Solution:

$$\begin{align*} (2^2 - 2) - &(3^2 - 3) + (4^2 - 4) \\ &= 2 - 6 + 12 \\ &= 8. \end{align*}$$

Thus, D is the correct answer.

Solution:

Let $x$ be the number of students in Lara's high school. Then Portia's high school has $3x$ students.

Therefore, $$\begin{align*} 3x + x &= 2600 \\ x &= 650. \end{align*}$$ Then $3x = 1950.$

Thus, C is the correct answer.

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Written solution:

Let $x$ and $y$ be the two numbers. WLOG, let $x$ be divisible by $10.$ Then the units digit of $x$ is $0.$

If we erase the units digit, then we are essentially dividing $x$ by $10.$ The problem statement also gives us that $\dfrac{x}{10} = y.$

Therefore, $$\begin{align*} \dfrac{x}{10} + x &= 17,402 \\ \dfrac{11x}{10} &= 17,402 \\ x &= 15,820. \end{align*}$$ Then $$x - \dfrac{x}{10} = 14,238.$$

Thus, D is the correct answer.

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Written solution:

The distance travelled every second forms an arithmetic sequence: $$5, 5 + 7, 5 + 2 \cdot 7, \ldots$$

The sum of an arithmetic sequence is \text{# of terms} \cdot \dfrac{\text{first + last}}{2}. We know the number of terms is $30$ and the first term is $5.$ The last term is $$5 + 29 \cdot 7 = 208.$$

Plugging these values into the expression yields $$30 \cdot \dfrac{5 + 208}{2} = 15 \cdot 213 = 3195.$$

Thus, D is the correct answer.

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Written solution:

The sum of the scores of everyone in the class is $8k.$ The sum of the scores in the collection of $12$ is $12 \cdot 14 = 168.$

This means that the sum of the scores of everyone not in the collection is $8k - 168.$ There are also $k - 12$ people not in the collection. Therefore, the average is $$\dfrac{8k - 168}{k - 12}.$$

Thus, B is the correct answer.

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Written solution:

Let $2d$ be the distance from the fire tower, where $d > 0.$

Then Chantal hiked for $$\dfrac{d}{4} + \dfrac{d}{2} + \dfrac{d}{3} = \dfrac{13d}{12}$$ hours.

If Jean travelled $d$ miles in $\dfrac{13d}{12}$ hours, then his speed was $$d \div \dfrac{13d}{12} = \dfrac{12}{13}$$ miles per hour.

Thus, A is the correct answer.

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Written solution:

Note that the third condition ensures that purple snakes can't add.

We also know that all happy snakes can add, which means that happy snakes can't be purple as well.

Thus, D is the correct answer.

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Written solution:

We can express $\underline{1}.\overline{\underline{a} \ \underline{b}}$ as an infinite geometric sum: $$\underline{1}.\overline{\underline{a} \ \underline{b}} = 1 + .\underline{a} \ \underline{b} + .00 \ \underline{a} \ \underline{b} + \cdots$$ We can therefore use the formula for the sum of a geometric sum: $$S = \dfrac{\text{first term}}{1 - \text{ratio}} = \dfrac{.\underline{a} \ \underline{b}}{1 - \dfrac{1}{100}}$$ $$= \dfrac{100}{99} \left(. \underline{a}\ \underline{b}\right) = \dfrac{\underline{a}\ \underline{b}}{99}.$$ We also know that $$1. \underline{a}\ \underline{b} = 1 + \dfrac{\underline{a}\ \underline{b}}{100}$$

Then $$\begin{align*} 66\left(1 + \dfrac{\underline{a}\ \underline{b}}{100}\right) + .5 &= 66\left(1 + \dfrac{\underline{a}\ \underline{b}}{99}\right) \\ \dfrac{66}{100}\underline{a}\ \underline{b} + .5 &= \dfrac{66}{99}\underline{a}\ \underline{b} \\ \dfrac{1}{150}\underline{a}\ \underline{b} &= .5 \\ \underline{a}\ \underline{b} &= 75. \end{align*}$$

Thus, E is the correct answer.

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Written solution:

Expanding, we get $$\begin{gather*} x^2y^2 - 2xy + 1 + x^2 + 2xy + y^2 \\ = x^2y^2 + x^2 + y^2 + 1. \end{gather*}$$ Note that every square must be non-negative. Therefore, the minimum value is when all the terms except $1$ are $0,$ making the sum $1.$

This is attainable when $x = y = 0.$

Thus, D is the correct answer.

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Written solution:

Notice that if we multiply by $3 - 2 = 1,$ we end up with a bunch of difference of squares.

$$\begin{gather*} (3 - 2)(2 + 3) = 3^2 - 2^2 \\ (3^2 - 2^2)(2^2 + 3^2) = 3^4 - 2^4\\ \vdots \end{gather*}$$ This ends up giving us a final value of $3^{128} - 2^{128}.$

Thus, C is the correct answer.

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Written solution:

We can express this expression in base $10$ using the definition of bases: $$2b^3 + 2b + 1 - 2b^2 - 2b - 1$$$$= 2b^3 - 2b^2 = 2b^2(b - 1).$$

For this to be divisible by $3,$ either $b$ or $b - 1$ must be divisible by $3.$

The only answer choice that satisfies neither of these conditions is $8.$

Thus, E is the correct answer.

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Written solution:

Let the heights of the narrow wide cones be $h_1$ and $h_2$ respectively.

Then $$\dfrac{1}{3} \cdot 3^2 \pi h_1 = \dfrac{1}{3} \cdot 6^2 \pi h_2,$$ which gives us $$\dfrac{h_1}{h_2} = 4.$$ Also note that $\dfrac{3}{h_1}$ and $\dfrac{6}{h_2}$ must remain constant due to similar triangles.

After the marble is added to each cone, let $3x$ be the radius of the narrow cone and $6y$ be that of the wide cone.

Then the height of the narrow cone is $h_1x$ and that of the wide cone is $h_2y$ due to similar triangles.

Equating the final volumes, we get $$\dfrac{1}{3} (3x)^2 \pi (h_1x) = \dfrac{1}{3} (6y)^2 \pi (h_2y),$$ which gives us $$h_1x^3 = 4h_2y^3.$$ We know that $\dfrac{h_1}{h_2} = 4,$ so this equation gives us that $x = y.$

The desired ratio is $$\dfrac{h_1x - h_1}{h_2y - h_2} = \dfrac{h_1(x - 1)}{h_2(y - 1)} = 4.$$

Thus, E is the correct answer.

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Written solution:

Analyzing the side lengths, we can realize that $$\triangle ACD \text{ and } \triangle ABC$$ are both right triangles.

Therefore, we can treat $\triangle ACD$ as the base of the tetrahedron and $\overline{AB}$ as the altitude.

With this setup, we can use the formula for the volume of a pyramid, yielding $$\dfrac{1}{3} \cdot \dfrac{3 \cdot 4}{2} \cdot 2 = 4.$$

Thus, C is the correct answer.

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Written solution:

By Vieta's formulas, we get that the product of the roots is $16$ and that their sum is $10.$

Given that all the roots are positive integers, we can see that the roots are $$1, 1, 2, 2, 2, 2.$$

The function is therefore just $$(z - 1)^2(z - 2)^4 = (z^2 - 2z + 1)$$ $$(z^4 - 8z^3 + 24z^2 - 32z + 16).$$

Calculating just the $z^3$ term, we get $$-32z^3 - 48z^3 - 8z^3 = -88z^3.$$

Thus, A is the correct answer.

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Written solution:

Setting the equations equal to each other, we get $$Ax^2 + B = Cx^2 + D$$ $$x^2(A - C) = D - B$$ $$x^2 = \dfrac{D - B}{A - C} \geq 0$$ since squares are non-negative.

This means $D - B$ and $A - C$ must both have the same sign.

If we choose two distinct values for $(A, C)$ and $(B, D),$ there are $2$ ways to arrange them such that the numerator and denominator both have the same sign.

We have to divide by $2,$ however, since the two curves are not considered distinct.

Therefore, the total number of tuples is $$\dfrac{1}{2} \binom{6}{2} \binom{4}{2} \cdot 2 = 90.$$

Thus, C is the correct answer.

Solution:

The total number of numbers in the list is $$1 + 2 + \cdots + 200$$$$= \dfrac{200 \cdot 201}{2} = 20100.$$

We want to find the median $k$ such that $\dfrac{k(k + 1)}{2}$ is near $\dfrac{20100}{2}.$

Multiplying by $2,$ we want $k(k + 1)$ near $20100.$ Note that $\sqrt{20100} \approx 142.$

Plugging in $k = 142$ yields $$\dfrac{1}{2} \cdot 142 \cdot 143 = 10153.$$

$10153 - 142 < 10050,$ which shows that $142$ is our desired median ($142$ is the $10049$th and $10050$th number).

Thus, C is the correct answer.

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Written solution:

First, we can show that $\triangle BPC \sim \triangle BDA.$

Let $\angle DBC = \alpha.$ Then $\angle DCB = 180 - 2\alpha$ since $\triangle DCB$ is isosceles.

Since $\overline{AB} \parallel \overline{CD},$ we get that $\angle ABD = \alpha.$ Then since $\triangle BPC$ and $\triangle BDA$ are right triangles, they are similar.

Using this fact, we get $$2 = \dfrac{BD}{BP} = \dfrac{AB}{AC} = \dfrac{AB}{43}.$$ From this, we get that $AB = 86.$

We also get that $\triangle ABO \sim \triangle CDO$ from parallel sides and vertical angles.

Therefore $$2 = \dfrac{AB}{CD} = \dfrac{BO}{OD} = \dfrac{BP + 11}{BP - 11}.$$ Solving, we get $BP = 33$ and $BD = 66.$

Using the Pythagorean Theorem on $\triangle ADB,$ we get that $$AD = \sqrt{86^2 - 66^2} = 4\sqrt{190}.$$

Thus, D is the correct answer.

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Written solution:

Note that for any number of the form $p^e$ where $p$ is prime, $$f(p^e) = ef(p) = ep.$$ This can be seen by applying the function property multiple times.

Also note that $$\begin{align*}f(a) &= f\left(\dfrac{a}{b} \cdot b\right) \\&= f\left(\dfrac{a}{b}\right) + f(b),\end{align*}$$ which gives us $$f\left(\dfrac{a}{b}\right) = f(a) - f(b).$$

We can know calculate each answer choice one-by-one.

$$f\left(\dfrac{17}{32}\right) = f(17) - f(32)$$ $$= 17 - 5 \cdot 2 = 7.$$

$$\begin{gather*} f\left(\dfrac{11}{16}\right) = f(11) - f(16) \\ = 11 - 4 \cdot 2 = 3 \end{gather*}$$

$$\begin{gather*} f\left(\dfrac{7}{9}\right) = f(7) - f(9) \\ = 7 - 2 \cdot 3 = 1 \end{gather*}$$

$$\begin{gather*} f\left(\dfrac{7}{6}\right) = f(7) - f(6) \\ = 7 - f(2) - f(3) = 2 \end{gather*}$$

$$\begin{gather*} f\left(\dfrac{25}{11}\right) = f(25) - f(11) \\ = 2 \cdot 5 - 11 = -1 \end{gather*}$$

Thus, E is the correct answer.

Solution:

We can case on the signs of $x - y$ and $x + y.$

Case $1:$$$|x - y| = x - y,$$$$|x + y| = x + y$$

Substituting and simplifying, we get $$\begin{gather*} x^2 - 6x + y^2 = 0 \\ (x - 3)^2 + y^2 = 3^2. \end{gather*}$$ This is a circle with radius $3$ centered at $(3, 0).$

Case $2:$$$|x - y| = y - x,$$$$|x + y| = x + y$$

As above, we get $$\begin{gather*} x^2 + y^2 - 6y = 0 \\ x^2 + (y - 3)^2 = 3^2. \end{gather*}$$ This is a circle with radius $3$ centered at $(0, 3).$

Case $3:$ $$|x - y| = x - y,$$$$|x + y| = -x - y$$

Again, we get $$\begin{gather*} x^2 + y^2 + 6y = 0 \\ x^2 + (y + 3)^2 = 3^2. \end{gather*}$$ This is a circle with radius $3$ centered at $(0, -3).$

Case $4:$ $$|y - x| = x - y,$$$$|x + y| = -x - y$$

Finally, we get $$\begin{gather*} x^2 + 6x + y^2 = 0 \\ (x + 3)^2 + y^2 = 3^2. \end{gather*}$$ This is a circle with radius $3$ centered at $(-3, 0).$

These circles form the following graph:

The $4$ semicircles form $2$ full two circle with radius $3$ for a total area of $18\pi.$

The middle square also contributes $6^2 \pi = 36 \pi$ to the total area. This means that the area of the region is $54 \pi.$

Thus, E is the correct answer.

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Written solution:

Note that all the sequences are symmetric about $3.$ $$x_1, x_2, x_3, x_4, x_5$$ is a valid sequence if and only if $$6 - x_1, 6 - x_2, 6 - x_3,$$$$6 - x_4, 6 - x_5$$ is a valid sequence.

Therefore, we can just count all the sequences that begin with $1,$ $2,$ $31,$ and $32.$

• $1, 3, 2, 5, 4$

• $1, 4, 2, 5, 3$

• $1, 4, 3, 5, 2$

• $1, 5, 2, 4, 3$

• $1, 5, 3, 4, 2$

• $2, 1, 4, 3, 5$

• $2, 1, 5, 3, 4$

• $2, 3, 1, 5, 4$

• $2, 4, 1, 5, 3$

• $2, 4, 3, 5, 1$

• $2, 5, 1, 4, 3$

• $2, 5, 3, 4, 1$

• $3, 1, 4, 2, 5$

• $3, 1, 5, 2, 4$

• $3, 2, 4, 1, 5$

• $3, 2, 5, 1, 4$

This shows that there are $16$ valid sequences starting with the above numbers. Doubling this yields the total number of sequences, $32.$

Thus, D is the correct answer.

Solution:

Let $P,$ $Q,$ $R,$ $X,$ $Y,$ and $Z$ be the points at $\overleftrightarrow{AB}\cap\overleftrightarrow{CD},$$\overleftrightarrow{CD}\cap\overleftrightarrow{EF},$$\overleftrightarrow{EF}\cap\overleftrightarrow{AB},$$\overleftrightarrow{BC}\cap\overleftrightarrow{DE},$$\overleftrightarrow{DE}\cap\overleftrightarrow{FA},$ and $\overleftrightarrow{FA}\cap\overleftrightarrow{BC},$ respectively.

Since $ABCDEF$ is equiangular, all of its interior angles are $720^{\circ} \div 6 = 120^{\circ}.$

This means that the exterior angles are all $60^{\circ},$ making all the small triangles equilateral triangles. This also shows that $\triangle PQR$ and $\triangle XYZ$ are also equilateral.

We know from the problem statement that $$[PQR] = \dfrac{3}{4} \cdot PQ^2 = 192\sqrt{3},$$ $$[XYZ] = \dfrac{3}{4} \cdot YZ^2 = 324\sqrt{3}.$$ Solving gives us that $PQ = 16\sqrt{3}$ and $YZ = 36.$

Finally, the perimeter of $ABCDEF$ is $$AB + BC + CD + DE + EF$$ $$+ FA = AZ + PC + CD + DQ$$ $$+ YF + FA = (YF + FA + AZ)$$ $$+ (PC + CD + DQ) = YZ$$ $$+ PQ = 36 + 16\sqrt{3}.$$

With this, we have $(36 + 16 + 3 = 55).$

Thus, C is the correct answer.

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Written solution:

Let the sheets the roommate took be $a$ through $b.$ This is the same as taking pages $2a - 1$ through $2b.$

The sum of the pages taken is $$\dfrac{(2a - 1 + 2b)(2b - 2a + 2)}{2}.$$ The sum of the pages remaining is $$19(50 - (2b - 2a + 2)).$$ The sum of these equals the sum of all the pages, $\dfrac{50 \cdot 51}{2}.$

Therefore $$\dfrac{(2a - 1 + 2b)(2b - 2a + 2)}{2} +$$$$19(50 - (2b - 2a + 2)) = \dfrac{50 \cdot 51}{2}.$$

Then $$(2a + 2b - 39)(b - a + 1)$$$$= \dfrac{50 \cdot 13}{2} = 25 \cdot 13.$$

We can set $$2a + 2b - 39 = 25$$ and $$b - a + 1 = 13$$ to get $$a + b = 32$$ and $$b - a = 12.$$ This yields $$b = 22 \text{ and } a = 12.$$

The desired answer is $22 - 10 + 1 = 13.$

Thus, B is the correct answer.

Solution:

Let $M$ denote the center, $E$ an edge, and $C$ a corner.

The only ways Frieda can reach a corner in $4$ or less moves are $$EC, EEC, EEEC, \text{ and } EMEC.$$

We have to find the probability of each of these cases happening.

Case $1 : EC$

On the first hop, Frieda necessarily moves to an $E.$ From there, Frieda has a $\dfrac{1}{2}$ chance of going to a $C.$ This gives us a probability of $$1 \cdot \dfrac{1}{2} = \dfrac{1}{2}$$ for this case.

Case $2 : EEC$

Again, Frieda moves to an $E$ on her first move. On the second move, however, she has a $\dfrac{1}{4}$ chance of wrapping around to another $E.$ Then there is a $\dfrac{1}{2}$ chance of going to a $C$ from the $E.$ This is a probability of $$1 \cdot \dfrac{1}{4} \cdot \dfrac{1}{2} = \dfrac{1}{8}.$$

Case $3 : EEEC$

Using the same logic as above, this case yields a probability of $$1 \cdot \dfrac{1}{4} \cdot \dfrac{1}{4} \cdot \dfrac{1}{2} = \dfrac{1}{32}.$$

Case $ : EMEC

Finally, this case has a probability of $$1 \cdot \dfrac{1}{4} \cdot 1 \cdot \dfrac{1}{2} = \dfrac{1}{8.}$$

Adding the probabilities together yields a total probability of $$\dfrac{1}{2} + \dfrac{1}{8} + \dfrac{1}{32} + \dfrac{1}{8} = \dfrac{25}{32}.$$

Thus, D is the correct answer.

Solution:

Note that each of the equations yields two parallel lines.

$$(x + ay)^2 = 4a^2$$ results in the two lines $$x + ay - 2a = 0$$ and $$x + ay + 2a = 0.$$ Both of these lines have a slope of $-\dfrac{1}{a}.$