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

Solution 2a (perhaps more clear)

Ideally, $M+1213$ and $M+3773$ are close squares. We would want them to be consecutive squares (squares of consecutive numbers), but $M+1213$ and $M+3773$ are of the same parity (and so are their squares) while consecutive squares are always opposite parity, so this is impossible. The next-best option is $M+1213=N^2$ and $M+3773=(N+2)^2$ ; the latter expands into $M+3773=N^2+4N+4$ , and subtracting the former from this result, we see that $2560=4N+4\implies N=639$ . Then $M+1213=639^2\implies M+1213\equiv639^2\pmod{10}\implies M+3\equiv9^2\pmod{10}\implies M\equiv\boxed{\textbf{(E) }8}\pmod{10}$ .

~Technodoggo

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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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

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2024 AMC 10A Problems/Problem 15

Contents

Problem

Solution 1

Solution 2 (not rigorously proven)

Solution 2a (perhaps more clear)

See also