The number of solutions $(x, y)$ of the equation $3 x+y=100$, where $x$ and $y$ are positive integers, is
(A) 33
(B) 35
(C) 100
(D) 101
(E) 97
The number of solutions $(x, y)$ of the equation $3 x+y=100$, where $x$ and $y$ are positive integers, is
(A) 33
(B) 35
(C) 100
(D) 101
(E) 97

## Solution

Rewrite the given equation as $x=\frac{100-y}{3}$. Since $x$ must be a positive integer, $100-y$ must be divisible by 3 . Since $y$ must also be a positive integer, the only possible values of $y$ are $1,4,7,10,13$, $\ldots, 94$, and 97 . Thus, there are 33 possible values for $y$, and 33 solutions $(x, y)$ that meet the given conditions.

ANSWER: (A)