Difference between revisions of "2002 AMC 12B Problems/Problem 4"
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\qquad\mathrm{(E)}\ n > 84</math> | \qquad\mathrm{(E)}\ n > 84</math> | ||
− | == Solution == | + | == Solution 1== |
Since <math>\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}</math>, | Since <math>\frac 12 + \frac 13 + \frac 17 = \frac {41}{42}</math>, | ||
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From which it follows that <math>\frac{41}{42} + \frac 1n = 1</math> and <math>n = 42</math>. The only answer choice that is not true is <math>\boxed{\mathrm{(E)}\ n>84}</math>. | From which it follows that <math>\frac{41}{42} + \frac 1n = 1</math> and <math>n = 42</math>. The only answer choice that is not true is <math>\boxed{\mathrm{(E)}\ n>84}</math>. | ||
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+ | == 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> | ||
== 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.
Contents
[hide]Problem
Let be a positive integer such that is an integer. Which of the following statements is not true:
Solution 1
Since ,
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
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.