Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 15"
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== Problem == | == Problem == | ||
+ | If we express the sum | ||
− | <center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ } </math></center> | + | <center> <math> \frac 1{3\cdot 5\cdot 7\cdot 11} + \frac 1{3\cdot 5\cdot 7\cdot 13} + \frac 1{3\cdot 5\cdot 11\cdot 13} + \frac 1{3\cdot 7\cdot 11\cdot 13} + \frac 1{5\cdot 7\cdot 11\cdot 13} </math> </center> |
+ | |||
+ | as a rational number in reduced form, then the denominator will be | ||
+ | |||
+ | <center><math> \mathrm{(A) \ }15015 \qquad \mathrm{(B) \ }5005 \qquad \mathrm{(C) \ }455 \qquad \mathrm{(D) \ }385 \qquad \mathrm{(E) \ }91 </math></center> | ||
== Solution == | == Solution == | ||
+ | By changing the fractions to have a common [[denominator]] of <math>3\cdot 5\cdot 7\cdot 11\cdot 13</math>, it is easier to add them and simplify the sum. | ||
+ | |||
+ | Doing so yields: | ||
+ | |||
+ | <math> \frac 1{3\cdot 5\cdot 7\cdot 11} + \frac 1{3\cdot 5\cdot 7\cdot 13} + \frac 1{3\cdot 5\cdot 11\cdot 13} + \frac 1{3\cdot 7\cdot 11\cdot 13} + \frac 1{5\cdot 7\cdot 11\cdot 13} </math> | ||
+ | |||
+ | <math>=</math> | ||
+ | |||
+ | <math> \frac {13}{3\cdot 5\cdot 7\cdot 11\cdot 13} + \frac {11}{3\cdot 5\cdot 7\cdot 11\cdot 13} + \frac {7}{3\cdot 5\cdot 7\cdot 11\cdot 13} + \frac {5}{3\cdot 5\cdot 7\cdot 11\cdot 13} + \frac {3}{3\cdot 5\cdot 7\cdot 11\cdot 13}</math> | ||
+ | |||
+ | <math>=</math> | ||
+ | |||
+ | <math> \frac {39}{3\cdot 5\cdot 7\cdot 11\cdot 13} = \frac {3\cdot13}{3\cdot 5\cdot 7\cdot 11\cdot 13} = \frac{1}{5\cdot 7\cdot 11} = \frac{1}{385}</math> | ||
+ | |||
+ | So the answer is <math>385 \Rightarrow D</math> | ||
+ | ---- | ||
+ | |||
+ | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 14|Previous Problem]] | ||
+ | * [[University of South Carolina High School Math Contest/1993 Exam/Problem 16|Next Problem]] | ||
+ | * [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]] | ||
− | + | [[Category:Intermediate Number Theory Problems]] | |
− |
Latest revision as of 22:11, 31 July 2006
Problem
If we express the sum
as a rational number in reduced form, then the denominator will be
Solution
By changing the fractions to have a common denominator of , it is easier to add them and simplify the sum.
Doing so yields:
So the answer is