**Solution:**

Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.

By Euclid’s division lemma, we have

a = bq + r; 0 ≤ r < b

For a = n and b = 3, we have

n = 3q + r …(i)

Where q is an integer and 0 ≤ r < 3, i.e. r = 0, 1, 2.

Putting r = 0 in (i), we get

n = 3q

So, n is divisible by 3.

n + 1 = 3q + 1

so, n + 1 is not divisible by 3.

n + 2 = 3q + 2

so, n + 2 is not divisible by 3.

Putting r = 1 in (i), we get

n = 3q + 1

so, n is not divisible by 3.

n + 1 = 3q + 2

so, n + 1 is not divisible by 3.

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

so, n + 2 is divisible by 3.

Putting r = 2 in (i), we get

n = 3q + 2

so, n is not divisible by 3.

n + 1 = 3q + 3 = 3(q + 1)

so, n + 1 is divisible by 3.

n + 2 = 3q + 4

so, n + 2 is not divisible by 3.

**Thus for each value of r such that 0 ≤ r < 3 only one out of n, n + 1 and n + 2 is divisible by 3.**