Show that any positive odd integer is of the form 6q+1 or 6q+3 or 6q+5, where q is some integer.

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.

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