Solution:

The exponent in the first term is $1+2+3 = 6,$ so the first term is $2^6 = 64.$ The parenthesized sum is $2 + 4 + 8 = 14.$ Therefore the value is $64 - 14 = 50.$

Thus, the correct answer is B.

Solution:

Because $\sqrt{a^2+b^2}$ is never negative, equality requires $a + b \ge 0.$ Squaring both sides gives $a^2 + b^2 = (a+b)^2 = a^2 + 2ab + b^2,$ which simplifies to $2ab = 0,$ i.e. $ab = 0.$

Conversely, if $ab = 0$ then $a^2 + b^2 = (a+b)^2,$ and if additionally $a + b \ge 0$ then $\sqrt{a^2+b^2} = |a+b| = a+b.$ So both conditions together are exactly what is needed.

Thus, the correct answer is D.

Solution:

The larger number ends in $0,$ and erasing that digit divides it by $10$ to give the smaller number. So the larger number is $10$ times the smaller. Writing the smaller number as $x,$ the sum is $x + 10x = 11x = 17{,}402,$ giving $x = 1{,}582.$

The two numbers are $1{,}582$ and $15{,}820,$ whose difference is $15{,}820 - 1{,}582 = 14{,}238.$

Thus, the correct answer is D.

Solution:

A purple snake cannot subtract, and any snake that cannot subtract also cannot add. So every purple snake cannot add.

Every happy snake can add. Since purple snakes cannot add, no happy snake can be purple; that is, happy snakes are not purple.

Thus, the correct answer is D.

Solution:

Let $n = \overline{ab}$ be the two-digit integer. Then $1.\overline{ab} = 1 + \dfrac{n}{99}$ while the terminating value is $1.ab = 1 + \dfrac{n}{100}.$ The correct product minus the student's product is $$66\left(\frac{n}{99} - \frac{n}{100}\right) = 66 \cdot \frac{n}{9900} = \frac{n}{150}.$$

Setting $\dfrac{n}{150} = 0.5$ gives $n = 75.$

Thus, the correct answer is E.

Solution:

Let $r$ be the number of red cards and $t$ the total. From $\dfrac{r}{t} = \dfrac13$ we get $t = 3r.$ After adding $4$ black cards, $\dfrac{r}{t+4} = \dfrac14,$ so $t + 4 = 4r.$

Substituting $t = 3r$ gives $3r + 4 = 4r,$ so $r = 4$ and $t = 12.$

Thus, the correct answer is C.

Solution:

Expanding, $$(xy-1)^2 + (x+y)^2 = x^2y^2 - 2xy + 1 + x^2 + 2xy + y^2 = x^2y^2 + x^2 + y^2 + 1.$$ This factors as $(x^2 + 1)(y^2 + 1).$

Each factor is at least $1,$ so the product is at least $1,$ with equality when $x = y = 0.$

Thus, the correct answer is D.

Solution:

Working modulo $2,$ the terms $D_0, D_1, D_2, \ldots$ have parities $$E, E, O, O, O, E, O, E, E, O, O, O, E, O, \ldots$$ which repeat with period $7$ starting from $D_0$ (indeed $D_7, D_8, D_9$ have the same parities $E, E, O$ as $D_0, D_1, D_2$).

Since $2021 \equiv 5,$ $2022 \equiv 6,$ and $2023 \equiv 0 \pmod 7,$ the parities match those of $D_5, D_6, D_0,$ namely $E, O, E.$

Thus, the correct answer is C.

Solution:

Since $3 - 2 = 1,$ multiplying the product by $3 - 2$ does not change it. Then $$(3-2)(3+2) = 3^2 - 2^2,$$ and multiplying by the next factor $(3^2 + 2^2)$ gives $3^4 - 2^4,$ and so on. Each step doubles the exponent.

After using all seven factors, the product telescopes to $3^{128} - 2^{128}.$

Thus, the correct answer is C.

Solution:

The liquid in each cone forms a smaller cone similar to the container. Let the narrow liquid cone have radius $3$ and height $h_1,$ and the wide one radius $6$ and height $h_2.$ Equal volumes give $\tfrac13\pi\cdot 9\cdot h_1 = \tfrac13\pi\cdot 36\cdot h_2,$ so $h_1 = 4h_2.$

Dropping the marble raises the volume by the same amount $\Delta V = \tfrac43\pi$ in each cone, and both start with the same volume $V.$ Because a cone's volume scales as the cube of its height, the new height is $h\sqrt[3]{1 + \Delta V/V},$ so each rise equals $h\left(\sqrt[3]{1 + \Delta V/V} - 1\right).$ This factor is identical for the two cones, so the rises are in the ratio $h_1 : h_2 = 4 : 1.$

Thus, the correct answer is E.

Solution:

Reflecting the path at each bounce turns it into a single straight segment. Reflect the start $(3, 5)$ across the $y$-axis to $(-3, 5),$ and reflect the target $(7, 5)$ across the $x$-axis to $(7, -5).$ The total travel distance equals the straight-line distance between these two images: $$\sqrt{(-3 - 7)^2 + (5 - (-5))^2} = \sqrt{100 + 100} = 10\sqrt{2}.$$

Thus, the correct answer is C.

Solution:

By Vieta's formulas the six roots sum to $10$ (the negative of the $z^5$ coefficient) and multiply to $16.$ Six positive integers with sum $10$ and product $16$ must be $2, 2, 2, 2, 1, 1.$

So the polynomial is $(z - 1)^2 (z - 2)^4.$ Expanding, $$(z^2 - 2z + 1)(z^4 - 8z^3 + 24z^2 - 32z + 16) = z^6 - 10z^5 + 41z^4 - 88z^3 + 104z^2 - 64z + 16.$$ The coefficient of $z^3$ is $B = -88.$

Thus, the correct answer is A.

Solution:

Each listed number has modulus $2,$ so $z^5$ has modulus $32,$ and its real part is $32\cos(5\theta),$ where $\theta$ is the argument of $z.$ The arguments are $180^\circ,$ $150^\circ,$ $135^\circ,$ $120^\circ,$ and $90^\circ.$

Multiplying by $5$ gives $900^\circ \equiv 180^\circ,$ $750^\circ \equiv 30^\circ,$ $675^\circ \equiv 315^\circ,$ $600^\circ \equiv 240^\circ,$ and $450^\circ \equiv 90^\circ.$ The largest cosine is $\cos 30^\circ,$ from $z = -\sqrt3 + i,$ giving real part $16\sqrt3.$

Thus, the correct answer is B.

Solution:

For the first sum, $\log_{5^k} 3^{k^2} = \dfrac{k^2}{k}\log_5 3 = k\log_5 3,$ so $$\sum_{k=1}^{20} k\log_5 3 = \frac{20\cdot 21}{2}\log_5 3 = 210\log_5 3.$$

For the second sum, $\log_{9^k} 25^k = \log_9 25 = \log_3 5,$ independent of $k,$ so the sum is $100\log_3 5.$

Since $\log_5 3 \cdot \log_3 5 = 1,$ the product is $210 \cdot 100 = 21{,}000.$

Thus, the correct answer is E.

Solution:

Choosing $t$ tenors and $b$ basses is weighted by $\binom{6}{t}\binom{8}{b}.$ To keep only $t - b \equiv 0 \pmod 4,$ apply a roots of unity filter with $\omega = i:$ $$N + 1 = \frac14\sum_{j=0}^{3}(1 + i^{j})^6\,(1 + i^{-j})^8.$$

The $j = 0$ term is $2^6\cdot 2^8 = 16384.$ The $j = 2$ term has factor $(1 + i^2)^6 = 0.$ The $j = 1$ and $j = 3$ terms are $-128i$ and $128i,$ which cancel. So the sum is $16384,$ and $\dfrac{16384}{4} = 4096.$

This count includes the empty group, so $N = 4096 - 1 = 4095,$ and $N \equiv 95 \pmod{100}.$

Thus, the correct answer is D.

Solution:

The list has $1 + 2 + \cdots + 200 = \dfrac{200\cdot 201}{2} = 20100$ terms, so the median is the average of the $10050$th and $10051$st terms.

The value $n$ occupies positions up to $\dfrac{n(n+1)}{2}.$ Since $\dfrac{141\cdot 142}{2} = 10011$ and $\dfrac{142\cdot 143}{2} = 10153,$ positions $10012$ through $10153$ all equal $142.$ Both middle positions fall in this block, so the median is $142.$

Thus, the correct answer is C.

Solution:

Place $D = (0,0)$ with $B = (b, 0)$ on one axis and $A = (0, a)$ on the other, so that $AD \perp BD.$ Since $CD \parallel AB,$ write $C = t(b, -a)$ for some $t.$ Then $CD = t\sqrt{a^2+b^2}$ and $BC^2 = b^2(1-t)^2 + t^2a^2.$ Setting $BC = CD$ gives $t^2 = (1-t)^2,$ so $t = \tfrac12.$

Thus $C = \left(\tfrac{b}{2}, -\tfrac{a}{2}\right),$ and $CD = 43$ gives $a^2 + b^2 = 4\cdot 43^2 = 7396.$ The diagonal $AC$ meets $BD$ (the $x$-axis) at $O = \left(\tfrac{b}{3}, 0\right),$ while $P = \left(\tfrac{b}{2}, 0\right).$ Hence $OP = \tfrac{b}{6} = 11,$ so $b = 66.$

Then $a^2 = 7396 - 66^2 = 3040,$ so $AD = a = \sqrt{3040} = 4\sqrt{190}.$ With $m = 4$ and $n = 190,$ we get $m + n = 194.$

Thus, the correct answer is D.

Solution:

The functional equation makes $f$ completely additive: for $x = \prod p^{e_p},$ we have $f(x) = \sum e_p\, f(p) = \sum e_p\, p,$ where a prime in the denominator contributes a negative exponent (since $f(1/p) = -p$).

Evaluating: $f\!\left(\tfrac{17}{32}\right) = 17 - 5\cdot 2 = 7,$ $f\!\left(\tfrac{11}{16}\right) = 11 - 4\cdot 2 = 3,$ $f\!\left(\tfrac{7}{9}\right) = 7 - 2\cdot 3 = 1,$ $f\!\left(\tfrac{7}{6}\right) = 7 - 2 - 3 = 2,$ and $f\!\left(\tfrac{25}{11}\right) = 2\cdot 5 - 11 = -1.$ Only the last is negative.

Thus, the correct answer is E.

Solution:

Write the right side as $\cos\left(\tfrac{\pi}{2}\sin x\right) = \sin\left(\tfrac{\pi}{2} - \tfrac{\pi}{2}\sin x\right).$ Equal sines require either $$\frac{\pi}{2}\cos x = \frac{\pi}{2}(1 - \sin x) + 2\pi k \quad\text{or}\quad \frac{\pi}{2}\cos x = \pi - \frac{\pi}{2}(1 - \sin x) + 2\pi k.$$

The first reduces to $\cos x + \sin x = 1 + 4k;$ since $\cos x + \sin x \in [-\sqrt2, \sqrt2],$ only $k = 0$ works, giving $\cos x + \sin x = 1,$ with solutions $x = 0$ and $x = \tfrac{\pi}{2}$ in $[0, \pi].$ The second reduces to $\cos x - \sin x = 1,$ whose only solution in $[0, \pi]$ is $x = 0.$

The distinct solutions are $x = 0$ and $x = \tfrac{\pi}{2},$ for a total of $2.$

Thus, the correct answer is C.

Solution:

Let $V = (0, 0),$ focus $F = (0, f),$ and directrix $y = -f,$ where $f = FV.$ A point $A = (x, y)$ on the parabola satisfies $x^2 = 4fy$ and $AF = y + f = 20,$ so $y = 20 - f.$ Also $AV^2 = x^2 + y^2 = 4fy + y^2 = 441.$

Substituting $y = 20 - f:$ $$4f(20 - f) + (20 - f)^2 = 441 \;\Longrightarrow\; 3f^2 - 40f + 41 = 0.$$ By Vieta's formulas, the sum of the two possible values of $f$ is $\dfrac{40}{3}.$

Thus, the correct answer is B.

Solution:

The roots are $z = 1,$ $z = -1 \pm i\sqrt3,$ and $z = -2 \pm i\sqrt2,$ giving the points $(1, 0),$ $(-1, \pm\sqrt3),$ and $(-2, \pm\sqrt2).$ By symmetry about the $x$-axis, the ellipse has the form $Ax^2 + Cy^2 + Dx + F = 0.$

Substituting the points yields $5x^2 + 6y^2 + 9x - 14 = 0.$ Completing the square gives $$5\left(x + \tfrac{9}{10}\right)^2 + 6y^2 = \tfrac{361}{20},$$ so $a^2 = \tfrac{361}{100}$ (along $x$) and $b^2 = \tfrac{361}{120}.$ Then $e^2 = 1 - \dfrac{b^2}{a^2} = 1 - \dfrac{100}{120} = \dfrac16,$ so $e = \sqrt{\tfrac16}.$

With $m = 1$ and $n = 6,$ we get $m + n = 7.$

Thus, the correct answer is A.

Solution:

The numbers $\cos\tfrac{2\pi}{7},$ $\cos\tfrac{4\pi}{7},$ $\cos\tfrac{6\pi}{7}$ are the three roots of $8x^3 + 4x^2 - 4x - 1 = 0.$ Dividing by $8$ puts it in monic form: $$x^3 + \tfrac12 x^2 - \tfrac12 x - \tfrac18 = 0.$$

Matching coefficients, $a = \tfrac12,$ $b = -\tfrac12,$ $c = -\tfrac18.$ Therefore $abc = \tfrac12\cdot\left(-\tfrac12\right)\cdot\left(-\tfrac18\right) = \tfrac{1}{32}.$

Thus, the correct answer is D.

Solution:

Classify squares as center $C,$ edge-middle $E,$ or corner (absorbing). From $C,$ every hop lands on an $E$ square. From an $E$ square, two of the four neighbors are corners, one is the center, and one is another $E$ square, so $P(\text{corner}) = \tfrac12,$ $P(\text{center}) = \tfrac14,$ $P(E) = \tfrac14.$

Let $a_n$ be the probability of reaching a corner within $n$ hops starting from an edge square, and $c_n = a_{n-1}$ the probability starting from the center (the first hop always goes to an edge). Then $a_1 = \tfrac12$ and $a_n = \tfrac12 + \tfrac14 c_{n-1} + \tfrac14 a_{n-1}.$ Computing: $a_2 = \tfrac12 + \tfrac14\cdot\tfrac12 = \tfrac58,$ and $$a_3 = \tfrac12 + \tfrac14\cdot\tfrac12 + \tfrac14\cdot\tfrac58 = \tfrac{25}{32}.$$

Starting from the center with four hops available, the probability equals $a_3 = \tfrac{25}{32}$ (the first hop reaches an edge, leaving three hops).

Thus, the correct answer is D.

Solution:

In circle $\Omega,$ the chord $QR$ subtends the inscribed angle $\angle QPR = 60^\circ,$ so $QR = 2r\sin 60^\circ,$ giving $3\sqrt3 = r\sqrt3,$ hence $r = 3.$

Place $A = (-7, 0),$ $B = (7, 0),$ with $\Gamma: x^2 + y^2 = 49$ (upper half). Since $\Omega$ is tangent to $AB$ at $P = (p, 0),$ its center is $(p, 3).$ Subtracting the two circle equations gives the line $QR,$ and the distance from the center $(p,3)$ to $QR$ must equal $r\cos 60^\circ = \tfrac32.$ This yields $(p^2 - 31)^2 = 9p^2 + 81,$ so $p^2 = 16$ (the root $p^2 = 55$ places $P$ outside $AB$).

With $p^2 = 16,$ the distance from $P$ to line $QR$ is $\dfrac{49 - p^2}{\sqrt{4p^2 + 36}} = \dfrac{33}{10}.$ Thus $$[\triangle PQR] = \tfrac12\cdot QR\cdot d = \tfrac12\cdot 3\sqrt3\cdot\tfrac{33}{10} = \frac{99\sqrt3}{20}.$$ So $a = 99,$ $b = 3,$ $c = 20,$ and $a + b + c = 122.$