Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 15"
Line 1: | Line 1: | ||
== 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 == | ||
Line 7: | Line 12: | ||
== See also == | == See also == | ||
* [[University of South Carolina High School Math Contest/1993 Exam]] | * [[University of South Carolina High School Math Contest/1993 Exam]] | ||
+ | |||
+ | [[Category:Intermediate Number Theory Problems]] |
Revision as of 20:21, 22 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)