How to solve a equation with special conditions?

Question

I have this equation: $z = 11n + 13m$

Conditions: $z < 2015$

$z$, $n$ and $m$ must be natural numbers ($>0$).

How many options are possible for $z$?

Answer 1

We have 11n + 13m ≤ 2014; m, n natural with nm > 0. 2014 – 11 = 2003 and 2003/13 = 154 + 1/13. Then for n = 1, 2, 3, 4,……, 154 there are respectively the solutions (1, x) with 1 ≤ x ≤ 154;(2, x) with 1≤ x ≤ 153;......(153, x) with x = 1 and 2; (154, x) with x = 1.

Then this gives 154x155/2 = 11 935 solutions. On the other hand, m can range from 1-154 only because 13x155 = 2015 and this gives the solutions (x, 1) with 1 ≤ x ≤ 154; (x, 2) with 1≤ x ≤ 153; ………, (x,153) with x = 1 and 2 and (x, 154) with x = 1. We have to note carefully that all the solutions of the first group are repeated in the second group. Consequently there are 11 935 – X possible values for z, where X is the number of eventual solutions of the diophantine equation 11x +13y = 11z + 13w with (x, y) ≠ (z, w) in the allowed domain. We need to solve this equation. I allow myself to leave this final problem to other citizens who want to address the problem posted by linuscl.