Just compute each piece. We have $9901 \cdot 101 = 990100 + 9901$ $= 1000001,$ and $99 \cdot 10101 = 999999.$ Subtracting, $1000001 - 999999 = 2.$ Thus, A is the correct answer.

Subtract the two equations $1.5a + 800b = 69$ and $1.2a + 1100b = 69$ to kill the $69.$ That leaves $0.3a - 300b = 0,$ so $a = 1000b.$ Now substitute: $1500b + 800b = 2300b = 69,$ so $b = 0.03$ and $a = 30.$ Then $T = 30(4.2) + 0.03(4000)$ $= 126 + 120$ $= 246.$ Therefore, the answer is B.

Suppose $2$ is one of the five primes. Then the total is even and bigger than $2,$ so it's composite. That means all five primes must be odd. The five smallest odd primes give $3 + 5 + 7 + 11 + 13 = 39$ $= 3 \cdot 13,$ which isn't prime. We can't hit $41$ with five distinct odd primes, but $3 + 5 + 7 + 11 + 17 = 43$ is prime. So the smallest such prime is $43,$ and its digit sum is $4 + 3 = 7.$ Thus, B is the correct answer.

Each two-digit number is at most $99,$ so $k$ of them sum to at most $99k.$ We need $99k \ge 2024,$ which forces $k \ge 20.4,$ so $k \ge 21.$ And $21$ really works: twenty $99$'s plus one $44$ give $1980 + 44 = 2024.$ Therefore, the answer is B.

Factor $2024 = 2^3 \cdot 11 \cdot 23.$ The prime $23$ is the bottleneck: for $23$ to divide $n!,$ we need $n \ge 23.$ At $n = 23,$ the product $23!$ already has $23,$ $11,$ and plenty of factors of $2,$ so $2024 \mid 23!.$ The least value is $23.$ Thus, D is the correct answer.

Reversing all six letters flips the relative order of every pair, so all $\binom{6}{2} = 15$ pairs end up inverted. Each adjacent swap fixes exactly one inversion. So we need at least $15$ swaps, and bubbling each letter into place hits $15$ exactly. Therefore, the answer is D.

A positive product comes from three positive integers or one positive and two negative integers. Three positive integers have sum at least $3$. In the second case write the numbers as $-x,-y,z$, where $x,y,z$ are positive and $xyz=60$. A positive sum requires $z \gt x+y \ge 2\sqrt{xy}$, so $60/(xy) \gt 2\sqrt{xy}$, hence $xy \lt 10$. Checking factor pairs of $60$ with $xy \lt 10$, the least positive value of $z-x-y$ occurs at $(x,y,z)=(1,6,10)$ and equals $3$. Thus $(-1)(-6)(10)=60$, and the least positive sum is $3$. Thus, B is the correct answer.

From 1:00 to 2:45 is $105$ minutes. Amy, Bomani, and Charlie pack $4 + 3 + 3 = 10$ packages every $3$ minutes, so $\tfrac{10}{3}$ per minute, which is $\tfrac{10}{3} \cdot 105 = 350$ packages. That leaves $450 - 350 = 100$ for Daria, who packs $\tfrac54$ per minute and so needs $100 / \tfrac54 = 80$ minutes. She worked the last $80$ minutes, joining $105 - 80 = 25$ minutes after 1:00. That's 1:25 PM. Therefore, the answer is A.

Split the $6$ juniors into three unordered pairs. There are $\frac{6!}{2!^3 3!} = 15$ ways, and the same $15$ for the seniors. Each team is one junior-pair paired with one senior-pair, so we match the three junior-pairs to the three senior-pairs in $3! = 6$ ways. That's $15 \cdot 15 \cdot 6 = 1350$ sets of teams. Thus, B is the correct answer.

Just run it from $100:$ $100 \to 110 \to 120 \to 40$ $\to 50 \to 60 \to 20$ $\to 30 \to 10 \to 20$ $\to 30 \to 10 \to \cdots.$ After the $8$th step we're at $10,$ and from there it cycles $10, 20, 30$ with period $3.$ So step $8 + k$ is the $k$th entry of the cycle. For step $100,$ $k = 92,$ and $92 \equiv 2 \pmod 3,$ which lands on $30.$ Therefore, the answer is C.

Note $m = \sqrt{n^2 - 49} \ge 0$ has to be an integer, so $n^2 - 49 = m^2,$ which means $(n - m)(n + m) = 49.$ The factorizations of $49$ give $|n| = 25, m = 24$ or $|n| = 7, m = 0.$ So the ordered pairs $(m, n)$ are $(24, 25),$ $(24, -25),$ $(0, 7),$ $(0, -7).$ That's $4$ of them. Thus, D is the correct answer.

Apply the daily changes $+80, -90, -10, +60, -40$ to the starting $1700.$ The six scores are $1700, 1780, 1690,$ $1680, 1740, 1700.$ They add to $10290,$ so the average is $10290 / 6 = 1715.$ Therefore, the answer is E.

The dilation just scales about the origin, so it commutes with both the rotation and the reflection. That's $2$ pairs. The translation commutes with the reflection across the $x$-axis too, since either order sends $(x, y) \to (x + 2, -y).$ The other three pairs fail: the translation clashes with the rotation and with the dilation, and the rotation clashes with the reflection. So $3$ pairs commute. Thus, C is the correct answer.

The equilateral triangle has side $16\sqrt3$. Put $\ell$ on the $x$-axis with base vertex $V = (16\sqrt3, 0)$; the slanted side lies on $\sqrt3\,x + y = 48$. The circle sits on $\ell$, has radius $12$, and touches that side externally, so its center is $O = (20\sqrt3, 12)$. Let $T = (20\sqrt3, 0)$ be its tangency point on $\ell$, and let $P$ be the tangency point on the slanted side. The two tangent lengths from $V$ satisfy $VT = VP = 4\sqrt3$, so kite $VTOP$ has area $4\sqrt3 \cdot 12 = 48\sqrt3$. The angle at $V$ is $120^\circ$, so the removed sector has angle $60^\circ$ and area $\tfrac16 \pi (12)^2 = 24\pi$. The region has area $48\sqrt3 - 24\pi$, giving $a + b + c = 48 + 3 + 24 = 75$. Therefore, the answer is D.

Set $M + 1213 = y^2$ and $M + 3773 = x^2.$ Subtracting, $x^2 - y^2 = 2560,$ so $(x - y)(x + y) = 2560.$ The two factors share a parity, and their product is even, so both are even: write $x - y = 2s,$ $x + y = 2t,$ with $st = 640.$ To make $M$ as large as possible we want $y = t - s$ as large as possible, so $s$ as small as possible. Take $s = 1, t = 640,$ giving $y = 639.$ Then $M = 639^2 - 1213 = 407108,$ whose units digit is $8.$ Thus, E is the correct answer.

Every piece is similar to the whole rectangle, so they all share one aspect ratio. The areas $1, 2, 4, 8, 16$ (and $9, 18$) come in factor-of-$2$ steps, and cutting a rectangle of aspect ratio $\sqrt2$ across its long side gives two similar copies of half the area. That pins the ratio at $\sqrt2.$ The total area is $36 + 16 + 8 + 18$ $+ 1 + 9 + 4 + 2 + 32$ $+ 25 + 49 = 200.$ The enclosing rectangle satisfies $AB \cdot \tfrac{AB}{\sqrt2} = 200,$ so $AB^2 = 200\sqrt2$ and $AB = \sqrt{200\sqrt2} = 10\sqrt[4]{8}.$ Therefore, the answer is D.

Team A takes game $1$ at home with probability $\tfrac23,$ and each away game with probability $p.$ It can win the playoff three disjoint ways: win games $1, 2;$ win $1,$ lose $2,$ win $3;$ lose $1,$ win $2, 3.$ Adding those, $\tfrac23 p + \tfrac23(1 - p)p + \tfrac13 p^2 = \tfrac12.$ This cleans up to $2p^2 - 8p + 3 = 0,$ so $p = \tfrac{4 - \sqrt{10}}{2} = \tfrac12\!\left(4 - \sqrt{10}\right).$ Then $m = 4,$ $n = 10,$ and $m + n = 14.$ Thus, E is the correct answer.

In base $b,$ $2024_b = 2b^3 + 2b + 4$ $= 2(b^3 + b + 2),$ so $16 \mid 2024_b$ exactly when $8 \mid b^3 + b + 2.$ Test the residues modulo $8:$ this holds precisely for $b \equiv 3, 6, 7 \pmod 8.$ Counting the $b$ with $5 \le b \le 2024$ in those three classes gives $K = 758,$ whose digit sum is $7 + 5 + 8 = 20.$ Therefore, the answer is D.

Since $720^2 = ab,$ the common ratio $r = \tfrac{720}{a} = \tfrac{b}{720}$ is rational. Write $r = \tfrac{p}{q}$ in lowest terms with $p \gt q.$ Then $a = \tfrac{720q}{p}$ and $b = \tfrac{720p}{q}$ are integers, which forces $p \mid 720$ and $q \mid 720.$ To make $b$ smallest, we want the smallest ratio $\tfrac{p}{q} \gt 1$ with both $p, q \mid 720,$ which is $\tfrac{16}{15}.$ That gives $b = 720 \cdot \tfrac{16}{15} = 768$ (and $a = 675$). The digit sum is $7 + 6 + 8 = 21.$ Thus, E is the correct answer.