Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 15"
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If we express the sum | If we express the sum | ||
− | <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> | + | <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 | as a rational number in reduced form, then the denominator will be | ||
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== 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> | ||
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Latest revision as of 22:11, 31 July 2006
Problem
If we express the sum
![$\frac 1{3\cdot 5\cdot 7\cdot 11} + \frac 1{3\cdot 5\cdot 7\cdot 13} + \frac 1{3\cdot 5\cdot 11\cdot 13} + \frac 1{3\cdot 7\cdot 11\cdot 13} + \frac 1{5\cdot 7\cdot 11\cdot 13}$](http://latex.artofproblemsolving.com/c/f/0/cf021ead1590554d4996161c2d950d3c7f447e08.png)
as a rational number in reduced form, then the denominator will be
![$\mathrm{(A) \ }15015 \qquad \mathrm{(B) \ }5005 \qquad \mathrm{(C) \ }455 \qquad \mathrm{(D) \ }385 \qquad \mathrm{(E) \ }91$](http://latex.artofproblemsolving.com/8/c/3/8c3fe4240270899d2fdd51fbf04d42974d3c2c43.png)
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