2024 AMC 12A Problem 9

Attempt Problem 9 of the 2024 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2024 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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

Let

M+1213=a^2

and

M+3773=b^2;

subtract to get

b^2-a^2=2560

(b-a)(b+a)=2560

with both factors even; to maximize

M,

make

b-a

as small as possible

Write

M+1213=a^2

and

M+3773=b^2,

b^2-a^2=2560,

i.e.

(b-a)(b+a)=2560.

Both factors have the same parity, hence both even. To maximize

a

(and thus

M

), minimize

b-a:

take

b-a=2,\ b+a=1280,

a=639.

Then

M=639^2-1213

=408321-1213

=407108,

whose units digit is

8.

Thus, the correct answer is E.

Problem 9 in other years