Testing $x=-1$ shows the other choices need not be positive: $\dfrac{x}{|x|}=-1,$ $-x^2=-1,$ $-2^x=-\tfrac12,$ and $\sqrt[3]{x}=-1.$

Thus, D is the correct answer.

Solution:

The opponent can score at most $50\cdot10=500$ on the last $50$ shots. Since Chelsea leads by $50,$ she must score more than $500-50=450$ points on her remaining shots to guarantee victory.

Her $n$ bullseyes give $10n$ points, and her other $50-n$ shots give at least $4(50-n)$ points, so $$10n+4(50-n)\gt450.$$ This simplifies to $6n\gt250,$ i.e. $n\gt41\tfrac23.$

Therefore Chelsea needs at least $42$ bullseyes.

Thus, C is the correct answer.

Solution:

Note that $x$ is at most $999.$ This means that $x + 32$ has a maximum of $1031.$

Similarly, we have that the minimum value of $x + 32$ is $1000.$

The only palindrome in this range is $1001,$ so this is what $x + 32$ equals.

Then $$x + 32 = 1001$$$$x = 969.$$

The sum of the digits is then $$9 + 6 + 9 = 24.$$

Thus, E is the correct answer.

Solution:

The miniature tower holds $$\dfrac{100,000}{.1} = 1,000,000$$ times less water than the actual tower. Since this is the ratio for volumes, the ratio of heights is $$(1,000,000)^{1 / 3} = 100.$$ This means that the height of the miniature tower is $$\dfrac{40}{100} = .4.$$

Thus, C is the correct answer.

Solution:

Let $\angle BAE = \angle ACD = x.$ Note that $\angle CFE = 60^{\circ}$ since $\triangle CFE$ is equilateral.

We then have that $$\angle AFC = 180^{\circ} - \angle CFE = 120^{\circ}.$$

Then: $$\begin{align*} \angle FAC &= 180^{\circ} - 120^{\circ} - x\\ &=60^{\circ} - x \\ &= \angle EAC.\end{align*}$$

We then get that $$\begin{align*} \angle BAC &= \angle BAE + \angle EAC\\ &= x + 60^{\circ} - x \\&= 60^{\circ}. \end{align*}$$

Since $AB = 2 \cdot AC$ and $\angle BAC = 60^{\circ},$ we have that $\triangle ABC$ is a $30-60-90$ triangle.

Thus, C is the correct answer.

Solution:

Note that all the cut out solids intersect in the middle of the cube.

This region of intersection is a cube with side length $2.$ Then the volume of the cutout region is $$\begin{align*}3 \cdot 2 \cdot 2 \cdot 3 - 2 \cdot 2^3 &= 36 - 16 \\&= 20.\end{align*}$$

We have to subtract out the center region twice since it is included in all $3$ regions.

The remaining volume is then $$3^3 - 20 = 27 - 20 = 7.$$

Thus, A is the correct answer.

Solution:

Consecutive terms differ by a common difference $d.$ From the last two terms, $d=(3p+q)-(3p-q)=2q.$

From the first two terms, $9-p=d=2q,$ and from the second and third, $(3p-q)-9=d=2q.$ Solving this system gives $p=5,$ $q=2,$ and $d=4.$

The $2010$th term is $$p+2009d=5+2009\cdot4=8041.$$

Thus, A is the correct answer.

Solution:

Since $x=\log_b 7^7,$ we have $b^x=7^7.$

Then $$(7b)^x=7^x\cdot b^x=7^x\cdot7^7=7^{x+7}=8^x.$$

Because $x\gt0,$ it follows that $7b=8,$ so $b=\dfrac{8}{7}.$

Thus, C is the correct answer.

Solution:

If Brian is a frog, then he must be lying, which means that Mike must be a frog.

If Brian is a toad, then he must be telling the truth, which also means that Mike is a frog.

Therefore, Mike is a frog, which means that Mike is lying. This means that there is at most one toad.

Then, at least one of LeRoy and Chris is a frog. This means the other is telling the truth, which makes them a toad.

This means there is one toad, which makes there be $3$ frogs.

Thus, D is the correct answer.

Solution:

For $k=0,$ the graph of $x^2+y^2=0$ is the single point $(0,0)$ and $xy=0$ is the two axes, which meet at the origin, so the graphs intersect.

For $k\ne0,$ the circle has radius $|k|,$ and the hyperbola $xy=k$ has its two vertices nearest the origin at distance $\sqrt{2|k|}.$ The graphs meet exactly when $|k|\ge\sqrt{2|k|},$ that is $|k|\ge2.$

So they fail to intersect only when $|k|=1,$ namely $k=1$ and $k=-1,$ giving $2$ values.

Thus, C is the correct answer.

Solution:

Using the Angle Bisector Theorem, we have that $$\dfrac{AB}{3} = \dfrac{BC}{8}$$$$AB = \dfrac{3}{8} BC.$$

For $AB$ and $BC$ to be integers, we must have that $BC$ is a multiple of $8.$

To minimize the perimeter, we can set $BC = 8$ and $AB = 3.$ This, however, makes the triangle degenerate.

$BC$ must then be $16$ and $AB = 6.$ Since $AC = AD + DC = 11,$ the perimeter is $$16 + 6 + 11 = 33.$$

Thus, B is the correct answer.

Solution:

Let $p$ be the probability of heads. The chance of two heads and two tails in four flips is $$\binom{4}{2}p^2(1-p)^2=6p^2(1-p)^2=\frac16.$$

Thus $p^2(1-p)^2=\dfrac{1}{36},$ so $p(1-p)=\dfrac16.$

This gives $6p^2-6p+1=0,$ so $p=\dfrac{3\pm\sqrt3}{6}.$ Since $p\lt\dfrac12,$ we take $p=\dfrac{3-\sqrt3}{6}.$

Thus, D is the correct answer.

Solution:

There are two cases: Bernardo picks a $9$ or he doesn't.

Case 1: Bernardo picks a $9$

Since a number is fixed, there are $\binom{8}{2} = 28$ ways to choose the other two numbers.

There are a total of $\binom{9}{3} = 84$ ways to pick all three numbers. The probability is then $$\dfrac{28}{84} = \dfrac{1}{3}.$$

Note that if Bernardo picks a $9,$ he automatically has a greater number than Silvia.

This means that Bernardo always wins in this case.

Case 2: Bernardo doesn't pick a $9$

There is a $1 - \frac{1}{3} = \frac{2}{3}$ chance of this happening. Since both people are choosing from the same numbers, they have an equal chance of winning.

We still need to find the probability that the numbers are the same. There is a $$\dfrac{1}{\binom{8}{3}} = \dfrac{1}{56}$$ chance that Silvia chooses the same numbers as Bernardo. The probability that Bernardo gets a higher number is then $$\dfrac{1 - \frac{1}{56}}{2} = \dfrac{55}{112}.$$

The total probability of Bernardo getting a higher number is then $$\dfrac{1}{3} + \dfrac{2}{3} \cdot \dfrac{55}{112} = \dfrac{37}{56}.$$

Thus, B is the correct answer.

Solution:

Note that $\triangle ACE$ is equilateral. Using the Law of Cosines in $\triangle ABC,$ we get $AC^2=r^2+1^2-2r\cos120^\circ=r^2+r+1.$

The area of $\triangle ACE$ is then $$\dfrac{\sqrt3}{4} (r^2 + r + 1).$$

The three corner triangles $\triangle ABC,$ $\triangle CDE,$ and $\triangle EFA$ each have area $\frac12\cdot1\cdot r\cdot\sin120^\circ=\frac{r\sqrt3}{4}.$

Thus the hexagon has area $\dfrac{\sqrt3}{4}(r^2+r+1)+3\cdot\dfrac{r\sqrt3}{4}=\dfrac{\sqrt3}{4}(r^2+4r+1).$

The condition $[ACE]=70\%\cdot[ABCDEF]$ gives $$r^2+r+1=\dfrac{7}{10}(r^2+4r+1),$$ so $r^2-6r+1=0.$

By Vieta's formulas, the sum of the possible values of $r$ is $6.$

Thus, E is the correct answer.

Solution:

Every step increases $x+y$ by $1,$ which runs from $-8$ to $8,$ so each path passes through exactly one lattice point with $x+y=0.$

To stay out of the open square, that point $(t,-t)$ must have $|t|\ge2,$ so it is one of $(\pm2,\mp2),(\pm3,\mp3),(\pm4,\mp4).$

By symmetry consider the three points $(-4,4),(-3,3),(-2,2)$ and double. The number of paths from $(-4,-4)$ to $(-(4-j),4-j)$ is $\binom{8}{j},$ and the number continuing on to $(4,4)$ is also $\binom{8}{j}.$

Therefore the total is $$2\left(\binom80^2+\binom81^2+\binom82^2\right)=2(1+64+784)=1698.$$

Thus, D is the correct answer.

Solution:

Since there are $k + 1$ marbles in the $k$ th box, there is a $\dfrac{k}{k + 1}$ chance Isabella draws a white marble from it.

The probability of drawing a red marble is then $\dfrac{1}{k + 1}.$ To stop after drawing the $n$ th marble, the first $n - 1$ marbles must have been white.

This happens with a probability of $$\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \ldots \cdot \dfrac{n - 1}{n} \cdot \dfrac{1}{n + 1}.$$

Note that all the numerators cancel with the adjacent denominator, which means that this expression reduces to $\dfrac{1}{n(n + 1)}.$

We have to find the smallest $n$ such that $$\dfrac{1}{n(n + 1)} \lt \dfrac{1}{2010}$$$$$$$$n(n + 1) \gt 2010.$$

Guessing and checking gives us that the smallest $n$ that works is $45.$

Thus, A is the correct answer.

Solution:

Since $a_n=1+(n-1)d_1$ and $b_n=1+(n-1)d_2$ for integers $d_1,d_2,$ the value $n-1$ divides both $a_n-1$ and $b_n-1,$ hence divides $\gcd(a_n-1,b_n-1).$

The factor pairs of $2010$ with $2\le a_n\le b_n$ are $(2,1005),(3,670),(5,402),(6,335), (10,201),(15,134),$ and $(30,67).$

For every pair except $(15,134),$ the numbers $a_n-1$ and $b_n-1$ are relatively prime, forcing $n=2.$ For $(15,134),$ $\gcd(14,133)=7,$ so $n-1$ can equal $7,$ giving $n=8.$

The sequences $a_n=2n-1$ and $b_n=19n-18$ realize this, so the largest value is $8.$

Thus, C is the correct answer.

Solution:

Let $f(x)$ be the graph minus the line. It is nonnegative and vanishes at three points, each a double root, so $$f(x)=\big(x^3-Ax^2+Bx-C\big)^2.$$

Matching coefficients gives $-2A=-10\Rightarrow A=5,$ then $A^2+2B=29\Rightarrow B=2,$ then $-2C-2AB=-4\Rightarrow C=-8.$

Thus the cubic is $x^3-5x^2+2x+8=(x+1)(x-2)(x-4),$ with roots $-1,2,$ and $4.$ The largest is $4.$

Thus, A is the correct answer.

Solution:

The function $f$ is piecewise linear with breakpoints at $x=\tfrac1k.$ On the interval $\left[\tfrac1m,\tfrac1{m-1}\right]$ its slope is $$\sum_{k=m}^{119}k-\sum_{k=1}^{m-1}k=7140-(m-1)m,$$ where $7140=\tfrac{119\cdot120}{2}.$

This slope is zero when $(m-1)m=7140,$ i.e. $m=85,$ so the minimum occurs at the right endpoint $x=\tfrac1{84}.$

There, terms with $k\le84$ contribute $\tfrac{84-k}{84}$ and terms with $k\ge85$ contribute $\tfrac{k-84}{84},$ so $$f\left(\tfrac1{84}\right)=\frac{3486}{84}+\frac{630}{84}=41.5+7.5=49.$$

Thus, A is the correct answer.

Solution:

The number of trailing zeroes in $90!$ is $\left\lfloor\dfrac{90}{5}\right\rfloor+\left\lfloor\dfrac{90}{25}\right\rfloor=21.$ Let $N=\dfrac{90!}{10^{21}}.$

There are still more than two factors of $2$ left after removing $10^{21},$ so $N\equiv0 \pmod4.$

Let $A$ be the product of factors of $90!$ not divisible by $5,$ and let $B$ be the product of the factors divisible by $5.$ Grouping residues modulo $25$ gives $A\equiv1\pmod{25}$ and $\dfrac{B}{5^{21}}\equiv-1\pmod{25}.$

Therefore $\dfrac{90!}{5^{21}}\equiv-1\pmod{25}.$ Since $2^{21}\equiv2\pmod{25},$ $N=\dfrac{90!}{5^{21}\cdot2^{21}}\equiv-13\equiv12\pmod{25}.$

The number congruent to $0\pmod4$ and $12\pmod{25}$ is $12\pmod{100},$ so the last two nonzero digits form $12.$

Thus, A is the correct answer.

Solution:

Let $g(x)=\prod_{k=1}^8\sin(k\pi x);$ the domain of $f$ is where $g(x)\gt0.$ Since $\sin(k\pi(1-x))=(-1)^{k+1}\sin(k\pi x)$ and $\sum_{k=1}^8(k+1)$ is even, $g(1-x)=g(x),$ so it suffices to study $\big(0,\tfrac12\big)$ and double.

In $\big(0,\tfrac12\big)$ the zeros of $g$ are the fractions $\tfrac{k}{n}$ with $2\le n\le8,$ $1\le k\lt\tfrac n2,$ and $\gcd(k,n)=1.$ For $n=2,\ldots,8$ there are $0,1,1,2,1,3,2$ of them, totaling $10.$

These $10$ zeros split $\big(0,\tfrac12\big)$ into $11$ subintervals on which $g$ has constant sign. Near $0$ every factor is positive, so $g\gt0$ there, and the sign flips at each zero except $\tfrac14=\tfrac28$ and $\tfrac13=\tfrac26,$ where an even number of factors vanish.

Tracking the signs, exactly $6$ of the $11$ subintervals have $g\gt0.$ By symmetry there are $6$ more in $\big(\tfrac12,1\big),$ so $n=12.$

Thus, B is the correct answer.

Solution:

A convex cyclic quadrilateral is determined up to rotation and translation by its cyclic sequence of side lengths, and it exists exactly when the largest side is less than the sum of the others. With perimeter $32,$ this means each side is at most $15.$

First count ordered quadruples $(a,b,c,d)$ of positive integers with $a+b+c+d=32$ and each entry at most $15.$ Without the upper bound there are $\binom{31}{3}=4495;$ removing those with some entry at least $16$ subtracts $4\binom{16}{3}=2240,$ leaving $2255.$

Rotations of the quadrilateral correspond to cyclic permutations of $(a,b,c,d).$ By Burnside's lemma the number of distinct quadrilaterals is $$\frac{1}{4}\big(2255+f_1+f_2+f_3\big),$$ where $f_i$ counts quadruples fixed by rotating $i$ positions.

A one- or three-step rotation fixes only $(8,8,8,8),$ so $f_1=f_3=1.$ A two-step rotation fixes $(a,b,a,b)$ with $a+b=16$ and $1\le a,b\le15,$ giving $f_2=15.$

Hence the count is $$\frac{1}{4}(2255+1+15+1)=\frac{2272}{4}=568.$$

Thus, C is the correct answer.