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University of South Carolina High School Math Contest/1993 Exam/Problem 15

From AoPSWiki

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\cd...

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

Solution

By changing the fractions to have a common denominator of 3\cdot 5\cdot 7\cdot 11\cdot 13, it is easier to add them and simplify the sum.

Doing so yields:

\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\cd...

=

\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...

=

\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} ...

So the answer is 385 \Rightarrow D


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