Question

What are the solutions of `69t+28s=1` ?

Answer 1

`69` and `28` are co-prime and hence this equation has integer solutions:

`(s, t) = (69k-32, -28k+13)`

for any integer `k`

Explanation:

Hmmm... I'm not sure what you are referring to here.

The prime factorisations of `69` and `28` are:

`69 = 3*23`

`28=2^2*7`

Since `69` and `28` have no common factors larger than `1`, they are called co-prime.

One consequence is that there are integer pairs `(s, t)` which are solutions of:

`69t+28s=1`

Conversely, if `69` and `28` were not co-prime then this equation would have no integer solutions.

Note that:

`69/28 = 2` with remainder `13`

So modulo `28`, multiples of `69` (starting with `0`) are:

`0, 13, 26, 11, 24, 9, 22, 7, 20, 5, 18, 3, 16, 1`

So we have:

`13*69 = 897 = 32*28+1`

So:

`(s, t) = (-32, 13)` is a solution.

Any other solutions will be formed by adding `69k` to `s` and `-28k` to `t` for some integer `k`.

So the solutions are:

`(s, t) = (69k-32, -28k+13)`