Solution:

Let the volumes of the first and second containers be $F$ and $S$. The amount of water is both $\dfrac56F$ and $\dfrac34S$, so $\dfrac56F=\dfrac34S$.

Thus $\dfrac{F}{S}=\dfrac{3/4}{5/6}=\dfrac9{10}$. Since the first container is smaller, the ratio of the smaller container to the larger container is $\dfrac9{10}$. Thus, D is the correct answer.

Solution:

We need $n$ to not be prime, so $n$ can only be $15,21,27.$ Then, $n-2$ must be not prime, leaving just $27.$

Thus, the answer is E .

Solution:

Let the number of seniors be $s.$ Then, $500-s$ people aren't seniors. We know $60\%$ of seniors don't play an instrument. Then, the number of students who don't play an instrument can be represented as $$0.3(500-s)+0.6(s) = 0.3s+150$$ and $$0.468\cdot 500=234.$$ Thus, $$0.3s+150=234$$ $$s=280.$$ This makes the number of non-seniors equal to $220.$ Since $70\%$ of non-seniors play instruments, we have the total number as $$220\cdot 0.7=154.$$

Thus, the answer is B .

Solution:

Let $d=b-a.$

Then, we have $$(a,b,c)= (a,a+d,a+2d).$$ Thus, $$ax+(a+d)y=a+2d.$$ If we match the parts of $a$ and $d,$ we get $$ax+ay=a$$ and $$dy=2d$$ for all $a,d.$ Therefore, we have $$y=2,x+y=1$$ implying that $$x=-1.$$ This makes the pair $(-1,2).$

Thus, the answer is A .

Solution:

Choice A must be true since the reflection of the first quadrant is itself, so anything inside stays inside after a reflection.

Choice B must be true as a reflection keeps the same area.

Choice C must be true as a reflection will have a perpendicular slope to the line its reflected about, so its slope is $-\frac 11 = -1.$

Choice D must be true as they both have the slope of $-1.$

Choice E can be false as if $$A = (2,1),$$$$B=(3,2),$$ then $AB$ and its reflection both have slope $-1,$ making them parallel. Therefore, they can be not perpendicular.

Thus, the answer is E .

Solution:

We can rewrite the left side as $$(n+1)n!+(n+2)(n+1)n!$$$$=((n+2)^2-1)n!,$$ so $$((n+2)^2-1) n! = 440 n!$$ Therefore, $$(n+2)^2=441,$$ so $n=19.$ The sum of its digits is $10.$

Thus, the answer is C .

Solution:

Let the number of cents he has $c.$ Then, $c$ is a multiple of $12,14,$ and $15.$ Thus, it must be a multiple of $420.$

Let $c=420k$ for some $k.$ Also, $c=20n,$ so $420k=20n,$ making $n=21k.$ Since $k$ is a whole number, the minimum possible value of $n$ is $21.$

Thus, the answer is B .

Solution:

The altitude of the triangle is $\sqrt 3$ using $30-60-90$ triangles, so the total base is $2\sqrt 3.$ The total amount of the base on each side that isn't in the white region is $2\sqrt 3 -2,$ so the amount from each triangle is $\sqrt 3 -1.$

This makes $8$ total triangles with base $\sqrt 3-1$ and altitude $\sqrt 3,$ so the combined area is $$8\cdot \dfrac{(\sqrt 3)(\sqrt 3-1)}2$$$$= 4\cdot (3-\sqrt 3)$$$$= 12-4\sqrt 3.$$

Thus, the answer is B .

Solution:

If $x$ was an integer, then $$\begin{align*}\lfloor|x|\rfloor - |\lfloor x \rfloor| &= |x| -|x|\\&=0.\end{align*}$$

If $x$ was positive, then $$\begin{align*}\lfloor|x|\rfloor - |\lfloor x \rfloor| &= \lfloor x \rfloor - \lfloor x \rfloor \\&= 0.\end{align*}$$

Now, we must look at negative non-integers. If we have a negative non-integer, then $\lfloor|x|\rfloor$ would negate $x$ and then round down, while $|\lfloor x \rfloor|$ would round down then negate it, effectively negating it and rounding up.

The first one rounds down and the second one rounds up, the second one is $1$ larger than the first, making $f=-1.$

Therefore, the range is $\{-1,0\}.$

Thus, the answer is A .

Solution:

The area condition fixes the distance from $C$ to line $AB.$ If that distance is $h,$ then $$\frac{10h}{2}=100,$$ so $h=20.$ Thus $C$ must lie on a line parallel to $AB$ at distance $20.$

The perimeter condition gives $AC+BC=40,$ so $C$ lies on an ellipse with foci $A,B$ and major axis $2a=40,$ hence $a=20$ and $c=5.$ Its semi-minor axis is $$b=\sqrt{a^2-c^2}=\sqrt{400-25}=\sqrt{375}\approx19.36.$$

The greatest possible distance from a point of the ellipse to line $AB$ is exactly $b\approx19.36,$ which is less than the required $20.$ So no point $C$ satisfies both conditions.

Thus, the answer is A .

Solution:

Let $x$ and $y$ be the numbers of green marbles in Jars $1$ and $2$, respectively. Then the total numbers of marbles in the jars are $10x$ and $9y$.

The jars contain the same number of marbles, so $10x=9y$. Also $x+y=95$. Solving gives $x=45$ and $y=50$.

Jar $1$ has $9x=405$ blue marbles, and Jar $2$ has $8y=400$ blue marbles. The difference is $5$. Thus, A is the correct answer.

Solution:

First, $2019 = 5613_7.$ Therefore, the $4$th digit from the left is at most $5.$ If that digit was $4$ and every other digit was maximized, then we get $4666_7,$ with a digit sum of $22.$ If that digit was $5,$ we can only have a sum greater than $22$ by making $5666_7$ which is too large.

Therefore, the largest digit sum is $22.$

Thus, the answer is C .

Solution:

Since there are $5$ numbers, the median is the $3$rd largest number. That would be $6$ if $x< 6,$ $8$ if $x> 8,$ and $x$ if $6 \leq x \leq 8.$

In addition, the mean is equal to $$\dfrac{4+6+8+17+x}5$$$$= 7+\dfrac x5.$$

If the mean is $6,$ then $$7+\frac x5=6,$$ so $x=-5.$

If the mean is $8,$ then $$7+\frac x5=8,$$ so $x=5,$ which is not in the required range.

If the mean is $x,$ then $$\begin{align*}7+\frac x5&=x\\7&=\frac{4x}5\\x&=8.75,\end{align*}$$ which is not in the required range.

Therefore, the only possible $x$ is $-5,$ making the sum $-5.$

Thus, the answer is A .

Solution:

We know $19!$ is a multiple of $5^3$ and $2^3,$ so it is a multiple of $1000.$ Therefore, $H=0$

We know it is a multiple of $9,$ so its digit sum must be a multiple of $9.$ As such, $$1+2+1+6+T+5+1+$$$$0+0+4+0+M+8$$$$+3+2+0+0+0$$$$=33+M+T$$ is a multiple of $9.$ With this in mind, we know that $$M+T \equiv 3 \mod 9,$$ leaving just $3$ and $12.$

We also know it is a multiple of $11,$ which means that when alternating between adding and subtracting digits, we get $$1-2+1-6+T-5+1-$$$$0+0-4+0-M+8-$$$$3+2-0+0-0$$$$= T-M-7$$ is a multiple of $11,$ so $$T-M \equiv 7 \mod 11.$$

The only way to satisfy both is $$T=4,M=8,H=0.$$ Their sum is $12.$

Thus, the answer is C .

Solution:

Let the two shared side lengths be $a\le b$. One triangle can have legs $a$ and $b$, while the other has side lengths $a$, $\sqrt{b^2-a^2}$, and hypotenuse $b$.

The product of the two non-shared third sides is $\sqrt{a^2+b^2}\sqrt{b^2-a^2}$, whose square is $b^4-a^4$.

Using the areas, $\dfrac{ab}{2}=2$ and $\dfrac{a\sqrt{b^2-a^2}}{2}=1$. Hence $a^2b^2=16$ and $a^2(b^2-a^2)=4$, so $a^4=12$. Then $b^4=\dfrac{256}{12}=\dfrac{64}{3}$, and $b^4-a^4=\dfrac{28}{3}$. Thus, A is the correct answer.

Solution:

Let $AC=4.$ From this, we get $$CD=4,ED=3,BE=3.$$

Notice that $$\angle EDB = \angle EBD,$$$$\angle CAD = \angle CDA,$$ so $$\angle EDB + \angle CDA = 90 ^\circ.$$ Thus, $\angle EDC = 90^\circ.$ From the Pythagorean Theorem, we get $$EC = \sqrt{3^2+4^2}$$$$= \sqrt{25}$$$$=5.$$ Therefore, $BC = 8.$ This makes $$\tan BAC = \frac 84 = 2.$$

Now, we can get $$BD = 2\cdot 3\cos B$$$$= 6\sin A$$ and $$AD = 2\cdot 4\cos A$$$$= 8\cos A .$$ Thus, $$\frac{AD}{BD} = \frac{8 \cos A}{6\sin A}$$$$= \frac 4{3 \tan A}$$ $$=\frac 4{3\cdot 2}$$$$= \frac 23.$$

Thus, the answer is A .

Solution:

Given that the two balls were tossed into separate bins, the probability that the ball in the higher-numbered bin is red is $\frac 12.$ Thus we must find $$\frac{P(\text{Balls in different bins})}2$$$$= \frac{1-P(\text{Balls in same bins})}2$$ by complementary counting.

The probability that both balls are in bin $k$ is $$2^{-k} \cdot 2^{-k} = 4^{-k}.$$

The probability that they are both in the same bin is therefore $$\sum_{k=1}^\infty 4^{-k}.$$ Using the geometric sequence formula, we get this to be $$\frac 14 \cdot \dfrac{1}{1-\frac 14} = \frac 13.$$

Therefore, our answer is $$\frac{1-\frac 13}2 = \frac 13 .$$

Thus, the answer is C .

Solution:

Suppose he starts $x$ miles from home and then goes in the direction of the gym before coming back. The current distance between him and the gym is $2-x,$ so he would have a distance of $\frac{2-x}4$ miles from the gym after walking. This would put him $$2-\frac {2-x}4 = 1.5 + \frac x4$$ miles from home.

Then, when he walks home, he becomes $$\frac{1.5 + \frac x4}4 = \frac 38 + \frac x {16}$$ miles from home.

If Henry is at one of the points, then his position would remain the same, so $$x= \frac 38 +\frac x{16}.$$ As such, we know that $$16=6x+x$$$$x=0.4.$$ Thus, one of the steady state points is $0.4$ miles from the house.

Using similar logic, but starting $x$ miles from the gym and then going home, and then to the gym, we get that another point is $1.6$ miles from the house.

Therefore, the points are $0.4$ and $1.6$ miles, so their difference is $1.2$ miles.

Thus, the correct answer is C .

Solution:

First, note that $100,000=2^55^5.$

Therefore, any element of $S$ must be of the form $2^a5^b$ with $0 \leq a$ and $b \leq 5.$

Suppose I have distinct $x,y \in S$ with $$x = 2^a5^b,$$$$y=2^c5^d.$$ Then, $$xy = 2^{a+c}5^{b+d}.$$ Thus, $$0 \leq a+c,b+d \leq 10.$$ This means that there are $$(10+1)(10+1)=121$$ divisors. However, there are some divisors that can only exist if $x=y,$ where $$(a,b)=(c,d).$$

If $a+c=0,$ then $a=0,c=0$ must be true.

If $a+c=10,$ then $a=5,c=5$ must be true.

With any other value of $a+c,$ we can have $a \neq c.$

Similar structure holds for $b+d.$ Thus, if $$a+c,b+d \in \{0,10\},$$ then $$(a,b)=(c,d),$$ thus making $x=y.$

This means we have to eliminate $4$ choices, leaving $$121-4=117.$$

Thus, the answer is C .

Solution:

Firstly, notice $FD = 1,$ so the arc ${XZ}$ must have length $$2 \arccos{\dfrac 12} = \dfrac {2\pi} 3 .$$ Since the area under semicircles is equal to the area of the arc minus the area of $FXZ,$ that area is $$\dfrac \theta {2} r^2 - \dfrac{FX\cdot FZ \sin XFZ}2$$$$= \dfrac{2\cdot 2^2 \pi}{3\cdot 2} - \dfrac{2\cdot 2 \frac {\sqrt{3}} 2}2$$$$= \frac 43 \pi - \sqrt 3 .$$ Then, the gray area above $EG$ is a semicircle $$\frac 12 r^2 \pi= \frac 12 \cdot 4 \pi = 2\pi.$$ Finally, the gray area consists of four of the following shapes. \t\t

The squares have side length $1$ so it has area $1.$ The quarter circle has area $$\frac \pi 4 r^2 = \frac \pi 4.$$ Therefore, the total amount of gray is $1- \frac \pi 4.$ We multiply this by $4$ since there are $4$ of these shapes, yielding an area of $4- \pi.$

The total area is $$\frac 43 \pi - \sqrt 3 + 2 \pi + 4 -\pi$$$$= \frac 73 \pi - \sqrt 3 + 4.$$ This makes our answer $$7+3+3+4 = 17.$$

Thus, the answer is E .

Solution:

If we flip heads first, then the only way to see the second tails before the next heads is with $HTT,$ which ends without two heads.

Therefore, we must start with tails. Then we need a heads to ensure that we don't end on two tails, and then another tails to ensure a second tails is seen.

This means we must start as $THT.$

Our sequence must also be alternating except the last two coins, or else there would be two consecutive flips that are the same causing it to stop. This means our sequence must be $THTH \cdots THH.$ This means that there is exactly one sequence of size $n$ for every odd $n$ at least $5,$ each with a probability of $\left(\frac 12\right) ^n.$ Thus, the total probability is $$\frac 12^5 + \frac 12^7 + \cdots$$ $$= \frac 12^5\left( 1+ \frac 14 + \frac 14^2 \cdots \right)$$$$= \dfrac 1{32} \cdot \dfrac{1}{1-\frac 14}$$ $$=\dfrac 1{32} \cdot \dfrac 43$$$$= \dfrac 1{24}.$$

Thus, the answer is B .

Solution:

There are only two possible money configurations up to order: $(1,1,1)$ and $(2,1,0)$. From $(1,1,1)$, the next state is again $(1,1,1)$ exactly when all three players pass dollars in the same cyclic direction, which has probability $2\left(\dfrac12\right)^3=\dfrac14$.

From $(2,1,0)$, only the players with money give a dollar. The next state is $(1,1,1)$ exactly when the player with $2$ dollars gives to the player with $0$ dollars and the player with $1$ dollar gives to the player with $2$ dollars, also with probability $\dfrac14$.

Therefore, regardless of the state after $2018$ rings, the probability that the state after the next ring is $(1,1,1)$ is $\dfrac14$. Thus, B is the correct answer.

Solution:

Let $P$ be the intersection point of the two tangents. Since tangent lengths from the same point are equal, $PA=PB$, so $P$ lies on the perpendicular bisector of $\overline{AB}$.

The midpoint of $A(6,13)$ and $B(12,11)$ is $(9,12)$, and the slope of $AB$ is $-\dfrac13$, so the perpendicular bisector is $y=3x-15$. Its intersection with the $x$-axis is $P=(5,0)$.

The tangent line through $P$ and $A$ has slope $13$, so the radius to $A$ has slope $-\dfrac1{13}$. Intersecting $y-13=-\dfrac1{13}(x-6)$ with $y=3x-15$ gives center $\left(\dfrac{37}{4},\dfrac{51}{4}\right)$.

Thus $r^2=\left(\dfrac{37}{4}-6\right)^2+\left(\dfrac{51}{4}-13\right)^2=\dfrac{85}{8}$, so the area is $\dfrac{85\pi}{8}$. Thus, C is the correct answer.

Solution:

Let $a_n=x_n-4$. Then $a_0=1$, and simplifying the recurrence gives $$a_{n+1}=\frac{a_n(a_n+9)}{a_n+10}.$$ As long as $0<a_n\le1$, this implies $$\frac9{10}a_n\le a_{n+1}\le\frac{10}{11}a_n.$$

By induction, $\left(\dfrac9{10}\right)^n\le a_n\le\left(\dfrac{10}{11}\right)^n$. For $n=80$, $\left(\dfrac{10}{9}\right)^{80}<2^{20}$, so $\left(\dfrac9{10}\right)^{80}>2^{-20}$, and therefore $m>80$.

Also $\left(\dfrac{11}{10}\right)^8>2$, so $\left(\dfrac{10}{11}\right)^{160}<2^{-20}$, which gives $m\le160$. Hence $81\le m\le160$, so $m$ lies in $[81,242]$. Thus, C is the correct answer.

Solution:

Our sequence starts with a $0$ then has sequences of $110$ and $10$ in some order, where they each come after a $0.$

Let the number of $110$ be $x$ and let the number of $10$ be $y.$ Then the number of terms in the sequence is $$3x+2y+1=19,$$ making $$3x+2y=18.$$ The only possible ordered pairs are $$\begin{align*}(x,y) = &(6,0),\\&(4,3),\\&(2,6),\\&(0,9).\end{align*}$$ Then, the number of ways to order the $x+y$ blocks is $$\binom{x+y}x,$$ since we choose which $x$ of the $x+y$ positions hold a $110$ block.

Therefore, the total number of ways is $$\binom 66 + \binom 74 + \binom 82 + \binom 90$$$$=1+35+28+1$$$$=65.$$