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The square of any odd integer, when divided by 8, leaves remainder

A3
B5
C0
D1
Answer & 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 $\equiv 1 \pmod 8$.
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