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 1

Solving for $x$ in the equation yields $x =rfthe meaning of theta 0$. 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 $\boxed{\textbf{(D)}}$.

Solution 2

We will prove that $y$ is an odd number by contradiction. If $y$ is even, then we know that $y = 2m$ where $m$ is some integer. However, this immediately assumes that $\text{ even } + \text{ even } = \text{ odd }$ which is impossible. therefore $y$ must ben odd. then we can easily prove that $x$ .....

Solution 3

We can solve for first solution by applying extended euclids division algorithm or we can apply hit and trial for the first solution to get $x_0$ = $380$ and $y_0$ =$1$ . then the general solution of the given diophanitine equation will be $x$ = $x_0$ +$3t$ and $y$ = $y_0$ - $2t$. Since we need only positive integer solutions So we solve $380$ + $3t$ $>$ $0$ and $1$-$2t$ $>$ $0$ to get $t$ $>$ $0$ (applying Greatest integer function) also we can clearly see that $t_{(min)}$ $=$ $0$ so,t $<$ $GIF$($383$/$3$). That implies $t$ ranges from $0$ to $127$. Hence,the correct answer is $127$, $\boxed{\textbf{(D)}}$.

~Geometry-Wizard

See Also

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