The square of any odd integer, when divided by 8, leaves remainder
- 3
- 5
- 0
- 1
Show answer and solution
Correct answer: D. 1
An odd integer is $2k+1$; its square is $4k(k+1)+1$.
Since $k(k+1)$ is even, $4k(k+1)$ is a multiple of 8, so the square is $\equiv 1 \pmod 8$.