Solution :
By Euclid’s division algorithm, we have
a = bq + r ……….(i)
On putting, b = 6 in (1), we get
a = 6q + r [0 \(\le\) r < 6]
If r = 0, a = 6q, 6q is divisible by 6 \(\implies\) 6q is even.
If r = 1, a = 6q + 1, 6q + 1 is not divisible by 2.
If r = 2, a = 6q + 2, 6q + 2 is divisible by 2 \(\implies\) 6q + 2 is even.
If r = 3, a = 6q + 3, 6q + 3 is not divisible by 2.
If r = 4, a = 6q + 4, 6q + 4 is divisible by 2 \(\implies\) 6q + 4 is even.
If r = 5, a = 6q + 5, 6q + 5 is not divisible by 2.
Since, 6q, 6q + 2, 6q + 4 are even.
Hence, the remaining integers are 6q + 1, 6q + 3, 6q + 5 are odd.
