2022 AMC 12A Problems/Problem 23

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 $1, 2, 3, \ldots, n$ . For how many integers with $1\le{n}\le{22}$ is $k_n<L_n$ ?

$\textbf{(A) }0 \qquad\textbf{(B) }3 \qquad\textbf{(C) }7 \qquad\textbf{(D) }8\qquad\textbf{(E) }10$

Solution 1

AIMING FOR A COMPREHENSIVE WRITTEN SOLUTION.

Solution 2

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 $\boxed{\textbf{(D) }8}$ .

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)

Video Solution

https://youtu.be/pZAez5A8tWA

~MathProblemSolvingSkills.com

2022 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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2022 AMC 12A Problems/Problem 23

Contents

Problem

Solution 1

Solution 2

Video Solution

See Also