Consider the integral points (x, y, z)(x,y,z) on the plane $35 x + 55 y + 77 z = 1$. How many are contained within a cube with side length 30 centered at (0, 0, 0)?
https://brilliant.org/problems/points-in-a-box-2/#!/solution-comments/
The solution by the author is not very comprehensive. I understand it vaguely until the end where the author ends with "part two: solving inequalities". How would you go about turning this into an inequality?
Thanks to everyone answering (-:
(Extension: Would it be possible to do this with a 3D space and 4D hypercube? Just wondering.)
Hint
You have the system
$$35x + 55y + 77z = 1$$
$$-15 \leq x,y,z \leq 15$$
Let $x+15 = X, y+15=Y, z+15 = Z$
The system becomes
$$35X + 55Y + 77Z = 2506$$
$$0 \leq X,Y,Z \leq 30$$
Can you solve it now? It's a basic combinatorial problem you can solve using stars and bars method