2024 AMC 10A Problem 15

Question

Let

M

be the greatest integer such that both

M + 1213

and

M + 3773

are perfect squares. What is the units digit of

M?

1

2

3

6

8

Set

M + 1213 = y^2

and

M + 3773 = x^2;

subtract to get

(x - y)(x + y) = 2560

Both factors are even; the greatest

M

comes from the smallest

x - y,

making

y

as large as possible

Set

M + 1213 = y^2

and

M + 3773 = x^2.

Subtracting,

x^2 - y^2 = 2560,

so

(x - y)(x + y) = 2560.

The two factors share a parity, and their product is even, so both are even: write

x - y = 2s,

x + y = 2t,

with

st = 640.

To make

M

as large as possible we want

y = t - s

as large as possible, so

s

as small as possible. Take

s = 1, t = 640,

giving

y = 639.

Then

M = 639^2 - 1213 = 407108,

whose units digit is

8.

Thus, E is the correct answer.

Problem 15 in other years