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 $y$ = $y_0$ - $2t$ . 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 $127$ , $\boxed{\textbf{(D)}}$ .

Solution by Geometry-Wizard

1963 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 30	Followed by Problem 32
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1963 AHSME Problems/Problem 31

Contents

Problem

Solution 1

Solution 2

Solution 3

See Also