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

Problem

If we express the sum

$\frac 1{3\cdot 5\cdot 7\cdot 11} + \frac 1{3\cdot 5\cdot 7\cdot 13} + \frac 1{3\cdot 5\cdot 11\cdot 13} + \frac 1{3\cdot 7\cdot 11\cdot 13} + \frac 1{5\cdot 7\cdot 11\cdot 13}$

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:

$\frac 1{3\cdot 5\cdot 7\cdot 11} + \frac 1{3\cdot 5\cdot 7\cdot 13} + \frac 1{3\cdot 5\cdot 11\cdot 13} + \frac 1{3\cdot 7\cdot 11\cdot 13} + \frac 1{5\cdot 7\cdot 11\cdot 13}$

$=$

$\frac {13}{3\cdot 5\cdot 7\cdot 11\cdot 13} + \frac {11}{3\cdot 5\cdot 7\cdot 11\cdot 13} + \frac {7}{3\cdot 5\cdot 7\cdot 11\cdot 13} + \frac {5}{3\cdot 5\cdot 7\cdot 11\cdot 13} + \frac {3}{3\cdot 5\cdot 7\cdot 11\cdot 13}$

$=$

$\frac {39}{3\cdot 5\cdot 7\cdot 11\cdot 13} = \frac {3\cdot13}{3\cdot 5\cdot 7\cdot 11\cdot 13} = \frac{1}{5\cdot 7\cdot 11} = \frac{1}{385}$

So the answer is $385 \Rightarrow D$

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

Problem

Solution