Difference between revisions of "2024 AMC 10B Problems/Problem 18"
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+ | {{duplicate|[[2024 AMC 10B Problems/Problem 18|2024 AMC 10B #18]] and [[2024 AMC 12B Problems/Problem 14|2024 AMC 12B #14]]}} | ||
+ | |||
+ | ==Problem == | ||
+ | How many different remainders can result when the <math>100</math>th power of an integer is | ||
+ | divided by <math>125</math>? | ||
+ | |||
+ | <math>\textbf{(A) } 1 \qquad\textbf{(B) } 2 \qquad\textbf{(C) } 5 \qquad\textbf{(D) } 25 \qquad\textbf{(E) } 125</math> | ||
+ | |||
==Solution 1== | ==Solution 1== | ||
First note that the totient function of <math>125</math> is <math>100</math>. We can set up two cases, which depend on whether a number is relatively prime to <math>125</math>. | First note that the totient function of <math>125</math> is <math>100</math>. We can set up two cases, which depend on whether a number is relatively prime to <math>125</math>. | ||
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~lprado | ~lprado | ||
+ | |||
+ | ==Solution 2== | ||
+ | We split the cases into: | ||
+ | |||
+ | 1. If x is not a multiple of 5: | ||
+ | we get x^100\equiv 1 (mod 125) | ||
+ | |||
+ | 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== | ||
+ | {{AMC10 box|year=2024|ab=B|num-b=17|num-a=19}} | ||
+ | {{AMC12 box|year=2024|ab=B|num-b=13|num-a=15}} | ||
+ | {{MAA Notice}} |
Revision as of 01:24, 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
[hide]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 x^100\equiv 1 (mod 125)
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.