Question:

If $\gcd(m,n)=1$ and $1^2-2^2+3^2-4^2+\cdots+(2023)^2=1012m^2n$, then $m^2-n^2$ is equal to

240

200

220

180

Step-by-Step Solution

Key Concept: $1^2-2^2+\cdots-(2022)^2+(2023)^2=-(3+7+\cdots+4043)+(2023)^2=-1011\cdot2023+(2023)^2=2023\cdot1012$.

$m=17,n=7$. $m^2-n^2=240$.

Correct Answer: 1