Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 9"
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== Problem == | == Problem == | ||
+ | Suppose that <math>x</math> and <math>y</math> are integers such that <math>y > x > 1</math> and <math>y^2 - x^2 = 187</math>. Then one possible value of <math>xy</math> is | ||
− | <center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ } </math></center> | + | <center><math> \mathrm{(A) \ }30 \qquad \mathrm{(B) \ }36 \qquad \mathrm{(C) \ }40 \qquad \mathrm{(D) \ }42 \qquad \mathrm{(E) \ }54 </math></center> |
== Solution == | == Solution == | ||
+ | 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>. | ||
− | + | ---- | |
− | * [[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]] |
Latest revision as of 11:16, 16 August 2006
Problem
Suppose that and
are integers such that
and
. Then one possible value of
is
![$\mathrm{(A) \ }30 \qquad \mathrm{(B) \ }36 \qquad \mathrm{(C) \ }40 \qquad \mathrm{(D) \ }42 \qquad \mathrm{(E) \ }54$](http://latex.artofproblemsolving.com/9/a/d/9add595fce11ea3c0ed2d84d445ec04ea079e723.png)
Solution
We have . Now,
, so we have as one possibility
and
. Thus,
is a possible solution and the answer is
.