1963 AHSME Problems/Problem 31: Difference between revisions

Latest revision as of 13:07, 14 January 2024

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

1963 AHSC (Problems • Answer Key • Resources)
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@@ Line 9: / Line 9: @@
 \textbf{(E)}\ 0     </math>
-==Solution==
+==Solution 1==
-Solving for <math>x</math> in the equation yields <math>x = \frac{763-3y}{2}</math>.  In order for <math>x</math> to be an integer, <math>763-3y</math> must be even.  Since <math>763</math> is odd, <math>3y</math> must be odd as well, so <math>y</math> must be an odd number.
+Solving for <math>x</math> in the equation yields <math>x =rfthe meaning of theta 0</math>.  Solving the inequality results in <math>y \le 254 \frac{1}{3}</math>.  From the two conditions, <math>y</math> can be an odd number from <math>1</math> to <math>253</math>, so there are <math>127</math> solutions where <math>x</math> and <math>y</math> are integers.  The answer is <math>\boxed{\textbf{(D)}}</math>.
+==Solution 2==
+We will prove that <math>y</math> is an odd number by contradiction. If <math>y</math> is even, then we know that <math>y = 2m</math> where <math>m</math> is some integer. However, this immediately assumes that <math>\text{ even } + \text{ even } = \text{ odd }</math> which is impossible. therefore <math>y</math> must ben odd. then we can easily prove that <math>x</math> .....
-In order for <math>x</math> to be positive, <math>763-3y \ge 0</math>.  Solving the inequality results in <math>y \le 254 \frac{1}{3}</math>.  From the two conditions, <math>y</math> can be an odd number from <math>1</math> to <math>253</math>, so there are <math>127</math> solutions where <math>x</math> and <math>y</math> are integers.  The answer is <math>\boxed{\textbf{(D)}}</math>.
+==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 <math>x_0</math> = <math>380</math> and <math>y_0</math> =<math>1</math> . then the general solution of the given diophanitine equation will be <math>x</math> = <math>x_0</math> +<math>3t</math> and <math>y</math> = <math>y_0</math> - <math>2t</math>. Since we need only positive integer solutions So we solve <math>380</math> + <math>3t</math> <math>></math> <math>0</math> and <math>1</math>-<math>2t</math> <math>></math> <math>0</math> to get <math>t</math> <math>></math> <math>0</math> (applying Greatest integer function) also we can clearly see that <math>t_{(min)}</math> <math>=</math> <math>0</math> so,t <math><</math> <math>GIF</math>(<math>383</math>/<math>3</math>). That implies <math>t</math> ranges from <math>0</math> to <math>127</math>. Hence,the correct answer is <math>127</math>, <math>\boxed{\textbf{(D)}}</math>.
+~Geometry-Wizard
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