2002 AMC 12B Problems/Problem 4
- The following problem is from both the 2002 AMC 12B #4 and 2002 AMC 10B #7, so both problems redirect to this page.
Contents
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
~superagh
Solution 3(similiar to solution2)
Cross multiplying and adding the fraction we get the fraction to be equal to , This value has to be an integer. This implies,
.
=>
but , hence but does not divides , -(1)
=>
but , -(2)
from (1) and (2) we get that n=42. Comparing this with the options, we see that option E is the incorrect statement and hence E is the answer.
~rudolf1279
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.