2024 AMC 12A Problems

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

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

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

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

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.

Subtract equations in pairs, e.g. $(ab+c)-(bc+a)$ factors as $(a-c)(b-1)=13$

$13$ is prime, so $(a-c,\ b-1)$ has few options; test them for integer solutions

Subtracting the second equation from the first gives $(a-c)(b-1)=13.$ Hence $$\begin{gathered} (a-c,b-1)\in\{(1,13),(13,1),\\ (-1,-13),(-13,-1)\}. \end{gathered}$$ The corresponding values of $b$ are $14,2,-12,0.$ Substituting $a=c+(a-c)$ into $ab+c=100$ eliminates the first two cases because they would require $15c=86$ or $3c=74.$ The case $b=-12$ gives $c=-8$ and $a=-9,$ which satisfies all three equations. The case $b=0$ gives $(a,c)=(87,100),$ which fails $ca+b=60.$ Thus the unique integer solution is $(-9,-12,-8),$ and $ab+bc+ca$ $=108+96+72=276.$ Thus, the correct answer is D.

The diagonal makes angle $\arctan\dfrac{1}{2+\sqrt3}=15^\circ$ with the long side, since $\dfrac{1}{2+\sqrt3}=2-\sqrt3$

The two diagonals meet at $30^\circ;$ track the unused diagonal as five new cards are added

A diagonal makes an angle $\theta$ with a long side, where $\tan\theta=\dfrac{1}{2+\sqrt3}$ $=2-\sqrt3=\tan15^\circ.$ Thus the acute angle between the two diagonals of a card is $2\theta=30^\circ.$

Each new card shares one diagonal with the previous card, and its other diagonal is the next line obtained by turning $30^\circ$ clockwise. All these equal diagonals have the same midpoint and are diameters of one common circle. The line through the original card’s other diagonal, which contains $B,$ is $30^\circ$ counterclockwise from $AC.$ As an unoriented line, this is the same as $150^\circ$ clockwise from $AC.$ Five additions advance the unused diagonal by $5\cdot30^\circ=150^\circ,$ so the sixth card is the first new card with a vertex at $B.$

Thus, the correct answer is A.

In $\triangle ACD,$ $AC^2=3^2+5^2$ $-2\cdot3\cdot5\cos120^\circ$

Find $AB$ using $\angle ABC=60^\circ,$ then apply Ptolemy: $AC\cdot BD=AB\cdot CD+BC\cdot DA$

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.

Let $x=\dfrac{AP}{AB}$ and $y=\dfrac{AQ}{AC}$ be uniform on $[0,1];$ the area ratio is $xy$

Compute $P(xy\ge\tfrac12)=\displaystyle\int_{\frac{1}{2}}^{1}\left(1-\tfrac{1}{2x}\right)dx$ and subtract from $1$

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 $$\begin{aligned} &\int_{\frac{1}{2}}^{1}\left(1-\frac{1}{2x}\right)dx \\ &=\frac12-\frac{\ln2}{2}\approx0.153. \end{aligned}$$ Therefore $P(xy\lt\tfrac12)\approx1-0.153=0.847,$ which lies in $\left(\tfrac34,\tfrac78\right].$ Thus, the correct answer is D.

Compute a few terms and conjecture the closed form $a_n=n+\dfrac1n$

Then $a_n^2=n^2+2+\dfrac1{n^2},$ and $\displaystyle\sum_{n=1}^{100}\dfrac1{n^2}$ lies strictly between $1$ and $2$

The recurrence rearranges to $a_n=1+\tfrac{n-1}{n}(a_{n-1}+1).$ Computing early terms $2,\tfrac52,\tfrac{10}3,\tfrac{17}4,\ldots$ suggests $a_n=n+\tfrac1n.$ This follows by induction: substituting $a_{n-1}=n-1+\tfrac1{n-1}$ into the recurrence gives $a_n=1+\tfrac{n-1}{n}(n+\tfrac1{n-1})$ $=n+\tfrac1n.$ Then $a_n^2=n^2+2+\tfrac1{n^2},$ so $$\begin{aligned} &\sum_{n=1}^{100}a_n^2 \\ &=\sum_{n=1}^{100}n^2+200 \\ &\quad {}+\sum_{n=1}^{100}\frac1{n^2} \\ &=338350+200+S, \end{aligned}$$ where $S\gt1$ and $S\lt1+\int_1^\infty x^{-2}\,dx=2.$ Hence the sum is between $338551$ and $338552,$ and its floor is $338551.$ Thus, the correct answer is B.