2024 AMC 10A Problem 15

1. Name the squares. Write $M + 1213 = a^2$ and $M + 3773 = b^2$, where $a$ and $b$ are whole numbers and $b$ is bigger than $a$.
2. Subtract to get rid of $M$:
$$b^2 - a^2 = 3773 - 1213 = 2560.$$
3. Factor the difference of squares:
$$(b - a)(b + a) = 2560.$$
4. What makes $M$ large? $M = a^2 - 1213$, so we want $a$ as large as possible. The two factors $b - a$ and $b + a$ multiply to a fixed $2560$. To make $a$ big we want the numbers $a$ and $b$ close together and large, which means the gap $b - a$ small and the sum $b + a$ large.
5. How small can the gap be? Try $b - a = 1$: then $b + a = 2560$. But adding the two equations gives $2b = 2561$, which is not a whole number. In fact $b - a$ and $b + a$ are either both odd or both even (they differ by $2a$). Their product $2560$ is even, so both must be even. The smallest even gap is $2$.
6. Solve. $b - a = 2$ and $b + a = 2560 \div 2 = 1280$. So $b = 641$ and $a = 639$.
7. Units digit of $M$. $M = 639^2 - 1213$. The units digit of $639^2$ is the units digit of $9 \cdot 9 = 81$, which is $1$. A number ending in $1$ minus a number ending in $3$ ends in $8$ (think $11 - 3 = 8$). (Indeed $M = 408321 - 1213 = 407108$.)

The answer is $\boxed{\textbf{(E)}\ 8}$.

Two squares that differ by a fixed amount always lead to (difference)(sum) $=$ that amount, so there are only a few possible pairs. The largest squares come from the pair with the smallest gap. The usual catch is odd and even: the difference and the sum of two whole numbers are both odd or both even, so not every factor pair is allowed.