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Show 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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Let q be quotient and r be the remainder. 

On applying Euclid’s algorithm, i.e. dividing n by 3, 

we have 

n = 3q + r 0 ≤ r ˂ 3 

⇒ n = 3q + r r = 0, 1 or 2 

⇒ n = 3q or n = (3q + 1) or n = (3q + 2) 

Case 1: If n = 3q, then n is divisible by 3. 

Case 2: If n = (3q+1), then (n+2) = 3q + 3 = 3(q + 1), which is clearly divisible by 3. 

In this case, (n+2) is divisible by 3. 

Case 3: If n = (3q+2), then (n+4) = 3q + 6 = 3(q + 2), which is clearly divisible by 3. 

In this case, (n+4) is divisible by 3. 

Hence, one and only one out of n, (n+2) and (n+4) is divisible by 3.

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