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2024 AMC 10A Problems/Problem 15

Revision as of 20:52, 8 November 2024 by I am suk at math 2 (talk | contribs) (→‎Solution 2 (not rigorously proven))
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 squares) 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, \[m=N^2-1213 \space \mod \space 10\] \[m=8 \space \mod \space 10\] =$\boxed{\textbf{(E) }8}$ ~lptoggled

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
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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