Question

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.

Answer 1

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.