Solution:

The ratio is equal to the ratio of the recipricols of how much they are filled. This makes the ratio $$\dfrac{\left(\frac 65\right)}{\left(\frac 43\right)} = \dfrac 9{10} .$$

Thus, the answer is D.

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 she has be $c.$ 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:

If the perimeter was $50,$ then $AC+BC=40.$ This means that their average is $20,$ so either both of them are $20$ or at least one of them is less than $20.$

If the area was $100$ and the altitude from $C$ was $h,$ then $$\frac{10h}2=100$$ $$h=20.$$ The altitude can't be less than a side, leaving $$AC=20$$$$BC=20.$$ However, this would also yield an altitude less than $20$ as neither $AC$ nor $BC$ are perpendicular with $AB,$ making the altitude less than the lengths of the legs.

This leaves no way to make such a triangle.

Thus, the answer is A.

Solution:

Let $x$ be the number of green marbles in Jar $1$ and let $y$ be the number of green marbles in Jar $2.$

Then, the total number of marbles is $$10x+9y=95,$$ implying $$x \equiv 5 \mod 5.$$ The only possible $x$ is $x=5,$ making $y=5.$ There are $$9x=9\cdot 5=45$$ green marbles in Jar $1$ and $$8y=8\cdot 5=40$$ green marbles in Jar $2.$ Therefore, the difference is $5.$

Thus, the answer is A.

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 $3$rd largest numbers. 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 isn't 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 isn't in the required range.

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

Thus, the answer is A.

Solution:

We know $p$ is a multiple of $5^3$ and $2^3,$ so its 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 its 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 congruent sides be $a,b$ such that $a \leq b.$ None of the triangles can have hypotenues $a$ since $a \leq b.$

Then, in one triangle (say, $T_1$), we have side lengths $$a,b, \sqrt{a^2+b^2},$$ and in $T_2,$ we have $$a,\sqrt{b^2-a^2},b.$$ Thus, the product of the other sides would be $$\sqrt{(b^2-a^2)(b^2+a^2)}$$$$= \sqrt{b^4-a^4} ,$$ making its square $$b^4-a^4.$$

The area of the smaller triangle $T_2$ is $$\frac{a\sqrt{b^2-a^2}}2=1$$ and the area of the larger triangle $T_1$ is $$\frac{ab}2=2$$ Thus, $$a^2(b^2-a^2)=a^2b^2-a^4$$$$=4,$$ and $$a^2b^2=16.$$ That implies $a^4=12.$ We can also get $a^4b^4=256,$ so $$b^4= \frac{256}{12}=\frac{64}3.$$

Therefore, $$b^4-a^4 = \frac{64}3-12$$$$= \frac{28}3.$$

Thus, the answer is A.

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 at the two balls were tossed into seperate balls, the probability that 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 diffence is $1.2$ miles.

Thus, 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 stucture 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.

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 all odd $n$ greater than $5,$ 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:

Suppose that they are sitting in a circle, where they can move the money clockwise or counterclockwise.

Also suppose that at some point, they each have one dollar. Then, they each get a dollar afterwards if they all send the dollar in the same direction. This has a probability of $$2\cdot \left(\frac 12\right)^2 = \frac 14$$ since there are $2$ directions and a probability of one half that they chose that direction. Otherwise, two people send it to the same person, which has a probability of $\frac 34.$

Suppose that we are at a point where one person has two dollars, one person has one dollar, and the other person has no money.

If the person with one dollar sends it to the person with two dollars, then after the person with two dollars sends money to someone, he has $2$ dollars still, and another person has one.

If both the person with one dollar and the person with two dollars send it to the other person, we will still have a $2-1-0$ configuration.

However, if the person with one dollar sends it to the person with no money and the person with two dollars sends it to the person with one dollar, we get a $1-1-1$ configuration. This happens with probability $\frac 14.$

Therefore, the only possible configurations are $1-1-1$ or $2-1-0,$ and each has a $\frac 14$ probability that the next one is $1-1-1.$

As such, whatever the configuration is after $2018$ rings, the probability of the next one being $1-1-1$ is $\frac 14.$

Thus, the answer is B.

Solution:

If the radius has radius $r$ and the distance to some point outside the circle is $d,$ then the distance from the point to some tangent point must be $$\sqrt{d^2-r^2},$$ making it a constant value. Therefore, it is equidistant from $A$ and $B,$ so it must lie on its perpendicular bisector. The midpoint betweem $A$ and $B$ is $(9,12)$ and the slope between them is $-\frac 13.$

Consequently, we know that the slope of the perpendicular bisector is $3.$ Thus, the perpendicular bisector is on the line $$y-12=3(x-9)$$$$y=3x-15.$$ Thus, its intersection with the $x$ axis is $(5,0).$

Now, one of the tangent lines has points $(5,0)$ and $(6,13),$ so their slope is $13.$ Since the line perpendicular to this going through $(6,13)$ goes through the center, the center is the intersection of the lines $$y-13= -\frac 1{13}(x-6)$$ and $$y=3x-15.$$ Using substitution, we get $$3x-15-13= -\frac 1{13}(x-6)$$$$x = \frac {37}4.$$ We then get $$y= \frac{51}4.$$ Now, the radius is the distance from $(\frac{37}4, \frac{51}4)$ and $(6,13),$ which is $$\sqrt{(\frac{37}4-6)^2+(\frac{51}4-13)}.$$ This would simplify to $$r=\sqrt{\frac{85}8}.$$

Therefore, the area is $$\pi r^2 = \frac{85}8 \pi.$$

Thus, the answer is C.

Solution:

Let $a_n+4=x_n.$ Then $a_0 = 1$ and we have to find the least $m$ such that $$a_m \leq \dfrac 1{2^{20}}.$$

Also, $$a_{n+1} +4$$$$= \frac{(a_n+4)^2+5(a_n+4)+4}{a_n+10},$$ so $$a_n = \dfrac{a_n^2 +13 a_n +40}{a_n+10}-4$$ $$= \dfrac{a_n(a_n+9)}{(a_n+10)} .$$ Thus, $$\dfrac{a_{n+1}}{a_n} =\dfrac{a_n+9}{a_n+10}.$$ If $0 \leq a_n \leq 1,$ then $$\dfrac{a_n+9}{a_n+10}$$ is less than $1$ and greater than $0,$ making $$0 \leq a_{n+1} \leq a_n \leq 1.$$ Thus, $$\dfrac{a_n+9}{a_n+10}$$ is between $\frac 9{10}$ and $\frac {10}{11},$ so $$\frac {9}{10} \leq \dfrac{a_{n+1}}{a_n} \leq \frac{10}{11}.$$ This means $$\frac {9}{10}^k \leq a_k \leq \frac{10}{11}^k.$$

To find the least $m,$ we must put $a_k = \dfrac 1{2^{20}},$ making $$\frac {9}{10}^k \leq \frac{1}{2^{20}} \leq \frac{10}{11}^k.$$

We can take the reciprocal, leaving $$\frac {10}{9}^k \geq 2^{20} \geq \frac{11}{10}^k.$$

Since $(1+ \frac 19)^9 \geq 2,$ we know $$9 \log_2\left( \frac{10}{9}\right) \geq 1.$$

Thus, $\log_2( \frac{10}{9}) \geq \frac 19 .$ As such, $$\frac k9 \leq k \log_2\left( \frac{10}{9}\right) \leq 20$$$$k \leq 180 .$$