Difference between revisions of "2002 AMC 10B Problems/Problem 7"
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<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> | <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> | ||
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| + | ==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>. | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2002|ab=B|num-b=6|num-a=8}} | ||
| + | |||
| + | [[Category:Introductory Number Theory Problems]] | ||
Revision as of 13:02, 27 December 2008
Problem
Let 
be a positive integer such that 
is an integer. Which of the following statements is not true?
Solution
Writing the first four fractions with a common denominator, we have 
, hence 
is a solution. Thus, our answer is 
.
See Also
| 2002 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
