Difference between revisions of "2002 AMC 12B Problems/Problem 4"

 
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#REDIRECT [[2002 AMC 12B Problems/Problem 5]]
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== Problem ==
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Let <math>n</math> be a positive [[integer]] such that <math>\frac 12 + \frac 13 + \frac 17 + \frac 1n</math> is an integer. Which of the following statements is '''not ''' true:
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<math>\mathrm{(A)}\ 2\ \text{divides\ }n
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\qquad\mathrm{(B)}\ 3\ \text{divides\ }n
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\qquad\mathrm{(C)}</math> <math>\ 6\ \text{divides\ }n
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\qquad\mathrm{(D)}\ 7\ \text{divides\ }n
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\qquad\mathrm{(E)}\ n > 84</math>
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== Solution ==
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Since <math>\frac 12 + \frac 13 + \frac 17  = \frac {41}{42}</math>,
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<cmath>0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2</cmath>
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From which it follows that <math>\frac{41}{42} + \frac 1n = 1</math> and <math>n = 42</math>. Thus the answer is <math>\mathrm{(E)}</math>.
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== See also ==
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{{AMC12 box|year=2002|ab=B|num-b=3|num-a=5}}
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[[Category:Introductory Algebra Problems]]

Revision as of 17:12, 18 January 2008

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

\[0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2\]

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 (ProblemsAnswer KeyResources)
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
All AMC 12 Problems and Solutions