1963 AHSME Problems/Problem 31

Problem

The number of solutions in positive integers of $2x+3y=763$ is:

$\textbf{(A)}\ 255 \qquad \textbf{(B)}\ 254\qquad \textbf{(C)}\ 128 \qquad \textbf{(D)}\ 127 \qquad \textbf{(E)}\ 0$

Solution

Solving for $x$ in the equation yields $x = \frac{763-3y}{2}$. In order for $x$ to be an integer, $763-3y$ must be even. Since $763$ is odd, $3y$ must be odd as well, so $y$ must be an odd number.

In order for $x$ to be positive, $763-3y \ge 0$. Solving the inequality results in $y \le 254 \frac{1}{3}$. From the two conditions, $y$ can be an odd number from $1$ to $253$, so there are $127$ solutions where $x$ and $y$ are integers. The answer is $\boxed{\textbf{(D)}}$.

See Also

1963 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 30
Followed by
Problem 32
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