Difference between revisions of "2022 AMC 12A Problems/Problem 23"
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− | Therefore, the answer is <math>\boxed{\textbf{(D) 8 | + | Therefore, the answer is <math>\boxed{\textbf{(D) }8}</math>. |
<math>\textbf{NOTE: Detailed analysis of this problem}</math> (particularly the motivation and the proof of the lemma above) <math>\textbf{can be found in my video solution:}</math> | <math>\textbf{NOTE: Detailed analysis of this problem}</math> (particularly the motivation and the proof of the lemma above) <math>\textbf{can be found in my video solution:}</math> | ||
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
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== Video Solution == | == Video Solution == | ||
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~MathProblemSolvingSkills.com | ~MathProblemSolvingSkills.com | ||
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== See Also == | == See Also == |
Revision as of 17:49, 3 January 2023
Problem
Let and be the unique relatively prime positive integers such that Let denote the least common multiple of the numbers . For how many integers with is ?
Solution 1
AIMING FOR A COMPREHENSIVE WRITTEN SOLUTION.
Solution 2
We will use the following lemma to solve this problem.
Denote by the prime factorization of . For any , denote , where and are relatively prime. Then if and only if for any , is not a multiple of .
Now, we use the result above to solve this problem.
Following from this lemma, the list of with and is
Therefore, the answer is .
(particularly the motivation and the proof of the lemma above)
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~MathProblemSolvingSkills.com
See Also
2022 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.