# Number System-eighth grade-3

### From Homeworkwiki

**Problem: Prove that when divided by 8, the square of an odd number always gives 1 as remainder.**

**Solution:** Let the odd number be (2n + 1)

Its square is (2n + 1)^{2} = (4n^{2} + 4n + 1)

= 4n(n + 1) + 1

We know that n(n + 1) is divisible by 2, whether n is even or odd

Therefore, 4n(n + 1) is always divisible by 8

Therefore, When 4n(n + 1) + 1 is divisible by 8, we get 1 as remainder.