Difference between revisions of "2024 AMC 10B Problems/Problem 18"
(→Solution 2) |
|||
| Line 22: | Line 22: | ||
1. If x is not a multiple of 5: | 1. If x is not a multiple of 5: | ||
| − | we get x^100\equiv 1 | + | we get <math>x^{100} \equiv 1 \pmod{125}</math> |
2. If x is a multiple of 125: | 2. If x is a multiple of 125: | ||
Revision as of 01:39, 14 November 2024
- The following problem is from both the 2024 AMC 10B #18 and 2024 AMC 12B #14, so both problems redirect to this page.
Contents
Problem
How many different remainders can result when the 
th power of an integer is
divided by 
?
Solution 1
First note that the totient function of is . We can set up two cases, which depend on whether a number is relatively prime to .
If 
is relatively prime to , then 
because of Euler's Totient Theorem.
If is not relatively prime to , it must be have a factor of 
. Express as 
, where 
is some integer. Then 
.
Therefore, 
can only be congruent to 
or 
. Our answer is 
.
~lprado
Solution 2
We split the cases into:
1. If x is not a multiple of 5:
we get 
2. If x is a multiple of 125: Clearly the only remainder provides 0
Therefore, the remainders can only be 1 and 0, which gives the answer (B)2
~mitsuihisashi14
See also
| 2024 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 | ||
| 2024 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 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. 
