Question

show that any +ve odd integer is of the form 6q+1or6q+3or6q+5 where q is some integer

Answer 1

Let 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 0 ≤ 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 = 2k1 + 1, where k1 is a positive integer 
6q + 3 = (6q + 2) + 1 = 2 (3q + 1) + 1 = 2k2 + 1, where k2 is an integer 
6q + 5 = (6q + 4) + 1 = 2 (3q + 2) + 1 = 2k3 + 1, where k3 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