2024 AMC 10A Problem 15

Write $M + 1213 = a^2$ and $M + 3773 = b^2$ with $0 \le a < b$. Subtracting,
$$b^2 - a^2 = (b - a)(b + a) = 2560 = 2^9\cdot 5.$$
The factors $b - a$ and $b + a$ have the same parity, and their product is even, so both are even.

Since $M = a^2 - 1213$, we want $a$ as large as possible. From $a = \frac{(b+a) - (b-a)}{2}$, $a$ is largest when the gap $b - a$ is smallest. The smallest even gap is $b - a = 2$, giving $b + a = 1280$, so $a = 639$ and $b = 641$.

Then $M = 639^2 - 1213$. For the units digit: $639^2$ ends in $9^2 = 81 \to 1$, and subtracting a number ending in $3$ from one ending in $1$ leaves a units digit of $8$. (Indeed $M = 408321 - 1213 = 407108$.)

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

Two squares with a fixed difference are governed by $(b-a)(b+a) = \text{constant}$, so there are only finitely many candidates and the extreme ones come from the extreme factorizations: the closest squares (largest $a$) use the smallest allowed gap. The parity restriction on the factor pair is the standard subtlety.

• Trying b − a = 1, which makes a non-integer; the two factors must have the same parity, so the smallest usable gap is 2.
• Computing the units digit as $1 - 3 = -2$ and reading it as $2$; borrowing gives $11 - 3 = 8$.