Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.By Euclid’s division lemma, we havea = bq + r; 0 ≤ r < bFor a = n and b = 3, we haven = 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{tex}n = 3q{/tex}∴ n is divisible by 3.{tex}n + 1 = 3q + 1{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 2{/tex}∴ n + 2 is not divisible by 3.Putting r = 1 in (i), we get{tex}n = 3q + 1{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 2{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 3 = 3(q + 1){/tex}∴ n + 2 is divisible by 3.Putting r = 2 in (i), we get{tex}n = 3q + 2{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 3 = 3(q + 1){/tex}∴ n + 1 is divisible by 3.{tex}n + 2 = 3q + 4{/tex}∴ 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.

Let the three consecutive positive integers be n, n + 1 and n + 2, where n is any integer.By Euclid’s division lemma, we havea = bq + r; 0 ≤ r < bFor a = n and b = 3, we haven = 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{tex}n = 3q{/tex}∴ n is divisible by 3.{tex}n + 1 = 3q + 1{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 2{/tex}∴ n + 2 is not divisible by 3.Putting r = 1 in (i), we get{tex}n = 3q + 1{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 2{/tex}∴ n + 1 is not divisible by 3.{tex}n + 2 = 3q + 3 = 3(q + 1){/tex}∴ n + 2 is divisible by 3.Putting r = 2 in (i), we get{tex}n = 3q + 2{/tex}∴ n is not divisible by 3.{tex}n + 1 = 3q + 3 = 3(q + 1){/tex}∴ n + 1 is divisible by 3.{tex}n + 2 = 3q + 4{/tex}∴ 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.