Skip to main content

2024 AMC 12A Problem 9

Problem 9 of 25EasierAlgebraNumber TheoryProblem-Solving Techniques

Let MM be the greatest integer such that both M+1213M+1213 and M+3773M+3773 are perfect squares. What is the units digit of M?M?

Answer choices

Show solution

Solution

Write M+1213=a2M+1213=a^2 and M+3773=b2,M+3773=b^2, so b2−a2=2560,b^2-a^2=2560, i.e. (b−a)(b+a)=2560.(b-a)(b+a)=2560. Both factors have the same parity, hence both even. To maximize aa (and thus MM), minimize b−a:b-a: take b−a=2, b+a=1280,b-a=2,\ b+a=1280, so a=639.a=639. Then M=6392−1213M=639^2-1213 =408321−1213=408321-1213 =407108,=407108, whose units digit is 8.8. Thus, the correct answer is E.
AoPS wiki

Tagged: difference of squares · perfect square · optimization

More practice