How many ordered pairs of positive integers (x, y) satisfy 3x + 5y = 301?

We need 301 − 5y to be a positive multiple of 3.

301 ≡ 1 (mod 3) and 5y ≡ 2y (mod 3), so 2y ≡ 1 (mod 3) → y ≡ 2 (mod 3).

Also 5y < 301 → y ≤ 60; the largest y ≡ 2 (mod 3) with 301 − 5y > 0 is 59 (x = 2). Smallest is y = 2 (x = 97).

y ∈ {2, 5, 8, …, 59}: count = (59 − 2)/3 + 1 = 20.

Tip: Once one solution (x = 97, y = 2) is found, others follow by x decreasing by 5 while y increases by 3.