Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m
Line 38: Line 38:
  
 
{{AMC12 box|year=2022|ab=A|num-b=22|num-a=24}}
 
{{AMC12 box|year=2022|ab=A|num-b=22|num-a=24}}
 +
 +
[[Category:Intermediate Number Theory Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 16:11, 15 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) $\textbf{can be found in my video solution:}$

https://youtu.be/4RHmsoDsU9E

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_12A_Problems/Problem_23&oldid=181545"