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.
