Difference between revisions of "2002 AMC 10B Problems/Problem 7"

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== Problem ==
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#REDIRECT[[2002 AMC 12B Problems/Problem 4]]
 
 
Let <math>n</math> be a positive integer such that <math>\frac {1}{2} + \frac {1}{3} + \frac {1}{7} + \frac {1}{n}</math> is an integer. Which of the following statements is ''not'' true?
 
 
 
<math> \mathrm{(A) \ } 2\text{ divides }n\qquad \mathrm{(B) \ } 3\text{ divides }n\qquad \mathrm{(C) \ } 6\text{ divides }n\qquad \mathrm{(D) \ } 7\text{ divides }n\qquad \mathrm{(E) \ } n>84 </math>
 
 
 
==Solution==
 
Writing the first four fractions with a common denominator, we have <math>\frac{41}{42}+\frac{1}{n}</math>, hence <math>n=42</math> is a solution. Thus, our answer is <math>\boxed{(E)}</math>.
 
 
 
==See Also==
 
{{AMC10 box|year=2002|ab=B|num-b=6|num-a=8}}
 
 
 
[[Category:Introductory Number Theory Problems]]
 

Latest revision as of 16:40, 28 July 2011