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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.

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#### Solution

Consider any two positive integers **a** and **b** such that **a** is greater than **b**, then according to Euclid's division algorithm:

a = bq + r; where q and r are positive integers and 0 ≤ r < b

Let a = n and b = 3, then

a = bq + r ⇒ n = 3q + r; where 0 ≤ r < 3.

**r = 0** ⇒ n = 3q + 0 = 3q

**r = 1** ⇒ n = 3q + 1 and **r = 2** ⇒ n = 3q + 2

If n = 3q; **n is divisible by 3**

If n = 3q + 1; then n + 2 = 3q + 1 + 2 = 3q + 3; which is divisible by 3

⇒ **n + 2 is divisible by 3**

If n = 3q + 2; then n + 4 = 3q + 2 + 4

= 3q + 6; which is divisible by 3

⇒ **n + 4 is divisible by 3**

Hence, if **n** is any positive integer, then one and only one out of n, n + 2 or n + 4 is divisible by 3. ** **

**Hence the required result.**

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