How many ordered pairs of positive integers (x, y) satisfy 3x + 5y = 200?

x = (200 − 5y)/3 must be a positive integer.

200 ≡ 2 (mod 3) and 5y ≡ 2y (mod 3), so 2y ≡ 2 → y ≡ 1 (mod 3).

Also 5y < 200 → y ≤ 39. Valid y: 1, 4, 7, …, 37 → (37 − 1)/3 + 1 = 13 values (y = 40 would give x = 0, not positive).

Answer: 13.

Tip: once one solution (x, y) = (65, 1) is found, others follow by x → x − 5, y → y + 3.