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

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

Solution:

Let a be positive odd integer which gives q as quotient and r as remainder when it is divide by 6.
Then using Euclid’s division lemma a = bq + r, 0 ≤ r < b we get,
a = 6q + r, 0 ≤ r < 6,
So possible values of r are 0, 1, 2, 3, 4 and 5.
Now, when r = 0, a = 6q = 2 x 3q, which is divisible by 2 so 6q is even positive number.
when, r = 1, a = 6q + 1, not divisible by 2, hence 6q + 1 is odd positive number.
when, r = 2, a = 6q + 2 = 2(3q + 1 ) i.e., divisible by 2, hence 6q + 2 is even positive number.
when, r = 3, a = 6q + 3 = 3(2q + 1), not divisible by 2, so 6q + 3 is odd positive number.
when, r = 4, a = 6q + 4 = 2(3q + 2),  divisible by 2, hence 6q + 4 is even positive number.
when, r = 5, a = 6q + 5, not divisible by 2, so it is an odd positive integer.
Hence only odd positive integers are 6q + 1, 6q + 3, 6q + 5.
Therefore, any positive odd integer is of the form 6q + 1 or 6q + 3 or 6q + 5, where q is some integer.

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