Difference between revisions of "1963 AHSME Problems/Problem 31"

(Solution to Problem 31)
 
 
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\textbf{(E)}\ 0    </math>
 
\textbf{(E)}\ 0    </math>
  
==Solution==
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==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 evenSince <math>763</math> is odd, <math>3y</math> must be odd as well, so <math>y</math> must be an odd number.
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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 integersThe answer is <math>\boxed{\textbf{(D)}}</math>.
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==Solution 2==
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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.  
  
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>.
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then we can easily prove that <math>x</math> .....
  
 
==See Also==
 
==See Also==

Latest revision as of 18:42, 18 January 2021

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 $x$ .....

See Also

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