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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 9"
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== Solution ==
 
== Solution ==
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We have <math>(y+x)(y-x)=187</math>. Now, since <math>187=11 \cdot 17</math>. Therefore, <math>y+x=17</math> and <math>y-x=11</math>. Thus, <math>x=3, y=14</math> is a possible solution and the answer is <math>42</math>.
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We have <math>(y+x)(y-x)=187</math>. Now, <math>187=11 \cdot 17</math>, so we have as one possibility <math>y+x=17</math> and <math>y-x=11</math>. Thus, <math>x=3, y=14</math> is a possible solution and the answer is <math>42</math>.
  
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== See also ==
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* [[University of South Carolina High School Math Contest/1993 Exam]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 8|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 10|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]

Latest revision as of 11:16, 16 August 2006

Problem

Suppose that $x$ and $y$ are integers such that $y > x > 1$ and $y^2 - x^2 = 187$. Then one possible value of $xy$ is

$\mathrm{(A) \ }30 \qquad \mathrm{(B) \ }36 \qquad \mathrm{(C) \ }40 \qquad \mathrm{(D) \ }42 \qquad \mathrm{(E) \ }54$

Solution

We have $(y+x)(y-x)=187$. Now, $187=11 \cdot 17$, so we have as one possibility $y+x=17$ and $y-x=11$. Thus, $x=3, y=14$ is a possible solution and the answer is $42$.

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