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Show that one and only one out of n, n + 2 or n + 4 is divisible by 3, where n is any positive integer. 

(Or) 

Show that one and only one out of a, a + 2 and a + 4 is divisible by 3 where ‘a’ is any positive integer.

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Let ‘n’ be any positive integer. 

Then from Euclidean’s lemma n = bq + r (now consider b = 3) 

⇒ n = 3q + r (here 0 ≤ r < 3) which means the possible values of ‘r’ = 0 or 1 or 2

Now consider r = 0 then ‘n’ = 3q (divisible by 3) and n + 2 = 3q + 2 (not divisible by 3) n + 4 = 3q + 4 (not divisible by 3) 

Case – II: For r = 1 

n = 3q + 1 (not divisible by 3) n + 2 = 3q + 1 + 2 = 3q + 3 = 3(q + l) divisible by 3 

n + 4 = 3q + 1 + 4 = 3q + 5 not divisible by 3 

Case – III: For r = 2, 

n = 3q + 2 not divisible by 3 

n + 2 = 3q + 2 + 2 = 3q + 4, not divisible by 3 

n + 4 = 3q + 2 + 4 = 3q + 6 = 3(q + 2) divisible by 3 

So in all above three cases we observe, only one of either (n) or (n + 1) or (n + 4) is divisible by 3. 

Hence proved.

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