Prove that if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for some integer q, but not conversely.

Let, n = 6q + 5, when q is a positive integer
We know that any positive integer is of the form 3k, or 3k + 1, or 3k + 2
∴ q = 3k or 3k + 1, or 3k + 2
If q = 3k, then
n = 6q + 5
= 6(3k) + 5
= 18k + 5
= 18k + 3 + 2
= 3(6k + 1) + 2
= 3m + 2, where m is some integer
If q = 3k + 1, then
n = 6q + 5
= 6(3k + 1) + 5
= 18k + 6 + 5
= 18k + 11
= 3(6k + 3) + 2
= 3m + 2, where m is some integer
If q = 3k + 2, then
n = 6q + 5
= 6(3k + 2) + 5
= 18k + 12 + 5
= 18k + 17
= 3(6k + 5) + 2
= 3m + 2, where m is some integer
Hence, if a positive integer is of the form 6q + 5, then it is of the form 3q + 2 for some integer q.

Conversely
Let n = 3q + 2
We know that a positive integer can be of the form 6k + 1, 6k + 2, 6k + 3, 6k + 4 or 6k + 5
So, now if q = 6k + 1 then
n = 3(6k + 1) + 2
= 18k + 5
= 6(3k) + 5

= 6m + 5, where m is some integer
So, now if q = 6k + 2 then
n = 3(6k + 2) + 2
= 18k + 8
= 6 (3k + 1) + 2
= 6m + 2, where m is some integer
Now, this is not of the form 6m + 5
Hence, if n is of the form 3q + 2, then it necessarily won’t be of the form 6q + 5 always.

let A = 6q + 5, be any number, where q is any positive integer.

Part 1:

To show A is in the form of 3q + 2, where q is another integer A = 6q + 5 = 6q + 3 + 2 = 3(2q + 1) + 2 = 3q' + 2

As,

q is any positive integer, q' = 3q + 2 is also a positive integer and hence 6q + 5, is in form of 3q' + 5

Part 2:

To show converse is not true, i.e. if a no is in the form of 3q + 2, then it may or may not be in the form of 6q + 5

For example, consider:

8 = 3(2) + 2 is in the 3q + 2 form, but it can't be expand in 6q + 5 form.