Difference between revisions of "2022 AMC 12A Problems/Problem 23"

(Solution)
(Solution)
Line 29: Line 29:
 
Therefore, the answer is <math>\boxed{\textbf{(D) 8}}</math>.
 
Therefore, the answer is <math>\boxed{\textbf{(D) 8}}</math>.
  
\textbf{NOTE: Detailed analysis of this problem (particularly the motivation and the proof of the lemma above) can be found in my video solution:}
+
<math>\textbf{NOTE: Detailed analysis of this problem (particularly the motivation and the proof of the lemma above) can be found in my video solution:}</math>
  
 
https://youtu.be/4RHmsoDsU9E
 
https://youtu.be/4RHmsoDsU9E
  
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Revision as of 20:27, 11 November 2022

Problem

Let $h_n$ and $k_n$ be the unique relatively prime positive integers such that

\[\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} = \frac{h_n}{k_n} .\]

Let $L_n$ denote the least common multiple of the numbers $1,2,3,\cdots,n$. For how many integers $n$ with $1 \leq n \leq 22$ is $k_n < L_n$?

Solution

We will use the following lemma to solve this problem.


Denote by $p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m}$ the prime factorization of $L_n$. For any $i \in \left\{ 1, 2, \cdots, m \right\}$, denote $\sum_{j = 1}^{\left\lfloor \frac{n}{p_i^{\alpha_i}} \right\rfloor} \frac{1}{j} = \frac{a_i}{b_i}$, where $a_i$ and $b_i$ are relatively prime. Then $k_n = L_n$ if and only if for any $i \in \left\{ 1, 2, \cdots, m \right\}$, $a_i$ is not a multiple of $p_i$.


Now, we use the result above to solve this problem.

Following from this lemma, the list of $n$ with $1 \leq n \leq 22$ and $k_n < L_n$ is \[ 6, 7, 8, 18, 19, 20, 21, 22 . \]

Therefore, the answer is $\boxed{\textbf{(D) 8}}$.

$\textbf{NOTE: Detailed analysis of this problem (particularly the motivation and the proof of the lemma above) can be found in my video solution:}$

https://youtu.be/4RHmsoDsU9E

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)