There are $1012$ people to the left of this spot and $1009$ to the right. Add those two groups plus the person themselves: $1012 + 1009 + 1 = 2022.$ Or, just as fast, the two positions overlap on one person, so $1013 + 1010 - 1 = 2022.$ Thus, B is the correct answer.

Write $10! = 10 \cdot 9 \cdot 8 \cdot 7! = 720 \cdot 7!.$ But $720 = 6!$ too, so $10! = 6! \cdot 7! = 7! \cdot 6!.$ The two terms are the same. That makes $10! - 7! \cdot 6! = 0.$ Therefore, the answer is B.

Divide by $2$ to get $|x| \le 3.5\pi \approx 10.996.$ The integers that fit run from $-10$ up to $10,$ and there are $21$ of them. Thus, E is the correct answer.

Group $g$ holds $g$ balls, so the first $g$ groups swallow $\tfrac{g(g+1)}{2}$ of them. Now $\tfrac{63 \cdot 64}{2} = 2016$ and $\tfrac{64 \cdot 65}{2} = 2080,$ which puts ball $2024$ in group $64$ (balls $2017$ through $2080$). The bins cycle $A, B, C, D, E,$ so group $g$ lands in bin number $(g - 1) \bmod 5.$ For $g = 64$ that's $63 \bmod 5 = 3,$ bin $D.$ Therefore, the answer is D.

The full sum is $1 + 3 + \cdots + 99 = 50^2 = 2500.$ Flipping a term $t$ drops the total by $2t,$ so to go negative the flipped terms have to add up to more than $1250.$ The greedy move is to flip the biggest odd numbers: flipping the top $k$ gives $99 + 97 + \cdots = k(100 - k).$ We want $k(100 - k) \gt 1250.$ At $k = 14$ it's only $1204,$ but at $k = 15$ it jumps to $1275.$ So $15$ flips do it. Thus, B is the correct answer.

The perimeter $2(\ell + w)$ with $\ell w = 2024$ is smallest when $\ell$ and $w$ are as close together as possible. Factor $2024 = 2^3 \cdot 11 \cdot 23.$ The divisor pair nearest $\sqrt{2024} \approx 45$ is $44 \times 46,$ which gives perimeter $2(44 + 46) = 180.$ Therefore, the answer is B.

Pull out the common power: $7^{2024} + 7^{2025} + 7^{2026}$ $= 7^{2024}(1 + 7 + 49)$ $= 7^{2024} \cdot 57.$ And $57 = 3 \cdot 19,$ so the product is a multiple of $19.$ The remainder is $0.$ Thus, A is the correct answer.

Since $42 = 2 \cdot 3 \cdot 7$ has $(1+1)^3 = 8$ divisors, we can pair each divisor with its complement, so $N = 42^{8/2} = 42^4.$ Only the units digit matters, and $2^4 = 16$ ends in $6,$ so $42^4$ does too. Therefore, the answer is D.

The means tell us $a + b + c = 0$ and $a^2 + b^2 + c^2 = 30.$ Square the first: $(a+b+c)^2$ $= a^2 + b^2 + c^2$ $+ 2(ab + bc + ca),$ so $0 = 30 + 2(ab + bc + ca)$ and $ab + bc + ca = -15.$ Their mean is $-15 / 3 = -5.$ Thus, A is the correct answer.

Area ratios don't change under an affine map, so drop in convenient coordinates: $A = (0,0),$ $B = (1,0),$ $C = (1,1),$ $D = (0,1),$ which makes $E = (0, \tfrac12).$ Line $AC$ is $y = x,$ and line $EB$ runs from $(0, \tfrac12)$ to $(1, 0);$ they cross at $F = (\tfrac13, \tfrac13).$ The shoelace formula gives quadrilateral $CDEF$ area $\tfrac{5}{12}$ and triangle $CFB$ area $\tfrac13.$ So the ratio is $\tfrac{5}{12} : \tfrac13 = 5 : 4.$ Therefore, the answer is A.

Set $X = (0,0),$ $Y = (8,0),$ $W = (0,4),$ $Z = (8,4),$ so $M = (m, 0)$ and $A = (8, a).$ The right angle means $\overrightarrow{MW} \cdot \overrightarrow{MA} = 0,$ which gives $-m(8 - m) + 4a = 0,$ that is $m(8 - m) = 4a.$ Equal areas $[WXM] = 2m$ and $[WAZ] = 4(4 - a)$ force $m = 8 - 2a,$ so $a = \tfrac{8 - m}{2}.$ Substitute back and $(8 - m)(2 - m) = 0.$ Taking $M \ne Y$ leaves $m = 2$ and $a = 3.$ Then $[WMA] = 32$ $- [WXM]$ $- [MYA]$ $- [AZW]$ $= 32 - 4 - 9 - 4$ $= 15.$ Thus, C is the correct answer.

Give each student the set of languages they speak. The condition says no one's set sits inside another's. Everyone speaks the same number $k$ of languages, and two distinct $k$-element sets can never contain each other, so all we need is $100$ different $k$-subsets of the $n$ languages, i.e. $\binom{n}{k} \ge 100.$ With $n = 8$ the best we can manage is $\binom{8}{4} = 70,$ short of $100.$ But $\binom{9}{4} = 126 \ge 100.$ So $9$ languages are both enough and necessary. Therefore, the answer is A.

Since $1183 = 7 \cdot 169 = 7 \cdot 13^2,$ we have $\sqrt{1183} = 13\sqrt7.$ Square $\sqrt x + \sqrt y = 13\sqrt7$ to get $x + y + 2\sqrt{xy} = 1183.$ So $\sqrt{xy}$ is rational, which forces each of $x, y$ to be $7$ times a perfect square: $x = 7a^2,$ $y = 7b^2$ with $a + b = 13.$ Now $x + y = 7(a^2 + b^2),$ smallest when $a$ and $b$ are as equal as we can make them. Take $a = 6,$ $b = 7$ for $7(36 + 49) = 595.$ Thus, B is the correct answer.

$B$ is the square $|x| + |y| \le 8,$ with area $2 \cdot 8^2 = 128.$ The target condition $(x^2 + y^2 - 25)^2 \le 49$ unpacks to $|x^2 + y^2 - 25| \le 7,$ that is $18 \le x^2 + y^2 \le 32,$ an annulus of area $\pi(32 - 18) = 14\pi.$ Does it fit inside $B?$ The distance from the origin to an edge $x + y = 8$ is $\tfrac{8}{\sqrt2} = 4\sqrt2 = \sqrt{32},$ exactly the outer radius, so yes, the annulus sits inside the square. The probability is $\tfrac{14\pi}{128} = \tfrac{7}{64}\pi,$ giving $m + n = 71.$ Therefore, the answer is B.

The six fixed numbers total $24.8,$ so an integer mean needs $x + y + z = 9k - 24.8$ for some positive integer $k.$ A range of $7$ pins down the overall min and max (the fixed values already stretch from $1$ to $7$), and the median is the $5$th smallest of the nine numbers, which has to be a positive integer. Grind through where $x, y, z$ can sit and exactly three triples survive: $(0, 5, 6.2),$ $(0.1, 4, 7.1),$ and $(6, 6.2, 8).$ So there are $3.$ Thus, C is the correct answer.

Among $1, \ldots, 2024$ there are $1012$ even numbers and $1012$ odd. Each operation eats $4$ numbers and writes back $1,$ so the count falls by $3;$ to keep it high we want as few operations as possible. Every even number has to go, and a move that produces an odd result can clear at most $3$ evens at once (one odd plus three evens sums to odd). Clearing all $1012$ evens therefore takes at least $\lceil 1012/3 \rceil = 338$ moves. And that's achievable: $337$ moves of "one odd $+$ three evens $\to$ odd sum" wipe out $1011$ evens, then one move of "three odds $+$ one even $\to$ odd sum" gets the last. That leaves $2024 - 3 \cdot 338 = 1010$ numbers. Therefore, the answer is A.

If nobody ties, the $5$ snails finish in $5! = 120$ orders. Now allow exactly one tied group of size $k$ with $2 \le k \le 5.$ Choose the group in $\binom{5}{k}$ ways, then treat it as one block, leaving $6 - k$ blocks to arrange in $(6 - k)!$ ways. Summing over $k:$ $\binom{5}{2}4!$ $+ \binom{5}{3}3!$ $+ \binom{5}{4}2!$ $+ \binom{5}{5}1!$ $= 240 + 60 + 10 + 1$ $= 311.$ Add the no-tie count: $120 + 311 = 431.$ Thus, D is the correct answer.

Here $125 = 5^3$ and $\varphi(125) = 100.$ If $\gcd(n, 5) = 1,$ Euler's theorem gives $n^{100} \equiv 1 \pmod{125}.$ And if $5 \mid n,$ then $n^{100}$ carries a factor of $5^{100},$ hence of $125,$ so $n^{100} \equiv 0 \pmod{125}.$ That leaves only two possible remainders, $0$ and $1.$ Therefore, the answer is B.