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Prove that one and only one out of n, n + 2 and n + 4 is divisible by 3, where n is any positive integer. |
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Answer» According to Euclid’s division Lemma, Let the positive integer = n And b=3 n =3q+r, where q is the quotient and r is the remainder 0<r<3 implies remainders may be 0, 1 and 2 Therefore, n may be in the form of 3q, 3q+1, 3q+2 When n=3q n+2=3q+2 n+4=3q+4 Here n is only divisible by 3 When n = 3q+1 n+2=3q=3 n+4=3q+5 Here only n+2 is divisible by 3 When n=3q+2 n+2=3q+4 n+4=3q+2+4=3q+6 Here only n+4 is divisible by 3 So, we can conclude that one and only one out of n, n + 2 and n + 4 is divisible by 3. Hence Proved |
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