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Difference between revisions of "2022 AMC 12A Problems/Problem 23"
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<math>\textbf{(A) }0 \qquad\textbf{(B) }3 \qquad\textbf{(C) }7 \qquad\textbf{(D) }8\qquad\textbf{(E) }10</math>
 
<math>\textbf{(A) }0 \qquad\textbf{(B) }3 \qquad\textbf{(C) }7 \qquad\textbf{(D) }8\qquad\textbf{(E) }10</math>
  
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==Solution 1==
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==Solution==
  
 
We are given that <cmath>\sum_{i=1}^{n}\frac1i = \frac{1}{L_n}\sum_{i=1}^{n}\frac{L_n}{i} = \frac{h_n}{k_n}.</cmath> Since <math>k_n < L_n,</math> we need <math>\gcd\left(\sum_{i=1}^{n}\frac{L_n}{i}, L_n\right)>1.</math>
 
We are given that <cmath>\sum_{i=1}^{n}\frac1i = \frac{1}{L_n}\sum_{i=1}^{n}\frac{L_n}{i} = \frac{h_n}{k_n}.</cmath> Since <math>k_n < L_n,</math> we need <math>\gcd\left(\sum_{i=1}^{n}\frac{L_n}{i}, L_n\right)>1.</math>
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For all primes <math>p</math> such that <math>p\leq n,</math> let <math>v_p(L_n)=e\geq1</math> be the largest power of <math>p</math> that is a factor of <math>L_n.</math>
 
For all primes <math>p</math> such that <math>p\leq n,</math> let <math>v_p(L_n)=e\geq1</math> be the largest power of <math>p</math> that is a factor of <math>L_n.</math>
  
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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{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).</cmath>
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We construct the following table:
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We construct the following table for <math>v_p(L_n)=e:</math>
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<cmath>\begin{array}{c|c|c|c|c}  
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<cmath>\begin{array}{c|c|l|c}  
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& & & & \\ [-2.25ex]
+
\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]
−
\boldsymbol{i} & \boldsymbol{p_i} & \boldsymbol{e_i} & \boldsymbol{\dfrac{(e_i+1)^3}{p_i^{e_i}}} & \textbf{Max?} \\ [2.5ex]
 
 
\hline\hline  
 
\hline\hline  
−
& & & & \\ [-2ex]
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& & & \\ [-2ex]
−
1 & 2 & 0 & 1 & \\     
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(2,1) & [2,3] & L_n/2 & \\     
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& & 1 & 4 & \\   
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(2,2) & [4,7] & L_n/4 & \\     
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& & 2 & 27/4 &\\     
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(2,3) & [8,15] & L_n/8 & \\     
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& & 3 & 8 & \checkmark\\     
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(2,4) & [16,22] & L_n/16 & \\ [0.5ex]
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& & 4 & 125/16 & \\ [0.5ex]
 
 
\hline   
 
\hline   
−
& & & & \\ [-2ex]
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& & & \\ [-2ex]
−
2 & 3 & 0 & 1 &\\     
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(3,1) & [3,5] & L_n/3 & \\     
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& & 1 & 8/3 & \\  
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& [6,8] & L_n/3 + L_n/6 & \checkmark \\  
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& & 2 & 3 &  \checkmark\\   
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(3,2) & [9,17] & L_n/9 & \\  
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& & 3 & 64/27 & \\ [0.5ex]
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& [18,22] & L_n/9 + L_n/18 & \checkmark \\ [0.5ex]
 
\hline   
 
\hline   
−
& & & & \\ [-2ex]
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& & & \\ [-2ex]
−
3 & 5 & 0 & 1 &  \\     
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(5,1) & [5,9] & L_n/5 & \\     
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& & 1 & 8/5 & \checkmark\\   
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& [10,14] & L_n/5 + L_n/10 & \\
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& & 2 & 27/25 & \\ [0.5ex]
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& [15,19] & L_n/5 + L_n/10 + L_n/15 & \\
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& [20,22] & L_n/5 + L_n/10 + L_n/15 + L_n/20 & \checkmark \\ [0.5ex]
 
\hline   
 
\hline   
−
& & & & \\ [-2ex]
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& & & \\ [-2ex]
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4 & 7 & 0 & 1 &  \\     
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(7,1) & [7,13] & L_n/7 & \\     
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& & 1 & 8/7 & \checkmark\\  
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& [14,20] & L_n/7 + L_n/14 & \\
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& & 2 & 27/49 & \\ [0.5ex]
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& [21,22] & L_n/7 + L_n/14 + L_n/21 & \\ [0.5ex]
 
\hline   
 
\hline   
−
& & & & \\ [-2ex]
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& & & \\ [-2ex]
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\geq5 & \geq11 & 0 & 1 & \checkmark \\  
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(11,1) & [11,21] & L_n/11 &  \\
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& & \geq1 & \leq8/11 &   \\ [0.5ex]
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& \{22\} & L_n/11 + L_n/22 & \\ [0.5ex]
 +
\hline
 +
& & & \\ [-2ex]
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(13,1) & [13,22] & L_n/13 &  \\ [0.5ex]
 +
\hline
 +
& & & \\ [-2ex]
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(17,1) & [17,22] & L_n/17 & \\ [0.5ex]
 +
\hline
 +
& & & \\ [-2ex]
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(19,1) & [19,22] & L_n/19 & \\ [0.5ex]
 
\end{array}</cmath>
 
\end{array}</cmath>
 +
Note that:
 +
<ol style="margin-left: 1.5em;">
 +
  <li>If the <b>Sum</b> column has only one term, then it is never congruent to <math>0</math> modulo <math>p.</math></li><p>
 +
  <li>If <math>p</math> and <math>q</math> are positive integers such that <math>p\geq q,</math> then <math>L_p</math> is a multiple of <math>L_q.</math> Therefore, for a specific case, if the sum is congruent to <math>0</math> modulo <math>p</math> for the smallest element in the interval of <math>n,</math> then it is also congruent to <math>0</math> modulo <math>p</math> for all other elements in the interval of <math>n.</math></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
  
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==Solution 2==
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== Video Solution ==
−
 
 
 
We will use the following lemma to solve this problem.
 
We will use the following lemma to solve this problem.
  
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<b>Note: Detailed analysis of this problem (particularly the motivation and the proof of the lemma above) can be found in my video solution below.</b>
 
<b>Note: Detailed analysis of this problem (particularly the motivation and the proof of the lemma above) can be found in my video solution below.</b>
  
−
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
+
==Video Solution==
−
 
 
−
== Video Solution ==
 
 
https://youtu.be/4RHmsoDsU9E
 
https://youtu.be/4RHmsoDsU9E
  
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~MathProblemSolvingSkills.com
 
~MathProblemSolvingSkills.com
 +
 +
==Video Solution by Math-X==
 +
https://youtu.be/7yAh4MtJ8a8?si=oklkf-_wUpjjfAed&t=8018
 +
 +
~Math-X
  
 
== See Also ==
 
== See Also ==

Latest revision as of 08:52, 29 March 2024

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

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.

Video Solution

https://youtu.be/4RHmsoDsU9E

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

Video Solution

https://youtu.be/pZAez5A8tWA

~MathProblemSolvingSkills.com

Video Solution by Math-X

https://youtu.be/7yAh4MtJ8a8?si=oklkf-_wUpjjfAed&t=8018

~Math-X

See Also

2022 AMC 12A (Problems • Answer Key • Resources)
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Problem 22 		Followed by
Problem 24
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All AMC 12 Problems and Solutions

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