Answer: D

Solution:

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

Thus, D is the correct answer.

Answer: C

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.

Answer: D

Video solution:

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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.

Answer: D

Video solution:

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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.

Answer: B

Video solution:

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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.

Answer: A

Video solution:

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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.

Answer: D

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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.

Answer: E

Video solution:

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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.

Answer: D

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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.

Answer: C

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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.

Answer: E

Video solution:

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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.

Answer: E

Video solution:

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Written solution:

Let the initial liquid heights in the narrow and wide cones be $h_1$ and $h_2$. Since the liquid volumes are equal,

$$\frac13\pi(3)^2h_1=\frac13\pi(6)^2h_2,$$

so $h_1=4h_2$.

After the identical marbles are dropped in, each cone must contain the same final volume below the liquid surface: the original liquid volume plus the volume of one marble. If the new liquid-surface radii are $3x$ and $6y$, similarity gives new heights $h_1x$ and $h_2y$. Thus

$$\frac13\pi(3x)^2h_1x=\frac13\pi(6y)^2h_2y.$$

Using $h_1=4h_2$, this simplifies to $x^3=y^3$, so $x=y$. The rise ratio is therefore

$$h_1(x-1):h_2(y-1)=h_1:h_2=4:1.$$

Thus, E is the correct answer.

Answer: C

Video solution:

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Written solution:

Place $A=(0,0,0)$, $B=(2,0,0)$, $C=(0,3,0)$, and $D=(0,0,4)$. Then

$$BC=\sqrt{2^2+3^2}=\sqrt{13},\quad BD=\sqrt{2^2+4^2}=2\sqrt5,\quad CD=\sqrt{3^2+4^2}=5,$$

so this coordinate model matches all the given edge lengths. The tetrahedron is a rectangular-corner tetrahedron with perpendicular edge lengths $2,3,4$ from $A$, so its volume is

$$\frac16(2)(3)(4)=4.$$

Thus, C is the correct answer.

Answer: A

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

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

Thus, A is the correct answer.

Answer: C

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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.

Answer: C

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.

Answer: D

Video solution:

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Written solution:

Because $BC=CD$, the median from $C$ to $BD$ is perpendicular to $BD$. Thus $\triangle BPC$ is a right triangle. Let $\angle DBC=\alpha$. Since $AB\parallel CD$, we also have $\angle ABD=\alpha$, so $\triangle BPC\sim\triangle BDA$.

Since $P$ is the midpoint of $BD$, we have $BD/BP=2$. In the similarity, $BC$ corresponds to $AB$, so

$$\frac{AB}{BC}=2,\qquad AB=2\cdot43=86.$$

Also, $\triangle ABO\sim\triangle CDO$, so

$$\frac{BO}{OD}=\frac{AB}{CD}=2.$$

Since $OP=11$ and $P$ is the midpoint of $BD$, write $BP=PD=t$. Then $BO=t+11$ and $OD=t-11$, so

$$\frac{t+11}{t-11}=2.$$

This gives $t=33$, hence $BD=66$. Finally, $\triangle ABD$ is right, so

$$AD=\sqrt{AB^2-BD^2}=\sqrt{86^2-66^2}=4\sqrt{190}.$$

Thus $m+n=4+190=194$.

Thus, D is the correct answer.

Answer: E

Video solution:

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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 now 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.

Answer: E

Solution:

Consider the four sign cases for $x-y$ and $x+y$. In one case, for example, $|x-y|=x-y$ and $|x+y|=x+y$, so

$$x^2+y^2=6x\quad\Longrightarrow\quad (x-3)^2+y^2=9.$$

The other three cases similarly give circles of radius $3$ centered at $(0,3)$, $(-3,0)$, and $(0,-3)$. The relevant arcs form the boundary shown by these four congruent circle pieces.

The region consists of a central square of side length $6$, together with four semicircles of radius $3$. The square contributes area $36$, and the four semicircles have the area of two full radius-$3$ circles, namely $18\pi$.

Therefore the area is $36+18\pi$, so $m+n=36+18=54$.

Thus, E is the correct answer.

Answer: D

Video solution:

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Written solution:

A permutation is valid exactly when the four comparison signs between consecutive terms alternate. Thus the signs must be either up-down-up-down or down-up-down-up.

For the up-down-up-down pattern, a direct count by the middle value, or equivalently the standard alternating-permutation count for five distinct numbers, gives $16$ permutations. Replacing every entry $x$ by $6-x$ gives a bijection from these to the down-up-down-up permutations, so there are another $16$.

The total number of valid rearrangements is $16+16=32$.

Thus, D is the correct answer.

Answer: C

Solution:

Let the intersections of lines $AB,CD,EF$ form triangle $PQR$, and let the intersections of lines $BC,DE,FA$ form triangle $XYZ$. Because the hexagon is equiangular, all these outer triangles are equilateral.

For an equilateral triangle with side length $s$, the area is $\frac{\sqrt3}{4}s^2$. Hence

$$\frac{\sqrt3}{4}PQ^2=192\sqrt3,\qquad \frac{\sqrt3}{4}YZ^2=324\sqrt3.$$

So $PQ=16\sqrt3$ and $YZ=36$. The perimeter of the hexagon is the sum of the side lengths cut out of these two equilateral triangles, which is

$$PQ+YZ=16\sqrt3+36.$$

Thus $m+n+p=36+16+3=55$.

Thus, C is the correct answer.

Answer: B

Video solution:

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Written solution:

Suppose the borrowed sheets are sheets $a$ through $b$. Then the borrowed pages run from $2a-1$ through $2b$, so there are $2(b-a+1)$ borrowed pages.

The total sum of all page numbers is $50\cdot51/2=1275$. If the remaining pages have mean $19$, then

$$1275-\frac{(2a-1+2b)\cdot 2(b-a+1)}{2}=19\bigl(50-2(b-a+1)\bigr).$$

Simplifying gives

$$(2a+2b-39)(b-a+1)=325=25\cdot13.$$

The positive solution with consecutive sheets in the middle is

$$2a+2b-39=25,\qquad b-a+1=13.$$

Thus $a+b=32$ and $b-a=12$, so $a=10$ and $b=22$. Therefore $13$ sheets were borrowed.

Thus, B is the correct answer.

Answer: D

Solution:

Classify a square as $M$ for the center, $E$ for a non-corner edge square, and $C$ for a corner. Frieda starts at $M$, and the first hop always takes her to an $E$.

From an edge square, the probabilities of moving to $C,E,M$ are $\frac12,\frac14,\frac14$, respectively. From $M$, the next hop always goes to an $E$.

Now count the possible first-hit patterns within four hops:

$$EC:\quad 1\cdot\frac12=\frac12,$$

$$EEC:\quad 1\cdot\frac14\cdot\frac12=\frac18,$$

$$EEEC:\quad 1\cdot\frac14\cdot\frac14\cdot\frac12=\frac1{32},$$

$$EMEC:\quad 1\cdot\frac14\cdot1\cdot\frac12=\frac18.$$

Adding gives

$$\frac12+\frac18+\frac1{32}+\frac18=\frac{25}{32}.$$

Thus, D is the correct answer.

Answer: D

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 $-\dfrac{1}{a}.$

Similarly, $$(ax-y)^2 = a^2$$ results in the lines $$ax - y - a = 0$$ and $$ax - y + a = 0.$$ These lines have slope $a.$

Note that each pair of lines is perpendicular to the other pair of lines. This shows that the equations form a rectangle.

Recall that the formula for the distance $d$ between two parallel lines $$\begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases}$$ is $$d = \dfrac{\mid C_2 - C_1 \mid}{\sqrt{A^2 + B^2}}.$$

Using this formula, we get that the distance between the first pair of lines is $$\dfrac{4a}{\sqrt{a^2 + 1}}.$$ Similarly, the distance between the second pair of lines is $$\dfrac{2a}{\sqrt{a^2 + 1}}.$$

These are the side lengths of the rectangle. Multiplying yields the area $$\dfrac{8a^2}{a^2 + 1}.$$

Thus, D is the correct answer.

Answer: E

Solution:

Let the colors be $A, B,$ and $C.$ Note that we can assign the $3$ colors to them in $3! = 6$ ways, so we have to multiply by $6$ at the end.

Let $A$ be in the center of the grid.

$$\begin{array}{ccc} ? & ? & ? \\ ? & A & ? \\ ? & ? & ? \end{array}$$

The other $A$s have to either be along the diagonal or on the same side.

$$\begin{array}{ccc} ? & ? & A \\ ? & A & ? \\ A & ? & ? \end{array}$$

$$\begin{array}{ccc} A & ? & A \\ ? & A & ? \\ ? & ? & ? \end{array}$$

The first scenario can happen in $2$ ways since there are $2$ diagonals. The second has $4$ ways since there are $4$ sides.

Either way, the positions of the $B$s and $C$s is fixed.

$$\begin{array}{ccc} C & B & A \\ B & A & C \\ A & C & B \end{array}$$

$$\begin{array}{ccc} A & B & A \\ C & A & C \\ B & C & B \end{array}$$

This is a total of $4 + 2 = 6$ ways to arrange the $A, B,$ and $C$s. This gives us a total of $6 \cdot 6 = 36$ configurations.

Thus, E is the correct answer.