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

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 $\sum_{i=1}^{n}\frac{L_n}{i} \equiv \sum_{i=1}^{\left\lfloor\tfrac np\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p^e) \equiv \sum_{i=1}^{\left\lfloor\tfrac np\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p).$ We construct the following table: $\begin{array}{c|l|c|c} \textbf{Case} & \hspace{22.75mm}\textbf{Sum} & \textbf{Interval of }n & \boldsymbol{\stackrel{?}{\equiv}0\ (\operatorname{mod} \ p)} \\ [0.5ex] \hline\hline & & & \\ [-2ex] v_2(L_n)=1 & L_n/2 & [2,3] & \\ v_2(L_n)=2 & L_n/4 & [4,7] & \\ v_2(L_n)=3 & L_n/8 & [8,15] & \\ v_2(L_n)=4 & L_n/16 & [16,22] & \\ [0.5ex] \hline & & & \\ [-2ex] v_3(L_n)=1 & L_n/3 & [3,5] & \\ & L_n/3 + L_n/6 & [6,8] & \checkmark \\ v_3(L_n)=2 & L_n/9 & [9,17] & \\ & L_n/9 + L_n/18 & [18,22] & \checkmark \\ [0.5ex] \hline & & & \\ [-2ex] v_5(L_n)=1 & L_n/5 & [5,9] & \\ & L_n/5 + L_n/10 & [10,14] & \\ & L_n/5 + L_n/10 + L_n/15 & [15,19] & \\ & L_n/5 + L_n/10 + L_n/15 + L_n/20 & [20,22] & \checkmark \\ [0.5ex] \hline & & & \\ [-2ex] v_7(L_n)=1 & L_n/7 & [7,13] & \\ & L_n/7 + L_n/14 & [14,20] & \\ & L_n/7 + L_n/14 + L_n/21 & [21,22] & \\ [0.5ex] \hline & & & \\ [-2ex] v_{11}(L_n)=1 & L_n/11 & [11,21] & \\ & L_n/11 + L_n/22 & \{22\} & \\ [0.5ex] \hline & & & \\ [-2ex] v_{13}(L_n)=1 & L_n/13 & [13,22] & \\ [0.5ex] \hline & & & \\ [-2ex] v_{17}(L_n)=1 & L_n/17 & [17,22] & \\ [0.5ex] \hline & & & \\ [-2ex] v_{19}(L_n)=1 & L_n/19 & [19,22] & \\ [0.5ex] \end{array}$

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 $i \in \left\{ 1, 2, \ldots, m \right\}$ , 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

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

Video Solution

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