2024 AMC 10A Problems/Problem 15: Difference between revisions

Latest revision as of 00:55, 11 November 2025

The following problem is from both the 2024 AMC 10A #15 and 2024 AMC 12A #9, so both problems redirect to this page.

Problem

Let $M$ be the greatest integer such that both $M+1213$ and $M+3773$ are perfect squares. What is the units digit of $M$ ?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }6\qquad\textbf{(E) }8$

Solution 1

Let $M+1213=P^2$ and $M+3773=Q^2$ for some positive integers $P$ and $Q.$ We subtract the first equation from the second, then apply the difference of squares: $(Q+P)(Q-P)=2560.$ Note that $Q+P$ and $Q-P$ have the same parity, and $Q+P>Q-P.$

We wish to maximize both $P$ and $Q,$ so we maximize $Q+P$ and minimize $Q-P.$ It follows that $\begin{align*} Q+P&=1280, \\ Q-P&=2, \end{align*}$ from which $(P,Q)=(639,641).$

Finally, we get $M=P^2-1213=Q^2-3773\equiv1-3\equiv8\pmod{10},$ so the units digit of $M$ is $\boxed{\textbf{(E) }8}.$

~MRENTHUSIASM ~Tacos_are_yummy_1

Solution 2

Ideally, we would like for the two squares to be as close as possible. The best case is that they are consecutive squares (no square numbers in between them); however, since $M+1213$ and $M+3773$ (and thus their square roots) have the same parity, they cannot be consecutive squares (which are always opposite parities). The best we can do is that $M+1213$ and $M+3773$ have one square in between them.

Let the square between $M+1213$ and $M+3773$ be $x^2$ . So, we have $M+1213 = (x-1)^2$ and $M+3773 = (x+1)^2$ . Subtracting the two, we have $(M+3773)-(M+1213) = (x+1)^2 - (x-1)^2$ , which yields $2560 = 4x$ , which leads to $x = 640$ . Therefore, the two squares are $639^2$ and $641^2$ , which both have units digit $1$ . Since both $1213$ and $3773$ have units digit $3$ , $M$ will have units digit $\boxed{\textbf{(E) }8}$ .

~i_am_suk_at_math_2 (parity argument editing by Technodoggo)

Solution 3

Let $M+1213=N^2$ $\Rightarrow M+3773=(N+a)^2$

It is obvious that $a\neq1$ by parity

Thus, the minimum value of $a$ is 2 Which gives us, $(N+a)^2-N^2=M+3773-(M+1213)$ $4N+4=2560$ $N=639$ Plugging this back in, $M=N^2-1213 \space \mod \space 10$ $M=8 \space \mod \space 10$ Hence the answer $\boxed{\textbf{(E) }8}$ .

~lptoggled

- trevian1(minor edit)

Solution 4

Let $M+1213=n^2$ and $M+3773=(n+1)^2$ for some positive integer $n$ . We do this because, in order to maximize $M$ , the perfect squares need to be as close to each other as possible. We find that this configuration doesn't work, as when we subtract the equations, we have $2n+1=2560$ ; impossible. Then we try $M+3773=(n+2)^2$ . Now we would have $4n+4=2560$ which indeed works! $n=639$ .

Finally, we get $M=n^2-1213$ so the units digit of $M$ is $11-3=\boxed{\textbf{(E) }8}.$

~xHypotenuse

Note

We experiment with values of $M$ to find the reason for why $M$ is maximised when $M + 1213$ and $M + 3773$ are nearly consecutive perfect squares. If $M$ is very small, $M + 1213$ and $M + 3773$ are perfect squares that are far apart. Yet, as $M$ grows, the relative distance between $M + 1213$ and $M + 3773$ decreases, so for very nearly consecutive perfect squares, $M$ is very large.

~LeonQS

(I don't know if this is trivial - when I first read the solutions, I was confused to why this was true. Maybe I didn't get enough sleep.)

Video Solution (Fast! About 3 min solve!)

https://youtu.be/l3VrUsZkv8I

~MC

Video Solution (4 min solve)

https://youtu.be/YgJ23mepN0Q

~Education, the Study of Everything

Video Solution by Pi Academy

https://youtu.be/ABkKz0gS1MU?si=ZQBgDMRaJmMPSSMM

Video Solution 1 by Power Solve

https://youtu.be/FvZVn0h3Yk4

Video Solution by SpreadTheMathLove

https://youtu.be/CmIPAvwtWLA?si=ZCv3ypdDmCaV-aX3

Video Solution by Dr. David

https://youtu.be/XLoetj5obYE

Video solution by TheNeuralMathAcademy

https://www.youtube.com/watch?v=4b_YLnyegtw&t=2808s

2024 AMC 10A (Problems • Answer Key • Resources)
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