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 the problem as following, We can apply hit and trial for the first solution $x_0$ = 380 and $y_0$ =1. then the general solution of the given diophanitine equation will be x = $x_0$ +3t and y = $y$ - 2t. we know that we need only positive integer solutions. so 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, $\boxed{\textbf{(D)}}$ ..

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 30	Followed by Problem 32
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1963 AHSME Problems/Problem 31

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also