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, 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>. | 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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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] |
Revision as of 12:26, 23 July 2006
Problem
Suppose that and are integers such that and . Then one possible value of is
Solution
We have . Now, since . Therefore, and . Thus, is a possible solution and the answer is .