Small Hint:

$9901\cdot101=9901\cdot100+9901$

Big Hint:

$99\cdot10101=100\cdot10101-10101$

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.

Small Hint:

Set the two estimates equal: $1.5a+800b=1.2a+1100b$

Big Hint:

This gives $a=1000b,$ so $69=2300b$

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.

Small Hint:

The largest two-digit number is $99,$ so many numbers are needed

Big Hint:

$20\cdot99=1980,$ which falls short; can $21$ numbers reach exactly $2024?$

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.

Small Hint:

Factor $2024=2^3\cdot11\cdot23$

Big Hint:

$n!$ must contain the prime $23,$ so $n\ge23$

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.

Small Hint:

The $20$ numbers sum to $20\cdot45=900$

Big Hint:

If there are $k$ sixes, the remaining $20-k$ numbers sum to $900-6k=66(20-k)$

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.

Small Hint:

Integers may be negative; the product stays positive with two negative factors

Big Hint:

Writing the numbers as $-p,-q,r$ with $pqr=60,$ minimize $r-p-q$ while keeping it positive

Solution:

To obtain a small positive sum, use two negative integers $-p,-q$ and one positive integer $r,$ where $pqr=60.$ Up to order, the positive factor triples of $60$ are $$\begin{gathered} (1,1,60),(1,2,30),\\ (1,3,20),(1,4,15),\\ (1,5,12),(1,6,10),\\ (2,2,15),(2,3,10),\\ (2,5,6),(3,4,5). \end{gathered}$$ A positive value of $r-p-q$ is smallest when the largest factor is chosen as $r.$ The positive values from the list are $58,27,16,10,6,3,11,5;$ the last two triples give negative values. Thus the least positive sum is $10-1-6=3.$ (Three positive integers have sum at least $3,$ and three negative integers cannot have positive product.) Thus, the correct answer is B.

Small Hint:

The points are symmetric about the midpoint $M$ of $AC,$ so the vectors pair up

Big Hint:

The sum equals $2024\,\vec{BM};$ the median to the hypotenuse of a right triangle is half the hypotenuse

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.

Small Hint:

$\log(\sin3\theta)+\log(\cos2\theta)$ $=\log(\sin3\theta\cos2\theta),$ which must equal $0$

Big Hint:

A product of two numbers each at most $1$ equals $1$ only when both equal $1$

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.

Small Hint:

Let $M+1213=a^2$ and $M+3773=b^2;$ subtract to get $b^2-a^2=2560$

Big Hint:

$(b-a)(b+a)=2560$ with both factors even; to maximize $M,$ make $b-a$ as small as possible

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.

Small Hint:

$\tan\alpha=\dfrac34;$ apply the double-angle formula for $\tan2\alpha$

Big Hint:

$\tan\beta=\dfrac{7}{24}$ is the reciprocal of $\tan2\alpha$

Solution:

The smallest angle of the $3\text{-}4\text{-}5$ triangle has $\tan\alpha=\tfrac34.$ Then $$\begin{aligned} &\tan2\alpha=\frac{2\cdot\frac34}{1-\frac{9}{16}} \\ &=\frac{\frac{3}{2}}{\frac{7}{16}}=\frac{24}{7}. \end{aligned}$$ 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.

Small Hint:

$2024_b=2b^3+2b+4,$ so you need $2(b^3+b+2)$ to be divisible by $16$

Big Hint:

Reduce $b^3+b+2\pmod 8$ and count $b$ in each valid residue class of $[5,2024]$

Solution:

Here $2024_b=2b^3+2b+4$ $=2(b^3+b+2),$ so $2024_b$ is divisible by $16$ exactly when $b^3+b+2$ is divisible by $8.$ 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.

Small Hint:

The middle term gives $720^2=ab,$ so $ab=518400$

Big Hint:

To minimize $b,$ take the largest divisor $a\lt720$ of $518400$

Solution:

Since the terms are geometric, $720^2=ab,$ so $ab=518400=2^8\cdot3^4\cdot5^2.$ Because $b=\frac{518400}{a},$ minimizing $b$ means finding the smallest divisor of $518400$ greater than $720.$ There is no divisor strictly between $720$ and $768:$ if its exponent of $5$ is $2,1,$ or $0,$ then after dividing by $25,5,$ or $1,$ respectively, a number of the form $2^i3^j$ would have to lie in $(28.8,30.72),\ (144,153.6),$ or $(720,768);$ the allowed powers $i\le8,j\le4$ give none. Since $768=2^8\cdot3$ is a divisor, it is the least possible $b.$ Its paired divisor is $a=\frac{518400}{768}=675,$ and the digit sum of $b$ is $7+6+8=21.$ Thus, the correct answer is E.

Small Hint:

The axis of symmetry is the vertical line $x=c$ where the function attains its minimum

Big Hint:

Set the derivative $e^{x+1}-e^{-x}=0$ to find $c,$ then reflect $x=-1$ across it

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.

Small Hint:

If every row and every column is arithmetic, the entry has the form $\alpha+\beta i+\gamma j+\delta ij$

Big Hint:

Substitute the four known entries into that form and solve for $\alpha,\beta,\gamma,\delta$

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 $$\alpha+5\beta+5\gamma+25\delta=0,$$ $$\alpha+2\beta+4\gamma+8\delta=48,$$ $$\alpha+4\beta+3\gamma+12\delta=16,$$ $$\alpha+3\beta+\gamma+3\delta=12.$$

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.

Small Hint:

$p^2+4=(p-2i)(p+2i),$ so the product relates to $P(2i)$ and $P(-2i),$ where $P(x)=x^3+2x^2-x+3$

Big Hint:

$P(2i)$ and $P(-2i)$ are complex conjugates; multiply them to get a real number

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.

Small Hint:

Assign the three roles (all reds, all whites, the lone blue) to the players in $3!$ ways

Big Hint:

The black tokens must then split as $1,2,3$ among the players; divide by $\binom{12}{4,4,4}$

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.