Number System-eighth grade-3
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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 = (4n2 + 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.
