Suppose a be any positive integer and b = 6. Then, by Euclid’s algorithm, 𝒂 = 𝟔𝒒 + 𝒓, for some integer 𝒒 ≥ 𝟎, and 𝒓 = 𝟎,𝟏,𝟐,𝟑,𝟒,𝟓, because of 𝟎≤𝒓<𝟔.

Now substituting the value of r, we get,

If r = 0, then 𝒂 = 𝟔𝒒

Similarly, for 𝒓= 𝟏,𝟐,𝟑,𝟒 and 5, the value of a is 𝟔𝒒+𝟏,𝟔𝒒+𝟐,𝟔𝒒+𝟑,𝟔𝒒+𝟒 and 𝟔𝒒+𝟓, respectively.

If 𝒂 = 𝟔𝒒,𝒂=𝟔𝒒+𝟐 𝐚𝐧𝐝 𝒂 = 𝟔𝒒+𝟒, then a is an even number and divisible by 2.

A positive integer can be either even or odd. Therefore, any positive odd integer is of the form of 𝟔𝒒+𝟏,𝟔𝒒+𝟑 𝐚𝐧𝐝 𝟔𝒒+𝟓, where q is some integer.