Difference between revisions of "2018 AMC 10A Problems/Problem 7"
MRENTHUSIASM (talk | contribs) |
|||
(6 intermediate revisions by 3 users not shown) | |||
Line 13: | Line 13: | ||
==Solution 1 (Algebra)== | ==Solution 1 (Algebra)== | ||
− | Note that <cmath>4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(\ | + | Note that <cmath>4000\cdot \left(\frac{2}{5}\right)^n=\left(2^5\cdot5^3\right)\cdot \left(2\cdot5^{-1}\right)^n=2^{5+n}\cdot5^{3-n}.</cmath> |
Since this expression is an integer, we need: | Since this expression is an integer, we need: | ||
<ol style="margin-left: 1.5em;"> | <ol style="margin-left: 1.5em;"> | ||
Line 36: | Line 36: | ||
~Little ~MRENTHUSIASM | ~Little ~MRENTHUSIASM | ||
− | ==Video | + | ==Video Solution (HOW TO THINK CREATIVELY!)== |
https://youtu.be/vzyRAnpnJes | https://youtu.be/vzyRAnpnJes | ||
− | + | Education, the Study of Everything | |
+ | |||
+ | |||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/ZiZVIMmo260 | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/2vz_CnxsGMA | ||
+ | |||
+ | ~savannahsolver | ||
+ | |||
+ | ==Video Solution by OmegaLearn== | ||
+ | https://youtu.be/ZhAZ1oPe5Ds?t=1763 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
== See Also == | == See Also == |
Latest revision as of 04:26, 15 August 2024
- 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.