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2002 AMC 12B Problems/Problem 4

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Problem

Let $n$ be a positive integer such that $\frac 12 + \frac 13 + \frac 17 + \frac 1n$ is an integer. Which of the following statements is not true:

$\mathrm{(A)}\ 2\ \text{divides\ }n\qquad\mathrm{(B)}\ 3\ \text{divides\ }n\qquad\mathrm{(C)}$ $\ 6\ \text{divides\ }n \qquad\mathrm{(D)}\ 7\ \text{divides\ }n\qquad\mathrm{(E)}\ n > 84$

Solution

Since $\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}$ ,

$0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42}...$

From which it follows that $\frac{41}{42} + \frac 1n = 1$ and $n = 42$ . Thus the answer is $\mathrm{(E)}$ .

See also

2002 AMC 12B (Problems • Resources)
Preceded by Problem 3	Followed by Problem 5
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

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2002_AMC_12B_Problems/Problem_4"

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