Difference between revisions of "2002 AMC 12B Problems/Problem 4"
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== Solution 2 (no limits)== | == Solution 2 (no limits)== | ||
Since <math>\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}</math>, it is very clear that <math>n=42</math> makes the expression an integer. Because <math>n</math> is a positive integer, <math>\frac{1}{n}</math> must be less than or equal to <math>1</math> and greater than <math>\frac{41}{42}</math>. Thus the only integer the expression can take is <math>1</math>, making the only value for <math>n</math> <math>42</math>. Thus <math>\boxed{\mathrm{(E)}\ n>84}</math> | Since <math>\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}</math>, it is very clear that <math>n=42</math> makes the expression an integer. Because <math>n</math> is a positive integer, <math>\frac{1}{n}</math> must be less than or equal to <math>1</math> and greater than <math>\frac{41}{42}</math>. Thus the only integer the expression can take is <math>1</math>, making the only value for <math>n</math> <math>42</math>. Thus <math>\boxed{\mathrm{(E)}\ n>84}</math> | ||
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| + | ~superagh | ||
== See also == | == See also == | ||
Revision as of 17:26, 19 August 2020
- The following problem is from both the 2002 AMC 12B #4 and 2002 AMC 10B #7, so both problems redirect to this page.
Problem
Let 
be a positive integer such that 
is an integer. Which of the following statements is not true:
Solution 1
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 
. The only answer choice that is not true is 
.
Solution 2 (no limits)
Since , it is very clear that 
makes the expression an integer. Because is a positive integer, 
must be less than or equal to 
and greater than 
. Thus the only integer the expression can take is , making the only value for 
. Thus
~superagh
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 | ||
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
