2018 AMC 10A Problems/Problem 7
- The following problem is from both the 2018 AMC 10A #7 and 2018 AMC 12A #7, so both problems redirect to this page.
Contents
Problem
For how many (not necessarily positive) integer values of is the value of an integer?
Solution 1 (Algebra)
Note that Since this expression is an integer, we need:
- from which
- from which
Taking the intersection gives So, there are integer values of
~MRENTHUSIASM
Solution 2 (Observations)
Note that will be an integer if the denominator is a factor of . We also know that the denominator will always be a power of for positive values and a power of for all negative values. So we can proceed to divide by for each increasing positive value of until we get a non-factor of and also divide by for each decreasing negative value of . For positive values we get and for negative values we get . Also keep in mind that the expression will be an integer for , which gives us a total of for
Solution 3 (Brute Force)
The values for are and
The corresponding values for are and respectively.
In total, there are values for
~Little ~MRENTHUSIASM
Video Solution (HOW TO THINK CREATIVELY!)
Education, the Study of Everything
Video Solution
Video Solution
~savannahsolver
Video Solution by OmegaLearn
https://youtu.be/ZhAZ1oPe5Ds?t=1763
~ pi_is_3.14
See Also
2018 AMC 10A (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 |
2018 AMC 12A (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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.