Difference between revisions of "2022 AMC 12A Problems/Problem 23"
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Denote by <math>p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m}</math> the prime factorization of <math>L_n</math>. | Denote by <math>p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m}</math> the prime factorization of <math>L_n</math>. | ||
− | For any <math>i \in \left\{ 1, 2, \ | + | For any <math>i \in \left\{ 1, 2, \ldots, m \right\}</math>, denote <math>\sum_{j = 1}^{\left\lfloor \frac{n}{p_i^{\alpha_i}} \right\rfloor} \frac{1}{j} = \frac{a_i}{b_i}</math>, where <math>a_i</math> and <math>b_i</math> are relatively prime. |
Then | Then | ||
− | <math>k_n = L_n</math> if and only if for any <math>i \in \left\{ 1, 2, \ | + | <math>k_n = L_n</math> if and only if for any <math>i \in \left\{ 1, 2, \ldots, m \right\}</math>, <math>a_i</math> is not a multiple of <math>p_i</math>. |
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Revision as of 18:15, 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 .
Note: Detailed analysis of this problem (particularly the motivation and the proof of the lemma above) can be found in my video solution below.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution
~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.