Difference between revisions of "2002 AMC 12B Problems/Problem 4"
(2002 AMC 12B Problems/Problem 4 moved to 2002 AMC 12B Problems/Problem 5: I skipped a question.) |
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| − | + | == Problem == | |
| + | 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: | ||
| + | |||
| + | <math>\mathrm{(A)}\ 2\ \text{divides\ }n | ||
| + | \qquad\mathrm{(B)}\ 3\ \text{divides\ }n | ||
| + | \qquad\mathrm{(C)}</math> <math>\ 6\ \text{divides\ }n | ||
| + | \qquad\mathrm{(D)}\ 7\ \text{divides\ }n | ||
| + | \qquad\mathrm{(E)}\ n > 84</math> | ||
| + | |||
| + | == Solution == | ||
| + | Since <math>\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}</math>, | ||
| + | |||
| + | <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> | ||
| + | |||
| + | 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>. | ||
| + | |||
| + | == See also == | ||
| + | {{AMC12 box|year=2002|ab=B|num-b=3|num-a=5}} | ||
| + | |||
| + | [[Category:Introductory Algebra Problems]] | ||
Revision as of 18:12, 18 January 2008
Problem
Let 
be a positive integer such that 
is an integer. Which of the following statements is not true:
Solution
Since 
,
![\[0 < \lim_{n \rightarrow \infty} \left(\frac{41}{42} + \frac{1}{n}\right) < \frac {41}{42} + \frac 1n < \frac{41}{42} + \frac 11 < 2\]](http://latex.artofproblemsolving.com/0/a/1/0a117286e3f8f0f84a7cbc59b1204f101146fc51.png)
From which it follows that 
and 
. Thus the answer is 
.
See also
| 2002 AMC 12B (Problems • Answer Key • 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 | |
| All AMC 12 Problems and Solutions | |
