Prove that one of every 3 consecutive integers is divisible by 3.

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1. Answer:

   Using Euclid division lemma

   Step-by-step explanation:

   Let n, n+1 & n+2 be three consecutive positive integers.

   We know that n is of the form 3q,3q+1 or, 3q+2 (As per Euclid Division Lemma),

   So, we have the following

   Case I

   When n=3q;

   In this case, n is divisible by 3 but n+1 and n+2 are not divisible by 3.

   Case II

   When n=3q+1

   In this case, n+2=3q+1+2=3(q+1) is divisible by 3 but n and n+1 are not divisible by 3.

   Case III

   When n=3q+2

   In this case, n+1=3q+1+2=3(q+1) is divisible by 3 but n and n+2 are not divisible by 3.

   Hence one of n,n+1 and n+2 is divisible by 3.

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