Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 9"
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− | We have <math>(y+x)(y-x)=187</math>. Now, | + | 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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− | * [[University of South Carolina High School Math Contest/1993 Exam]] | + | |
+ | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 8|Previous Problem]] | ||
+ | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 10|Next Problem]] | ||
+ | * [[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 and are integers such that and . Then one possible value of is
Solution
We have . Now, , so we have as one possibility and . Thus, is a possible solution and the answer is .