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2024 AMC 10A Problems/Problem 15

Revision as of 16:09, 8 November 2024 by Tacos are yummy 1 (talk | contribs)

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

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

Solution 2

Let $M+1213=x^2$ and $M+3773=y^2$. Subtracting the two equations will give \[y^2-x^2=2560.\] Factoring the left hand side as a difference of squares gives \[(y-x)(y+x)=2560.\] The goal in the problem was to maximize $M$, which therefore means maximizing both $x$ and $y$ and maximizing $y+x$.

$y-x$ and $y+x$ are two factors of $2560$, therefore maximizing $y+x$ means making $y-x$ and $y+x$ as far as possible from each other. We start with $2560=1\cdot2560$, the most obvious choice that makes $y+x$ as large as possible, but solving $y-x=1$ and $y+x=2560$ will give non-integer solutions, therefore this case does not work.

The next most obvious choice would be $2560=2\cdot1280$, which does give integer solutions to $x$ and $y$. \[y-x=2\] \[y+x=1280\] gives the solutions $y=641,x=639$.

Since we're only focusing on the units digit of $M=x^2-1213=y^2-3773$, then we only have to focus on the units digits of $x^2$ and $1213$ or $y^2$ and $3773$. $1^2$ and $9^2$ give the same units digit, and $1213$ and $3773$ also have the same units digit. So the units digit of $M$ is $10-|1-3|=\boxed{\text{(E) }8}$

See also

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

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