- #1

- 1,039

- 2

## Homework Statement

Prove that for every integer n>=8, there exists nonnegative integers a and b, such that n =3a+5b

## Homework Equations

## The Attempt at a Solution

I'm trying to understand the proof of this. It goes as follows:

I am having a hard time figuring out what is going on in any of the three cases.

**Firstly, why did they pick 3q, 3q+1, and 3q+2?**

They could do this proof using n = 5q, 5q+1, 5q+2, 5q+3, 5q+4 also correct? It would just have more cases, and therefor, be more work?

Then, about the cases.

Case I: n = 3q

I see what they did here. Since they picked to represent their integer as 3q, they supposed that b = 0, and then it's easy to show there exists an a, since a = q at that point. And a >=3 because n = 3q needs to be greater than 8.

Case II: n = 3q+1

Here is where I am troubled.

**How do they make the jump to claim that n = 3(q − 3) + 10 ??**

What is the thought process behind this?

I would appreciate any explanation that you could provide. Thanks!