Congruence question; please help.

Question

Determine the largest integer $n$ which has at most three digits and equals the remainder when $n^2$ is divided by $1000$.

I've determined that the last digit can only be 1, 5, or 6, but if I try to continue, I meet some problems with $n^2$.

Welcome to Mathematics Stack Exchange. You mean like $625$? Are you familiar with modular arithmetic? — J. W. Tanner, Aug 26 '20 at 22:45
Same method(s) as in the linked dupe, except choose the largest solution. — Bill Dubuque, Aug 26 '20 at 22:54

score 0 · Accepted Answer · answered Aug 26 '20 at 22:53

You're looking for $n$ such that $1000$ divides $n^2-n=n(n-1)$.

That means $8$ divides $n(n-1)$, and $125$ divides $n(n-1)$.

That means $8$ divides $n$ or $n-1$, and $125$ divides $n$ or $n-1$.

There aren't many numbers $n$ of at most three digits such that $125$ divides $n$ or $n-1$.

Look for ones such that also $8$ divides $n$ or $n-1$, and you'll find what you're looking for.

1 Answers1