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

Let a be any positive integer and b = 6. Then, by Euclid’s algorithm,

a = 6q + r for some integer $$q \ne 0$$, and r = 0, 1, 2, 3, 4, 5 because $$0 \leq r < 6$$.

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also, 6q + 1 = 2 × 3q + 1 = $$2k_1 + 1$$, where $$k_1$$ is a positive integer

6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = $$2k_2 + 1$$, where $$k_2$$ is an integer

6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = $$2k_3 + 1$$, where $$k_3$$ is an integer

Clearly, 6q + 1, 6q + 3, 6q + 5 are of the form 2k + 1, where k is an integer.

Therefore, 6q + 1, 6q + 3, 6q + 5 are not exactly divisible by 2.

Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form 6q + 1, or 6q + 3,
or 6q + 5