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Difference between revisions of "2022 AMC 12A Problems/Problem 23"
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It is clear that <math>L_n\equiv0\pmod{p},</math> so we test whether <math>\sum_{i=1}^{n}\frac{L_n}{i}\equiv0\pmod{p}.</math> Note that <cmath>\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).</cmath>
 
It is clear that <math>L_n\equiv0\pmod{p},</math> so we test whether <math>\sum_{i=1}^{n}\frac{L_n}{i}\equiv0\pmod{p}.</math> Note that <cmath>\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).</cmath>
−
We construct the following table:
+
We construct the following table for <math>v_p(L_n)=e:</math>
−
<cmath>\begin{array}{c|l|c|c}  
+
<cmath>\begin{array}{c|c|l|c}  
−
\textbf{Case} & \hspace{22.75mm}\textbf{Sum} & \textbf{Interval of }n & \boldsymbol{\stackrel{?}{\equiv}0\ (\operatorname{mod} \ p)} \\ [0.5ex]
+
\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  
 
\hline\hline  
 
& & & \\ [-2ex]
 
& & & \\ [-2ex]
−
v_2(L_n)=1 & L_n/2 & [2,3] &  \\     
+
(2,1) & [2,3] & L_n/2 &  \\     
−
v_2(L_n)=2 & L_n/4 & [4,7] &  \\     
+
(2,2) & [4,7] & L_n/4 &  \\     
−
v_2(L_n)=3 & L_n/8 & [8,15] & \\     
+
(2,3) & [8,15] & L_n/8 & \\     
−
v_2(L_n)=4 & L_n/16 & [16,22] &  \\ [0.5ex]
+
(2,4) & [16,22] & L_n/16 &  \\ [0.5ex]
 
\hline   
 
\hline   
 
& & & \\ [-2ex]
 
& & & \\ [-2ex]
−
v_3(L_n)=1 & L_n/3 & [3,5] & \\     
+
(3,1) & [3,5] & L_n/3 & \\     
−
& L_n/3 + L_n/6 & [6,8] & \checkmark \\   
+
& [6,8] & L_n/3 + L_n/6 & \checkmark \\   
−
v_3(L_n)=2 & L_n/9 & [9,17] & \\  
+
(3,2) & [9,17] & L_n/9 & \\  
−
& L_n/9 + L_n/18 & [18,22] & \checkmark \\ [0.5ex]
+
& [18,22] & L_n/9 + L_n/18 & \checkmark \\ [0.5ex]
 
\hline   
 
\hline   
 
& & & \\ [-2ex]
 
& & & \\ [-2ex]
−
v_5(L_n)=1 & L_n/5 & [5,9] & \\     
+
(5,1) & [5,9] & L_n/5 & \\     
−
& L_n/5 + L_n/10 & [10,14] & \\
+
& [10,14] & L_n/5 + L_n/10 & \\
−
& L_n/5 + L_n/10 + L_n/15 & [15,19] & \\   
+
& [15,19] & L_n/5 + L_n/10 + L_n/15 & \\   
−
& L_n/5 + L_n/10 + L_n/15 + L_n/20 & [20,22] & \checkmark \\ [0.5ex]
+
& [20,22] & L_n/5 + L_n/10 + L_n/15 + L_n/20 & \checkmark \\ [0.5ex]
 
\hline   
 
\hline   
 
& & & \\ [-2ex]
 
& & & \\ [-2ex]
−
v_7(L_n)=1 & L_n/7 & [7,13] & \\     
+
(7,1) & [7,13] & L_n/7 & \\     
−
& L_n/7 + L_n/14 & [14,20] & \\
+
& [14,20] & L_n/7 + L_n/14 & \\
−
& L_n/7 + L_n/14 + L_n/21 & [21,22] & \\ [0.5ex]
+
& [21,22] & L_n/7 + L_n/14 + L_n/21 & \\ [0.5ex]
 
\hline   
 
\hline   
 
& & & \\ [-2ex]
 
& & & \\ [-2ex]
−
v_{11}(L_n)=1 & L_n/11 & [11,21] &  \\
+
(11,1) & [11,21] & L_n/11 &  \\
−
& L_n/11 + L_n/22 & \{22\} & \\ [0.5ex]
+
& \{22\} & L_n/11 + L_n/22 & \\ [0.5ex]
 
\hline
 
\hline
 
& & & \\ [-2ex]
 
& & & \\ [-2ex]
−
v_{13}(L_n)=1 & L_n/13 & [13,22] &  \\ [0.5ex]
+
(13,1) & [13,22] & L_n/13 &  \\ [0.5ex]
 
\hline
 
\hline
 
& & & \\ [-2ex]
 
& & & \\ [-2ex]
−
v_{17}(L_n)=1 & L_n/17 & [17,22] &  \\ [0.5ex]
+
(17,1) & [17,22] & L_n/17 &  \\ [0.5ex]
 
\hline
 
\hline
 
& & & \\ [-2ex]
 
& & & \\ [-2ex]
−
v_{19}(L_n)=1 & L_n/19 & [19,22] & \\ [0.5ex]
+
(19,1) & [19,22] & L_n/19 & \\ [0.5ex]
 
\end{array}</cmath>
 
\end{array}</cmath>
 +
Note that:
 +
<ol style="margin-left: 1.5em;">
 +
  <li></li><p>
 +
  <li></li><p>
 +
</ol>
 +
Together, there are <math>\boxed{\textbf{(D) }8}</math> such integers <math>n,</math> namely <cmath>6,7,8,18,19,20,21,22.</cmath>
 +
 +
~MRENTHUSIASM
  
 
==Solution 2==
 
==Solution 2==

Revision as of 02:01, 4 January 2023

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 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:

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

~MRENTHUSIASM

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

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

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