# 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?

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

## Solution

We are given that $$\sum_{i=1}^{n}\frac1i = \frac{1}{L_n}\sum_{i=1}^{n}\frac{L_n}{i} = \frac{h_n}{k_n}.$$ Since $k_n < L_n,$ we need $\gcd\left(\sum_{i=1}^{n}\frac{L_n}{i}, L_n\right)>1.$

For all primes $p$ such that $p\leq n,$ let $v_p(L_n)=e\geq1$ be the largest power of $p$ that is a factor of $L_n.$

It is clear that $L_n\equiv0\pmod{p},$ so we test whether $\sum_{i=1}^{n}\frac{L_n}{i}\equiv0\pmod{p}.$ Note that $$\sum_{i=1}^{n}\frac{L_n}{i} \equiv \sum_{i=1}^{\left\lfloor\tfrac{n}{p^e}\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p^e) \equiv \sum_{i=1}^{\left\lfloor\tfrac{n}{p^e}\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p).$$ We construct the following table for $v_p(L_n)=e:$ $$\begin{array}{c|c|l|c} \textbf{Case of }\boldsymbol{(p,e)} & \textbf{Interval of }\boldsymbol{n} & \hspace{22.75mm}\textbf{Sum} & \boldsymbol{\stackrel{?}{\equiv}0\ (\operatorname{mod} \ p)} \\ [0.5ex] \hline\hline & & & \\ [-2ex] (2,1) & [2,3] & L_n/2 & \\ (2,2) & [4,7] & L_n/4 & \\ (2,3) & [8,15] & L_n/8 & \\ (2,4) & [16,22] & L_n/16 & \\ [0.5ex] \hline & & & \\ [-2ex] (3,1) & [3,5] & L_n/3 & \\ & [6,8] & L_n/3 + L_n/6 & \checkmark \\ (3,2) & [9,17] & L_n/9 & \\ & [18,22] & L_n/9 + L_n/18 & \checkmark \\ [0.5ex] \hline & & & \\ [-2ex] (5,1) & [5,9] & L_n/5 & \\ & [10,14] & L_n/5 + L_n/10 & \\ & [15,19] & L_n/5 + L_n/10 + L_n/15 & \\ & [20,22] & L_n/5 + L_n/10 + L_n/15 + L_n/20 & \checkmark \\ [0.5ex] \hline & & & \\ [-2ex] (7,1) & [7,13] & L_n/7 & \\ & [14,20] & L_n/7 + L_n/14 & \\ & [21,22] & L_n/7 + L_n/14 + L_n/21 & \\ [0.5ex] \hline & & & \\ [-2ex] (11,1) & [11,21] & L_n/11 & \\ & \{22\} & L_n/11 + L_n/22 & \\ [0.5ex] \hline & & & \\ [-2ex] (13,1) & [13,22] & L_n/13 & \\ [0.5ex] \hline & & & \\ [-2ex] (17,1) & [17,22] & L_n/17 & \\ [0.5ex] \hline & & & \\ [-2ex] (19,1) & [19,22] & L_n/19 & \\ [0.5ex] \end{array}$$ Note that:

1. If the Sum column has only one term, then it is never congruent to $0$ modulo $p.$
2. If $p$ and $q$ are positive integers such that $p\geq q,$ then $L_p$ is a multiple of $L_q.$ Therefore, for a specific case, if the sum is congruent to $0$ modulo $p$ for the smallest element in the interval of $n,$ then it is also congruent to $0$ modulo $p$ for all other elements in the interval of $n.$

Together, there are $\boxed{\textbf{(D) }8}$ such integers $n,$ namely $$6,7,8,18,19,20,21,22.$$ ~MRENTHUSIASM

## Video 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 $i \in \left\{ 1, 2, \ldots, m \right\}$, 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 $i \in \left\{ 1, 2, \ldots, m \right\}$, $a_i$ 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 below.

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

## Video Solution

~MathProblemSolvingSkills.com