Solution:

We can simplify this expression as follows:

$$\begin{aligned} 3 +& \dfrac{1}{3 + \dfrac{1}{3 + \dfrac{1}{3}}} \\ &= 3 + \dfrac{1}{3 + \dfrac{1}{\dfrac{10}{3}}} \\ &= 3 + \dfrac{1}{3 + \dfrac{3}{10}} \\ &= 3 + \dfrac{1}{\dfrac{33}{10}} \\ &= 3 + \dfrac{10}{33} \\ &= \dfrac{109}{33}. \end{aligned}$$

Thus, D is the correct answer.

Solution:

We can set up a proportion to solve this problem:

$$\dfrac{15}{57} = \dfrac{x}{27}.$$

Cross multiplying, we get $$x = \dfrac{15}{57} \cdot 27 = \dfrac{135}{19} \approx 7.$$

Thus, B is the correct answer.

Solution:

Let $x, y,$ and $z$ be the three numbers. The conditions from the problem give us the following relations:

$$\begin{aligned} x+y+z&=96 &&\text{(1)} \\ x&=6z &&\text{(2)} \\ z&=y-40 &&\text{(3)}. \end{aligned}$$

Rearranging $(3),$ we get $y = z + 40.$ Plugging this new equation and $(2)$ into $(1),$ we get $$6z + z + 40 + z = 96$$ $$8z + 40 = 96$$ $$8z = 56 \Rightarrow z = 7.$$

From this, we get that $$x = 6 \cdot z = 6 \cdot 7 = 42$$ and $$y = z + 40 = 7 + 40 = 47.$$

Therefore, $y - x = 47 - 42 = 5.$

Thus, E is the correct answer.

Video solution:

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Written solution:

A car that gets $x$ miles per gallon uses $\frac{1}{x}$ gallons per mile.

Since $1$ mile is $\frac{1}{m}$ kilometers, this is $\frac{m}{x}$ gallons per kilometer.

Multiplying by $l$ liters per gallon gives $\frac{lm}{x}$ liters per kilometer, so for $100$ kilometers the fuel used is $\frac{100lm}{x}$ liters.

Thus, E is the correct answer.

Solution:

Consider the diagram:

Since $AP = QC = s,$ we know that $PB = BQ.$ This shows that $\triangle PBQ$ is an isosceles right triangle with hypotenuse $PQ=s.$ Using the Pythagorean theorem, we get that $PB = \dfrac{s}{\sqrt{2}}.$

We also know that $$1 = AB = AP + PB = s + \dfrac{s}{\sqrt{2}}.$$

This equation simplifies to $$1 = (1 + \dfrac{1}{\sqrt{2}})s$$ Which implies that $$s = \dfrac{1}{1 + \dfrac{1}{\sqrt{2}}} = \dfrac{\sqrt{2}}{\sqrt{2} + 1}.$$

We can rationalize this fraction to get

$$\dfrac{\sqrt{2}}{\sqrt{2} + 1} \cdot \dfrac{\sqrt{2} - 1}{\sqrt{2} - 1} = 2 - \sqrt{2}.$$

Thus, C is the correct answer.

Solution:

By the definition of square root, we get that $$\sqrt{(a - 1)^2} = |a - 1|.$$

Since $a < 0,$ we get that $a - 1 < 0,$ which means that $|a - 1| = 1 - a.$

The whole expression therefore simplifies to $$|a - 2 - (1 - a)| = |2a - 3|.$$ Since $a < 0,$ we know that $2a - 3 < 0.$ This means that $|2a - 3| = 3 - 2a.$

Thus, A is the correct answer.

Solution:

Prime factorize $18=2\cdot3^2,$ $45=3^2\cdot5,$ and $180=2^2\cdot3^2\cdot5.$

The lcm condition forces the exponent of $2$ in $n$ to be $2,$ the exponent of $5$ to be $1,$ the exponent of $3$ to be at most $2,$ and every other prime exponent to be $0.$ The gcd condition $\gcd(n,45)=15=3\cdot5$ further forces the exponent of $3$ in $n$ to be exactly $1.$

Therefore $n=2^2\cdot3\cdot5=60,$ and the sum of its digits is $6.$

Thus, B is the correct answer.

Solution:

The average of the $6$ numbers is $$\dfrac{1 + 7 + \cdots + X}{6} = \dfrac{20 + X}{6}.$$

This value can equal any of the terms in the set, so we can case on what it equals.

$$\dfrac{20 + X}{6} = 1 \iff X = -14$$

$$\dfrac{20 + X}{6} = 7 \iff X = 22$$

$$\dfrac{20 + X}{6} = 5 \iff X = 10$$

$$\dfrac{20 + X}{6} = 2 \iff X = -8$$

$$\dfrac{20 + X}{6} = X \iff X = 4$$

Adding up all the positive values for $X,$ we get $36.$

Thus, D is the correct answer.

Solution:

There are $5$ choices for the color of the bottom left rectangle. This forces there to be $4$ choices for the top left rectangle. The middle bottom rectangle touches both of the previous ones, so there are $3$ color options for this rectangle.

The rectangle in the top right is also limited to $3$ colors since it touches the two previous rectangles. Finally, the rectangle in the bottom right also has $3$ color options.

Multiplying these together, we get $5\cdot4\cdot3\cdot3\cdot3=540$ total colorings.

Thus, D is the correct answer.

Solution:

We can label $a$ and $b$ as the width and height as in the diagram. Then we get that $$a^2 + b^2 = 64$$ and $$(a - 2)^2 + (b - 2)^2 = 32.$$

The latter expression simplifies to $$a^2 + b^2 - 4a - 4b + 4 + 4 = 32,$$ which is the same as $$72 - 4(a + b) = 32.$$ From this we get $$a + b = 10.$$

Squaring this, we get $$a^2 + b^2 + 2ab = 100,$$ which gets us that $$2ab = 36,$$ which means that the area $(ab)$ is $18.$

Thus, E is the correct answer.

Solution:

We can rewrite $4096$ as $2^{12},$ so $\dfrac{1}{4096} = 2^{-12}.$ Then if we equate the given expressions, we get $$2^m \cdot 2^{-6} = 2 \cdot 2^{\frac{-12}{m}}.$$ Equating the exponents, we get $$m - 6 = 1 + \dfrac{-12}{m}.$$

Multiplying by $m,$ we get $$m^2 - 6m = m - 12$$ and so $$m^2 - 7m + 12 = 0$$$$(m-4)(m-3)$$$$m=4,~m=3$$

Therefore, we can see that the sum of the solutions is $7.$

Thus, C is the correct answer.

Solution:

For the first question, the truth-tellers will respond yes, the liars will respond yes, and the alternaters who decided to lie first will say yes. The alternaters who decide to tell the truth first will say no. Denote this as $$22 = t + l + a_l.$$

For the second questions, the liars will respond yes, and the alternaters who decided to lie first will say yes (they are forced to tell the truth for this question). The truth-tellers will respond no, and the alternaters who told the truth first would lie this round, responding no. Denote this as $$15 = l + a_l.$$

From this, we get that $t = 7.$ The principal only gives candy to children who always tell the truth in the first round, therefore only giving them $7$ candies total.

Thus, A is the correct answer.

Solution:

Consider the following diagram:

Let $Y$ be the intersection of $\overline{BD}$ and $\overline{AC}.$ By the Angle Bisector Theorem, $AB:AC=BP:PC=2:3,$ so write $AB=2x$ and $AC=3x.$

Reflection across the angle bisector $\overline{AP}$ sends ray $AB$ to ray $AC.$ Because $BY\perp AP,$ it sends $B$ to $Y.$ Thus $AY=AB=2x,$ and hence $YC=AC-AY=x.$

Since $AD\parallel BC,$ with $B,Y,D$ collinear and $A,Y,C$ collinear, we have $\triangle BYC\sim\triangle DYA.$ Therefore $$\frac{AD}{BC}=\frac{AY}{YC}=2.$$

Finally, $BC=BP+PC=5,$ so $AD=2BC=10.$

Thus, C is the correct answer.

Solution:

The numbers from $8$ through $14$ cannot be paired with one another, so they must be paired with the numbers from $1$ through $7.$ In particular, $7$ must be paired with $14,$ since no other available number is at least twice $7.$

Now let's look at what the other numbers can pair with. $8$ and $9$ can pair with any number $1-4.$ $10$ and $11$ can pair with any number $1-5,$ and $12$ and $13$ can pair with any number $1-6.$

$8$ can pair with $4$ numbers, but then $9$ only has $3$ options since $8$ took one. $10$ then has $3$ options, since $2$ choices are taken, but it has one more to choose from $(5).$ $11$ then has $2$ options, $12$ has $2$ options, and $13$ only has $1.$

Multiplying these together yields $$4 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 144.$$

Thus, E is the correct answer.

Solution:

Notice that $7^2 + 24^2$ and $15^2 + 20^2$ are both the same. This forces $AC = 25$ since otherwise $\angle B$ and $\angle D$ would both be acute or obtuse, violating the fact that their sum is $180^{\circ}.$

Also since $\angle B$ is right, we know that $AC$ is the diameter of the circle. The area of the circle is then $\dfrac{625}{4} \pi.$

To find the area of the quadrilateral, we can find the area of each of the triangles, which is $$\dfrac{1}{2}(7 \cdot 24 + 20 \cdot 15) =$$ $$84 + 150 = 234.$$

To find the area outside the quadrilateral, we subtract to get $$\dfrac{625}{4} \pi - 234 = \dfrac{625 \pi - 936}{4}.$$

Therefore, $$a + b + c = 625 + 936 + 4$$ $$= 1565.$$

Thus, D is the correct answer.

Solution:

Let $h, l,$ and $w$ be the dimensions of the old box. Then the volume of the new box is $$(h + 2)(l + 2)(w + 2).$$ Expanding, we get $$hlw + 2(hl + hw + lw)$$$$+ 4(l + h + w) + 8.$$ We can use Vieta's formulas to find the terms in this expression. We get that $$hlw = -\dfrac{D}{A} = \dfrac{3}{5},$$ $$hl + hw + lw = \dfrac{C}{A} = \dfrac{29}{10},$$ and $$l + h + w = -\dfrac{B}{A} = \dfrac{39}{10}.$$

Plugging these values into the expression, we get $$\dfrac{3}{5} + 2 \cdot \dfrac{29}{10} + 4 \cdot \dfrac{39}{10} + 8 = 30.$$

Thus, D is the correct answer.

Solution:

The repeating decimals satisfy $$\begin{aligned} 0.\overline{\underline{a}\underline{b}\underline{c}}&=\frac{100a+10b+c}{999},\\ 0.\overline a&=\frac a9,\quad 0.\overline b=\frac b9,\\ 0.\overline c&=\frac c9. \end{aligned}$$

Substitution and multiplication by $999$ give $$100a+10b+c=37(a+b+c),$$ or $$7a=3b+4c.$$

For each $a\in\{1,2,\ldots,9\},$ checking the nonzero digits $b,c$ in this linear equation gives $$\begin{gathered} (1,1,1),(2,2,2),(3,3,3),\\ (4,4,4),(4,8,1),(5,1,8),\\ (5,5,5),(5,9,2),(6,2,9),\\ (6,6,6),(7,7,7),(8,8,8),\\ (9,9,9). \end{gathered}$$ Thus there are $13$ integers.

Thus, D is the correct solution.

Solution:

Since we are working with angles and reflections, working with polar coordinates would make this problem easier to deal with.

Let $(r, \theta)$ be a polar coordinate. Rotating this by $k$ degrees counterclockwise maps the point to $(r, \theta + k^{\circ})$ and then reflecting it maps it to $(r, 180^{\circ} - \theta - k^{\circ}).$

Therefore, we have that $$T_k(r, \theta) = (r, 180^{\circ} - \theta - k^{\circ}).$$

From this, we can see that $$T_{k + 1}(T_k(r, \theta)) =$$$$T_{k + 1}(r, 180^{\circ} - \theta - k^{\circ}) =$$$$(r, \theta - 1^{\circ}).$$

Now, let's analyze what happens to the point $(1, 0^{\circ}).$

After $T_1,$ we get $(1, 179^{\circ}).$