Question

show that 1 & 1 out of n , (n+2) , (n+4) is divisible by 3 , where n is any positive integer

Answer 1

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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 . Euclid's division Lemma any natural number can be written as: . where r = 0, 1, 2,. and q is the quotient. ∵ Thus any number is in the form of 3q , 3q+1 or 3q+2.

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Answer 2

(n+2) (n+4) is divisible by 3

We applied Euclid Division algorithm on n and 3.

a = bq +r on putting a = n and b = 3

n = 3q +r , 0<r<3

i.e n = 3q -------- (1),n = 3q +1 --------- (2), n = 3q +2 -----------(3)

n = 3q is divisible by 3

or n +2 = 3q +1+2 = 3q +3 also divisible by 3

or n +4 = 3q + 2 +4 = 3q + 6 is also divisible by 3

Hence n, n+2 , n+4 are divisible by 3.