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

(Created page with "==Problem== Let <math>h_n</math> and <math>k_n</math> be the unique relatively prime positive integers such that \[ \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{...")
 
Line 13: Line 13:
 
We will use the following lemma to solve this problem.
 
We will use the following lemma to solve this problem.
  
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\begin{lemma}
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20:20, 11 November 2022 (EST)20:20, 11 November 2022 (EST)LEMMA20:20, 11 November 2022 (EST)20:20, 11 November 2022 (EST)
 
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, \cdots, 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.
 
For any <math>i \in \left\{ 1, 2, \cdots, 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, \cdots, m \right\}</math>, <math>a_i</math> is not a multiple of <math>p_i</math>.
 
<math>k_n = L_n</math> if and only if for any <math>i \in \left\{ 1, 2, \cdots, m \right\}</math>, <math>a_i</math> is not a multiple of <math>p_i</math>.
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\end{lemma}
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20:20, 11 November 2022 (EST)20:20, 11 November 2022 (EST)20:20, 11 November 2022 (EST)20:20, 11 November 2022 (EST)20:20, 11 November 2022 (EST)
  
 
Now, we use the result above to solve this problem.
 
Now, we use the result above to solve this problem.

Revision as of 20:20, 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.

20:20, 11 November 2022 (EST)20:20, 11 November 2022 (EST)LEMMA20:20, 11 November 2022 (EST)20:20, 11 November 2022 (EST) 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 $p_i$. 20:20, 11 November 2022 (EST)20:20, 11 November 2022 (EST)20:20, 11 November 2022 (EST)20:20, 11 November 2022 (EST)20:20, 11 November 2022 (EST)

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}}.

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