How many positive integer solutions does the equation 4x + 3y = 100 have?

Concept: For a linear Diophantine equation ax + by = c in two variables with gcd(a, b) = 1, integer solutions always exist, and once one solution (x₀, y₀) is found, every integer solution has the form x = x₀ + bt, y = y₀ − at for integer t. Restricting to POSITIVE integer solutions turns this into counting how many values of the step parameter t keep both x > 0 and y > 0 at the same time — a finite, bounded range.

1. Write y in terms of x: y = (100 − 4x) / 3.

2. For y to be an integer, 100 − 4x must be divisible by 3. Since 4 ≡ 1 (mod 3) and 100 ≡ 1 (mod 3), this requires x ≡ 1 (mod 3).

3. So x must come from the arithmetic sequence 1, 4, 7, 10, 13, 16, 19, 22, 25, … (first term 1, common difference 3).

4. Apply the positivity bound: y > 0 needs 4x < 100, i.e. x < 25 — so x = 25 (which makes y = 0) is excluded, leaving x ∈ {1, 4, 7, 10, 13, 16, 19, 22}.

5. That is 8 values of x, each giving a unique positive integer y — so the equation has 8 positive integer solutions.

Cross-check the boundary of the valid range directly: at x = 1 (the smallest term), y = (100 − 4) / 3 = 32, a positive integer; at x = 22 (the largest term counted), y = (100 − 88) / 3 = 4, a positive integer; but at the next candidate x = 25, y = (100 − 100) / 3 = 0, which is not positive and is correctly excluded — confirming the listed range x = 1 to x = 22 is exactly right.