2025 AMC 12A Problems/Problem 2

The following problem is from both the 2025 AMC 10A #2 and 2025 AMC 12A #2, so both problems redirect to this page.

Problem

Let $m$ and $n$ be $2$ integers such that $m>n$ . Suppose $m+n=20$ , $m^2+n^2=328$ , find $m^2-n^2$ .

$\textbf{(A)}~280\qquad\textbf{(B)}~292\qquad\textbf{(C)}~300\qquad\textbf{(D)}~320\qquad\textbf{(E)}~340$

Solution

Since $m^2-n^2=(m+n)(m-n)$ , we only need to find $m-n$ . Squaring $m+n=20$ gives $m^2+2mn+n^2=400$ , thus $2mn=400-328=72$ and $mn=36$ . Through trial and error, the only possible value of $(m,n)$ is $(18,2)$ , giving $m-n=16$ . Hence, $m^2-n^2=20*16=\boxed{\textbf{(D) }320}.$

~Frankensteinquixotecabin

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

Page

Toolbox

Search

2025 AMC 12A Problems/Problem 2

Problem

Solution

See also