2024 AMC 10A Problems/Problem 15

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 (not rigorously proven)

We assume that when the maximum values are achieved, the two squares are consecutive squares. However, since both have the same parity, the closest we can get to this is that they are 1 square apart.

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

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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2024 AMC 10A Problems/Problem 15

Contents

Problem

Solution 1

Solution 2 (not rigorously proven)

See also