Solution:

Directly, $9901\cdot101=990100+9901=1000001$ and $99\cdot10101=1010100-10101=999999.$ Their difference is $1000001-999999=2.$ Thus, the correct answer is A.

Solution:

From $1.5a+800b=1.2a+1100b$ we get $0.3a=300b,$ so $a=1000b.$ Then $69=1.5a+800b=1500b+800b=2300b,$ giving $b=0.03$ and $a=30.$ For $L=4.2,\ G=4000,$ $T=30(4.2)+0.03(4000)=126+120=246.$ Thus, the correct answer is B.

Solution:

Each number is at most $99,$ so $k$ numbers sum to at most $99k.$ Since $99\cdot20=1980\lt2024,$ at least $21$ numbers are required. With $21,$ we can use twenty $99$s and one $44$: $20\cdot99+44=1980+44=2024.$ Thus, the correct answer is B.

Solution:

Factoring, $2024=2^3\cdot11\cdot23.$ The factorial $n!$ contains the prime $23$ only when $n\ge23.$ At $n=23,$ the product $23!$ already includes $23,\ 11,$ and plenty of factors of $2,$ so $23!$ is a multiple of $2024.$ Thus, the correct answer is D.

Solution:

The full set sums to $20\cdot45=900.$ Removing $k$ sixes leaves $20-k$ numbers summing to $900-6k,$ with mean $66,$ so $900-6k=66(20-k)=1320-66k.$ Then $60k=420,$ giving $k=7.$ Thus, the correct answer is D.

Solution:

To keep the product positive we use two negative integers $-p,-q$ and one positive $r,$ with $pqr=60$ and sum $r-p-q.$ Trying $(p,q,r)=(1,6,10)$ gives product $60$ and sum $10-1-6=3.$ Checking the other factorizations shows no positive sum smaller than $3$ is attainable (for example all-positive triples give at least $3+4+5=12$). Thus, the correct answer is B.

Solution:

The points $P_k$ are symmetric about the midpoint $M$ of $AC,$ so pairing $P_k$ with its mirror gives $\vec{BP_k}+\vec{BP_{2025-k}}=2\,\vec{BM}.$ Hence the whole sum is $2024\,\vec{BM}.$ In a right triangle the median to the hypotenuse has length half the hypotenuse; here $AC=2,$ so $BM=1.$ The length of the sum is $2024\cdot1=2024.$ Thus, the correct answer is D.

Solution:

The equation means $\sin(3\theta)\cos(2\theta)=1$ with both factors positive (for the logs to be defined). Since $\sin(3\theta)\le1$ and $\cos(2\theta)\le1,$ their product is $1$ only if $\sin(3\theta)=1$ and $\cos(2\theta)=1$ simultaneously. But $\cos(2\theta)=1$ forces $\theta\in\{0,\pi,2\pi\},$ where $\sin(3\theta)=0\ne1.$ No angle works. Thus, the correct answer is A.

Solution:

Write $M+1213=a^2$ and $M+3773=b^2,$ so $b^2-a^2=2560,$ i.e. $(b-a)(b+a)=2560.$ Both factors have the same parity, hence both even. To maximize $a$ (and thus $M$), minimize $b-a:$ take $b-a=2,\ b+a=1280,$ so $a=639.$ Then $M=639^2-1213=408321-1213=407108,$ whose units digit is $8.$ Thus, the correct answer is E.

Solution:

The smallest angle of the $3\text{-}4\text{-}5$ triangle has $\tan\alpha=\tfrac34.$ Then $$\tan2\alpha=\frac{2\cdot\frac34}{1-\frac{9}{16}}=\frac{3/2}{7/16}=\frac{24}{7}.$$ The smallest angle of the $7\text{-}24\text{-}25$ triangle has $\tan\beta=\tfrac{7}{24}=\cot2\alpha =\tan\!\left(\tfrac{\pi}{2}-2\alpha\right).$ Hence $\beta=\tfrac{\pi}{2}-2\alpha.$ Thus, the correct answer is C.

Solution:

Here $2024_b=2b^3+2b+4=2(b^3+b+2),$ so $16\mid2024_b$ exactly when $8\mid b^3+b+2.$ Checking residues $\bmod 8,$ $b^3+b+2\equiv0$ precisely for $b\equiv3,6,7\pmod8.$

Counting $b$ in $[5,2024]:$ residue $3$ gives $11,\ldots,2019$ ($252$ values), residue $6$ gives $6,\ldots,2022$ ($253$ values), and residue $7$ gives $7,\ldots,2023$ ($253$ values). So $K=252+253+253=758,$ and its digit sum is $7+5+8=20.$

Thus, the correct answer is D.

Solution:

Since the terms are geometric, $720^2=ab,$ so $ab=518400=2^8\cdot3^4\cdot5^2.$ Because $b=518400/a,$ minimizing $b$ means maximizing the divisor $a$ subject to $a\lt720.$ The largest such divisor is $a=675=3^3\cdot5^2,$ giving $b=518400/675=768.$ Its digit sum is $7+6+8=21.$ Thus, the correct answer is E.

Solution:

The curve $y=e^{x+1}+e^{-x}-2$ is symmetric about the vertical line through its minimum. Setting the derivative $e^{x+1}-e^{-x}=0$ gives $x+1=-x,$ so $x=-\tfrac12.$ Reflecting $\left(-1,\tfrac12\right)$ across $x=-\tfrac12$ keeps the $y$-coordinate and sends $x=-1$ to $x=0.$ The image is $\left(0,\tfrac12\right).$ Thus, the correct answer is D.

Solution:

A grid whose rows and columns are all arithmetic has entries of the bilinear form $a(i,j)=\alpha+\beta i+\gamma j+\delta ij.$ The four givens yield $$\alpha+5\beta+5\gamma+25\delta=0,\quad \alpha+2\beta+4\gamma+8\delta=48,$$ $$\alpha+4\beta+3\gamma+12\delta=16,\quad \alpha+3\beta+\gamma+3\delta=12.$$

Solving gives $\delta=-5,\ \beta=5,\ \gamma=22,\ \alpha=-10.$ Then $a(1,2)=\alpha+\beta+2\gamma+2\delta=-10+5+44-10=29.$

Thus, the correct answer is C.

Solution:

Since $P(x)=(x-p)(x-q)(x-r),$ grouping $p^2+4=(p-2i)(p+2i)$ over all roots gives $$\prod(p^2+4)=P(2i)\,P(-2i).$$ Compute $P(2i)=-8i-8-2i+3=-5-10i$ and $P(-2i)=8i-8+2i+3=-5+10i.$ Their product is $(-5)^2+10^2=25+100=125.$ Thus, the correct answer is D.

Solution:

Treat all tokens as distinct; the total number of ways to deal $4$ to each player is $\binom{12}{4,4,4}=34650.$ For the favorable event, choose which player gets the reds, whites, and blue in $3!=6$ ways. The red player needs $1$ more token, the white player $2$ more, and the blue player $3$ more, all black; the $6$ black tokens split as $1,2,3$ in $\tfrac{6!}{1!\,2!\,3!}=60$ ways. So the probability is $\dfrac{6\cdot60}{34650}=\dfrac{360}{34650}=\dfrac{4}{385}.$ Then $m+n=4+385=389.$ Thus, the correct answer is C.

Solution:

Subtracting pairs gives $(a-c)(b-1)=13,\ (a-b)(c-1)=-27,$ and $(b-c)(a-1)=40.$ Because $13$ is prime, only a few cases arise; testing them yields $a=-9,\ b=-12,\ c=-8,$ which satisfy all three original equations. Then $ab+bc+ca=108+96+72=276.$ Thus, the correct answer is D.

Solution:

The diagonal of the card makes an angle $\theta$ with the long side where $\tan\theta=\dfrac{1}{2+\sqrt3}=2-\sqrt3=\tan15^\circ,$ so $\theta=15^\circ.$ All the cards share diagonals that are equal chords (diameters) of one common circle, and each newly added card is the previous one turned $15^\circ$ clockwise about the common center. A fresh vertex first coincides with $B$ once the accumulated rotation reaches $90^\circ,$ i.e. after $90^\circ/15^\circ=6$ cards. Since $15^\circ$ divides $90^\circ$ evenly, a vertex does land on $B.$ Thus, the correct answer is A.

Solution:

In $\triangle ACD,$ the law of cosines gives $AC^2=9+25-2(15)\cos120^\circ=34+15=49,$ so $AC=7.$

Since $ABCD$ is cyclic, $\angle ABC=180^\circ-120^\circ=60^\circ.$ In $\triangle ABC$ with $BC=3$ and $AC=7,$ the law of cosines gives $49=AB^2+9-3AB,$ so $AB=8.$ By Ptolemy, $AC\cdot BD=AB\cdot CD+BC\cdot DA=8\cdot3+3\cdot5=39,$ hence $BD=\tfrac{39}{7}.$ This is shorter than $AC=7.$

Thus, the correct answer is D.

Solution:

With $x=\tfrac{AP}{AB}$ and $y=\tfrac{AQ}{AC}$ uniform on $[0,1],$ the area ratio $\tfrac{[APQ]}{[ABC]}=xy.$ The complementary event $xy\ge\tfrac12$ requires $x\ge\tfrac12$ and $y\in[\tfrac{1}{2x},1],$ with probability $$\int_{1/2}^{1}\left(1-\frac{1}{2x}\right)dx=\frac12-\frac{\ln2}{2}\approx0.153.$$ Therefore $P(xy\lt\tfrac12)\approx1-0.153=0.847,$ which lies in $\left(\tfrac34,\tfrac78\right].$ Thus, the correct answer is D.

Solution:

The recurrence rearranges to $a_n=1+\tfrac{n-1}{n}(a_{n-1}+1).$ Computing early terms $2,\tfrac52,\tfrac{10}3,\tfrac{17}4,\ldots$ reveals $a_n=n+\tfrac1n,$ which checks out. Then $a_n^2=n^2+2+\tfrac1{n^2},$ so $$\sum_{n=1}^{100}a_n^2=\sum_{n=1}^{100}n^2+200+\sum_{n=1}^{100}\frac1{n^2}=338350+200+S,$$ where $1\lt S\lt2.$ Hence the sum is between $338550$ and $338552,$ and its floor is $338551.$ Thus, the correct answer is B.

Solution:

The toothpicks form a single simple closed curve. The constraint forces each of the eight middle-row cells to have exactly one of its four sides used, which severely limits how the loop threads through the grid: for each middle cell the loop must contribute one edge (a top, bottom, left, or right side), and consecutive choices must join up into one non-self-intersecting closed curve.

Working left to right and tracking how the loop's upper and lower portions enter and leave each column (equivalently, a transfer-matrix/casework count over the $8$ columns) enumerates all admissible loops. Carrying out this casework gives $146$ valid configurations.

Thus, the correct answer is C.

Solution:

With $a=\tan^2\tfrac{\pi}{16},\ b=\tan^2\tfrac{3\pi}{16},\ c=\tan^2\tfrac{5\pi}{16},\ d=\tan^2\tfrac{7\pi}{16},$ the expression is $ab+ac+bd+cd=(a+d)(b+c).$

Since $\tfrac{7\pi}{16}=\tfrac{\pi}{2}-\tfrac{\pi}{16},$ we have $d=\cot^2\tfrac{\pi}{16},$ so $a+d=\tan^2\tfrac{\pi}{16}+\cot^2\tfrac{\pi}{16}=\tfrac{4}{\sin^2(\pi/8)}-2=14+8\sqrt2.$ Likewise $b+c=\tfrac{4}{\sin^2(3\pi/8)}-2=14-8\sqrt2.$ Their product is $14^2-(8\sqrt2)^2=196-128=68.$

Thus, the correct answer is B.

Solution:

A disphenoid exists (as the tetrahedron formed by the face-plane midpoints of a box) exactly when the common face triangle is acute, and its total surface area is $4$ times one face's area. We want the smallest-area acute scalene integer triangle. The candidates $(2,3,4)$ and $(3,5,6)$ are obtuse, and $(3,4,5)$ is right (giving a degenerate flat figure), but $(4,5,6)$ is acute since $4^2+5^2\gt6^2.$ By Heron with $s=\tfrac{15}2,$ its area is $\sqrt{\tfrac{15}2\cdot\tfrac72\cdot\tfrac52\cdot\tfrac32}=\tfrac{15\sqrt7}{4}.$ The total surface area is $4\cdot\tfrac{15\sqrt7}{4}=15\sqrt7.$ Thus, the correct answer is D.

Solution:

Reflecting the graph of $y=f(x)$ over $y=x$ produces the graph of its inverse, so the graph is symmetric about $y=x$ exactly when $f$ equals its own inverse. For $f(x)=\tfrac{ax+b}{cx+d}$ this happens in two ways: when $a+d=0$ with $ad-bc\ne0$ (a genuine involution, including the slope$-1$ lines when $c=0$), or when $f$ is the identity $y=x$ ($b=c=0,\ a=d\ne0$).

For $a+d=0,$ set $d=-a;$ the determinant $-a^2-bc$ must be nonzero, so we need $a^2+bc\ne0,$ together with $(c,d)\ne(0,0).$ Counting $(a,b,c)$ with each in $[-5,5]$ gives $1282$ quadruples. The identity case adds $10$ more ($a=d\in\{\pm1,\ldots,\pm5\}$). The total is $1282+10=1292.$

Thus, the correct answer is B.