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Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 15"

 
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== Problem ==
 
== Problem ==
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If we express the sum
  
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<center><math> \mathrm{(A) \ } \qquad \mathrm{(B) \ } \qquad \mathrm{(C) \ } \qquad \mathrm{(D) \ } \qquad \mathrm{(E) \ }  </math></center>
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<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>
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as a rational number in reduced form, then the denominator will be
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<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 ==
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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.
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 +
Doing so yields:
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<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>
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<math>=</math>
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<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>
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<math>=</math>
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<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>
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So the answer is <math>385 \Rightarrow D</math>
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----
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 14|Previous Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam/Problem 16|Next Problem]]
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* [[University of South Carolina High School Math Contest/1993 Exam|Back to Exam]]
  
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== See also ==
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[[Category:Intermediate Number Theory Problems]]
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* [[University of South Carolina High School Math Contest/1993 Exam]]
 

Latest revision as of 22:11, 31 July 2006

Problem

If we express the sum

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

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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