Solution:

We can simplify this as follows. $$\begin{align*} &\left(\left((2+1)^{-1}+1\right)^{-1}+1\right)^{-1}+1 \\ &=\left(\left(\dfrac{1}{3}+1\right)^{-1}+1\right)^{-1}+1 \\ &=\left(\dfrac{3}{4}+1\right)^{-1}+1 \\ &=\dfrac{4}{7} + 1 \\ &=\dfrac{11}{7} \end{align*}$$

Thus, B is the correct answer.

Solution:

Let $x$ be the number of gallons of soda that Jacqueline has. Then Alice has $1.25x$ gallons, and Liliane has $1.5x$ gallons.

Therefore, the relationship can be found by dividing the amount of soda that each has to yield $\dfrac{1.5x}{1.25x} = 1.2,$ which mean Liliane has $20\%$ more soda.

Thus, A is the correct answer.

Solution:

We can divide $10!$ by $60,$ $60,$ and $24$ to get the number of days that it takes for a unit of blood to expire.

The first division cancels a $6$ and $10.$ The second division cancels $3, 4,$ and $5.$ The final division cancels $8$ and turns the $9$ into a $3.$

This leaves $2, 7,$ and a $3,$ which multiply to $42.$ There are $31$ days in January, so by February $1,$ the blood only has $42 - 31 = 11$ days left.

$11$ days from February $1$ would make the blood expire on February $12.$

Thus, E is the correct answer.

Solution:

The $3$ classes can occupy the following periods: $$(1, 3, 5),$$$$(1, 3, 6),$$$$(1, 4, 6),$$$$(2, 4, 6).$$

This means that there are $4$ ways to choose which periods the mathematics courses occur.

For each configuration, there are $3!$ ways to determine the order of the courses, for a total of $6 \cdot 4 = 24$ schedules.

Thus, E is the correct answer.

Solution:

Alice's statement tells us that $d \lt 6.$ Bob's statement tells us that $d \gt 5.$ Charlie's statement tells us that $d \gt 4.$

Combining all of these tells us that $5 \lt d$ and $d \lt 6,$ which means $d$ is in the interval $(5, 6).$

Thus, D is the correct answer.

Solution:

If $65\%$ of votes were like votes, then $35\%$ of votes are dislike votes. Then Sangho's score is $65\% - 35\% = 30\%$ the total number of votes.

We know that Sangho's score is $90,$ so the total number of votes is $90 \div 30\% = 300.$

Thus, B is the correct answer.

Solution:

We can rewrite the expression as $$(2^5 \cdot 5^3) \cdot \left(\dfrac{2}{5}\right)^n = 2^{5 + n} \cdot 5^{3 - n}.$$

For this to be an integer, the exponents must be positive. This means that $$\begin{align*} 5 + n \geq 0 &\Rightarrow n \geq -5 \\ 3 - n \geq 0 &\Rightarrow n \leq 3. \end{align*}$$

This gives us $5 + 3 + 1 = 9$ values for $n.$

Thus, E is the correct answer.

Solution:

Let $x$ be the number of $5$-cent coins that Joe has. Then the number of $10$-cent coins he has is $x + 3.$

Therefore, Joe has $$23 - x - (x + 3) = 20 - 2x$$ $25$-cent coins.

The total value of all these coins is $$5x + 10(x + 3) + 25(20 - 2x)$$$$= 530 - 35x.$$

We know that $$530 - 35x = 320 \Rightarrow x = 6.$$

This means that Joe has $20 - 2 \cdot 6 = 8$ $25$-cent coins. Therefore, he has $8 - 6 = 2$ more $25$-cent coins than $5$-cent coins.

Thus, C is the correct answer.

Solution:

We know that the side length of the smaller triangles is $\sqrt{\frac{1}{40}}$ times the length of the larger triangle from similar triangles.

Then the side length of $\triangle ADE$ is $4\sqrt{\frac{1}{40}}$ times the length of the side length of the larger triangle.

This makes the ratio of the areas $$\left(4\sqrt{\dfrac{1}{40}}\right)^2 = 16 \cdot \dfrac{1}{40} = \dfrac{2}{5}.$$

Therefore, the area of $\triangle ADE$ is $\frac{2}{5} \cdot 40 = 16.$ The area of the trapezoid is then $40 - 16 = 24.$

Thus, E is the correct answer.

Solution:

Note that the left hand side of the equation and the desired expression are conjugates. Multiplying them would remove the square roots.

Multiplying them yields $$49 - x^2 - 25 + x^2 = 24.$$

This means that the product of the values of the expressions is equal to $24.$ The desired value is therefore $24 \div 3 = 8.$

Thus, A is the correct answer.

Solution:

We can use stars and bars to find $n.$ It is the same as finding the number of ways to put $10$ balls into $7$ boxes, where each box has at least one ball.

The formula for such a scenario is $$\binom{n - 1}{k - 1},$$ where $n$ is the number of balls and $k$ is the number of boxes.

The desired answer is therefore $$\binom{9}{6} = \binom{9}{3} = 84.$$

Thus, E is the correct answer.

Solution:

The second equation says $|x|-|y|=1$ or $|x|-|y|=-1$, so it is enough to check the four linear possibilities $x=y\pm1$ and $x=-y\pm1$.

Combining these with $x+3y=3$ gives $(x,y)=\left(\dfrac32,\dfrac12\right)$, $(0,1)$, $(0,1)$ again, and $(-3,2)$.

These are three distinct ordered pairs, and each satisfies the original absolute-value equation. Thus, C is the correct answer.

Solution:

Note that the crease will the perpendicular bisector of $\overline{AB}.$ Let $\overline{DE}$ be the crease.

By $AA$ similarity, we know that $\triangle ADE \sim \triangle ACB.$ Therefore, $$\dfrac{BC}{AC} = \dfrac{DE}{AD}$$Plugging in values: $$\dfrac{3}{4} = \dfrac{DE}{\frac{5}{2}}.$$

Simplifying gets us that $$DE = \dfrac{15}{8}.$$

Thus, D is the correct answer.

Solution:

Let $a=3^{96}$ and $b=2^{96}$. The expression is $\dfrac{81a+16b}{a+b}=16+\dfrac{65a}{a+b}$, so it is less than $16+65=81$.

To show the floor is $80$, we also need the expression to be greater than $80$. This is equivalent to $81a+16b>80a+80b$, or $a>64b$.

Because $\dfrac{a}{b}=\left(\dfrac32\right)^{96}>2^6=64$, the expression is greater than $80$ and less than $81$. Thus, A is the correct answer.

Solution:

Let $X$ be the center of the large circle and let $Y,Z$ be the centers of the two smaller circles. Then $XY=XZ=13-5=8$ and $YZ=10$.

The radii to tangent points make $XA$ collinear with $XY$ and $XB$ collinear with $XZ$, so $\triangle XAB\sim \triangle XYZ$. Thus $\dfrac{AB}{YZ}=\dfrac{XA}{XY}=\dfrac{13}{8}$.

Hence $AB=10\cdot\dfrac{13}{8}=\dfrac{65}{4}$, so $m+n=65+4=69$. Thus, D is the correct answer.

Solution:

Let $P$ be the foot of the altitude from $B$ to $\overline{AC}.$ We also get that $AC = 29.$

This tells us that $$\dfrac{29 \cdot PB}{2} = \dfrac{20 \cdot 21}{2}$$$$PB = \dfrac{20 \cdot 21}{29},$$ by calculating the area in two ways. This value is between $14$ and $15.$

Note that as we move the line segment from $\overline{AB}$ to $\overline{PB},$ the line segment's length ranges from $AB$ to $PB.$

The integer values it covers therefore goes from $20$ to $15.$ Similarly, as the line segment moves from $\overline{PB}$ to $\overline{CB},$ its takes on the values from $15$ to $21.$

This gives us $13$ unique line segments that have an integer value length.

Thus, D is the correct answer.

Solution:

We proceed by casing on possible values for $S:$

$1$ cannot be the smallest element since that would mean that no other number can be in the set.

$2$ cannot be the smallest element since we would have to include every odd number except $1.$ This would make $3$ and $9$ violate the rule.

Let $3$ be the smallest element. Then we can include $7$ and $11.$ We can finally include either $4$ or $8$ and $5$ or $10.$

Either way, the maximum number of elements that we can include is $5,$ so $3$ cannot be the smallest element.

Starting with $4,$ we can include $6, 7, 9$ and $11.$ Finally, we can add either $5$ or $10,$ creating a $6$-element set.

Thus, C is the correct answer.

Solution:

Note that every number formed by this sum is either positive, negative, or zero.

The number of positive numbers equals the number of negative numbers due to symmetry (flip the $1$ s to $-1$ s and $-1$ s to $1$ s).

The only way for the sum to be $0$ is if all the coefficients are $0.$

The total number of numbers is $3^8 = 6561.$ Because each power of $3$ is larger than the sum of all previous powers of three, each combination of coefficients yields different numbers.

Therefore, there are $$\dfrac{6561 - 1}{2} + 1 = 3281$$ distinct nonnegative integers.

Thus, D is the correct answer.

Solution:

Since we only care about the units digit, we can turn the set $$\{11,13,15,17,19\}$$ into $$\{1,3,5,7,9\}.$$ Then we can case on the value of $m.$

$m = 1$

Any value of $n$ works. This occurs with a $\frac{1}{5}$ probability.

$m = 3$

Looking at powers of $3,$ we see that this sequence of units digits repeats: $$3, 9, 7, 1, \ldots$$

This means that $n$ must be a multiple of $4.$ There are $5$ such values. This means that $n$ works $\frac{5}{20} = \frac{1}{4}$ of the time. The total probability is $$\dfrac{1}{5} \cdot \dfrac{1}{4} = \dfrac{1}{20}.$$

$m = 5$

Powers of $5$ always end in $5,$ which means that this case will never work.

$m = 7$

The units digits repeat in this pattern: $$7, 9, 3, 1,\ldots$$ This means that $n$ must be a multiple of $4$ to work. As when $m = 3,$ this case works with a probability of $\frac{1}{20}.$

$m = 9$

The units digit alternates between $1$ and $9.$ This means that $n$ has to be even. This happens with a $\frac{1}{2}$ chance. The total probability is then $$\dfrac{1}{5} \cdot \dfrac{1}{2} = \dfrac{1}{10}.$$

The total probability is therefore $$\dfrac{1}{5} + 2 \cdot \dfrac{1}{20} + \dfrac{1}{10} = \dfrac{2}{5}.$$

Thus, E is the correct answer.

Solution:

Under the symmetries of the square, the $7\times7$ grid splits into $10$ orbits of squares. Once one square in each orbit is colored, symmetry forces the colors of all other squares in that orbit.

There are therefore $2^{10}$ symmetric colorings before the condition about using both colors. The all-black and all-white colorings are not allowed.

The total number of valid symmetric scanning codes is $2^{10}-2=1022$. Thus, B is the correct answer.

Solution:

Substitute $y=x^2-a$ into $x^2+y^2=a^2$. This gives $x^2+(x^2-a)^2=a^2$, so $x^2(x^2-(2a-1))=0$.

The factor $x^2=0$ always gives the single point $(0,-a)$. The other factor gives two additional real points exactly when $2a-1>0$.

There are exactly three intersection points when $a>\dfrac12$. Thus, E is the correct answer.

Solution:

From $\gcd(a,b)=24=2^3\cdot3$ and $\gcd(b,c)=36=2^2\cdot3^2$, the number $a$ is divisible by $2^3\cdot3$ but not by $3^2$.

From $\gcd(b,c)=36$ and $\gcd(c,d)=54=2\cdot3^3$, the number $d$ is divisible by $2\cdot3^3$ but not by $2^2$. Therefore $\gcd(d,a)=2\cdot3\cdot n$, where $n$ has no factor $2$ or $3$.

Since $70<6n<100$, we have $12<n<17$. The only possible integer $n$ with no factor $2$ or $3$ is $13$, so the factor must be $13$. Thus, D is the correct answer.

Solution:

Let $x$ be the side length of $S.$ Then we can split the field up into the following shapes.

We can express the area of the field in two ways: $$\dfrac{3 \cdot 4}{2} = x^2 + \dfrac{x(3 - x)}{2}$$$$+ \dfrac{x(4 - x)}{2} + \dfrac{2 \cdot 5}{2}.$$

Simplifying yields $$6 = \dfrac{7x}{2} + 5$$$$x = \dfrac{2}{7}.$$

The desired fraction is $$\dfrac{6 - x^2}{6} = \dfrac{6 - \frac{4}{49}}{6} = \dfrac{145}{147}.$$

Thus, D is the correct answer.

Solution:

Let $BC = a, BG = x, GC = y,$ and $h$ be the length of the altitude through $A.$

By the angle bisector theorem, we get that $$\dfrac{50}{x} = \dfrac{10}{y},$$ where $y = a - x.$ Substituting yields $BG = \frac{5a}{6}.$ We also know that $DF = \frac{5a}{12}$ due to similar triangles.

Note that the height of the trapezoid is $\frac{1}{2}h,$ and $\frac{ah}{2} = 120.$ The area of the trapezoid is $$\dfrac{5a}{8} \cdot \dfrac{h}{2} = \dfrac{5}{8} \cdot \dfrac{ah}{2} = 75.$$ Thus, D is the correct answer.

Solution:

We can use the formula for the sum of a geometric sequence to rewrite $A_n, B_n,$ and $C_n.$

$$\begin{gather*} A_n = a(11\cdots11) \\ =a(1 + 10 + 10^2 + \cdots + 10^{n - 1}) \\ =a \cdot \dfrac{10^n - 1}{9} \end{gather*}$$

Similarly, we get that $$B_n = b \cdot \dfrac{10^n - 1}{9}$$ and $$C_n = c \cdot \dfrac{10^{2n} - 1}{9}.$$

We can substitute these expressions into our condition to get $$c \cdot \dfrac{10^{2n} - 1}{9} - b \cdot \dfrac{10^n - 1}{9}$$ $$= a^2 \left(\dfrac{10^n - 1}{9}\right)^2.$$

Simplifying yields $$\begin{align*} c(10^n + 1) - b &= a^2 \cdot \dfrac{10^n - 1}{9} \\ 9c(10^n + 1) - 9b &= a^2 \cdot (10^n - 1) \\ (9c - a^2)10^n &= 9b - 9c - a^2. \end{align*}$$

From the last line, we see that $9c - a^2$ and $9b - 9c - a^2$ are constants.

For there to be at least $2$ unique values of $n$ that satisfy the equation, both sides must equal zero.

We can see this by realizing that this equation is linear with respect to $10^n.$ If both sides are non-zero, then there cannot exist $2$ unique solutions to a linear equation.

This tells us that $$9c - a^2 = 0$$ and $$9b - 9c - a^2 = 0.$$

The first equation gives us $$c = \dfrac{a^2}{9}.$$ Plugging this into the second equation gives us $$b = \dfrac{2a^2}{9}.$$

This tells us that $a$ must be divisible by $3.$ This gives us the following triples:$$(3, 2, 1), (6, 8, 4), (9, 18, 9).$$

The last triple is not allowed, so the maximum sum is $$6 + 8 + 4 = 18.$$ Thus, D is the correct answer.