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Difference between revisions of "2022 AMC 12A Problems/Problem 23"
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==Solution 1==
 
==Solution 1==
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AIMING FOR A COMPREHENSIVE WRITTEN SOLUTION.
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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>
 +
 
 +
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 for <math>v_p(L_n)=e:</math>
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<cmath>\begin{array}{c|c|l|c}
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\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]
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\hline\hline
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& & & \\ [-2ex]
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(2,1) & [2,3] & L_n/2 &  \\   
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(2,2) & [4,7] & L_n/4 &  \\   
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(2,3) & [8,15] & L_n/8 & \\   
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(2,4) & [16,22] & L_n/16 &  \\ [0.5ex]
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\hline 
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& & & \\ [-2ex]
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(3,1) & [3,5] & L_n/3 & \\   
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& [6,8] & L_n/3 + L_n/6 & \checkmark \\ 
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(3,2) & [9,17] & L_n/9 & \\
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& [18,22] & L_n/9 + L_n/18 & \checkmark \\ [0.5ex]
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\hline 
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& & & \\ [-2ex]
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(5,1) & [5,9] & L_n/5 & \\   
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& [10,14] & L_n/5 + L_n/10 & \\
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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]
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\hline 
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& & & \\ [-2ex]
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(7,1) & [7,13] & L_n/7 & \\   
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& [14,20] & L_n/7 + L_n/14 & \\
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& [21,22] & L_n/7 + L_n/14 + L_n/21 & \\ [0.5ex]
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\hline 
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& & & \\ [-2ex]
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(11,1) & [11,21] & L_n/11 &  \\
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& \{22\} & L_n/11 + L_n/22 & \\ [0.5ex]
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\hline
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& & & \\ [-2ex]
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(13,1) & [13,22] & L_n/13 &  \\ [0.5ex]
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\hline
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& & & \\ [-2ex]
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(17,1) & [17,22] & L_n/17 &  \\ [0.5ex]
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\hline
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& & & \\ [-2ex]
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(19,1) & [19,22] & L_n/19 & \\ [0.5ex]
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\end{array}</cmath>
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Note that:
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<ol style="margin-left: 1.5em;">
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  <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>
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  <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>
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</ol>
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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==
 +
Notice that <math>\sum_{i=1}^{n}\frac{1}{i} = \frac{\sum_{i=1}^{n}\frac{L_n}{i}}{L_n}</math>. Thus, in order for <math>L_n</math> to reduce to a smaller value, the numerator and denominator must share a common factor.
 +
 +
We can start by testing some values of <math>n</math>, and quickly observe that <math>n=6</math> is the first number that satisfies the desired conditions. In particular, <math>\sum_{i=1}^{6}\frac{1}{i} = \frac{\sum_{i=1}^{6}\frac{L_6}{i}}{L_6} = \frac{147}{60} = \frac{49}{20}</math>. Since a factor of three is shared, we are motivated to observe factors of three in the numerator <math>\sum_{i=1}^{6}\frac{60}{i}</math>. Notice that <math>v_3(L_6) = 1</math>, since <math>3^1</math> is the greatest power of three that occurs in <math>1,2,3,4,5,6</math>. Consider the sum in the numerator in mod 3. We have <math>(\frac{60}{1}, \frac{60}{2}, \frac{60}{3}, \frac{60}{4}, \frac{60}{5}, \frac{60}{6} )\rightarrow (0,0,2,0,0,1)</math>. Adding 0+0+2+0+0+1 we get 0 mod 3, which is why we are able to simplify <math>\frac{147}{60}</math>.
  
 +
Thus, we should be taking the numerator mod p, where p is a prime.
 +
 +
Quickly, we wish to show that the numerator and denominator can never share a factor of 2. Say we have least common multiple <math>L_n</math>. Let <math>v_2(L_n) = e</math>. In other words, let <math>2^e</math> be the largest power of 2 such that <math>2^e</math> divides <math>L_n</math>. Notice that among all numbers <math>1</math> through <math>n</math>, only one of them can be a multiple of <math>2^e</math>, being <math>2^e</math> itself. The next multiple of <math>2^e</math> would be <math>2 \cdot 2^e = 2^{e+1}</math>, contradicting the fact that <math>v_2(L_n) = e</math>. This means that in the sum <math>\sum_{i=1}^{n}\frac{L_n}{i}</math>, there will be <math>1</math> value that is <math>1</math> mod <math>2</math>, resulting from <math>\frac{L_n}{2^e}</math>, and the rest are all <math>0</math>s. Clearly, the numerator will then be <math>1</math> mod <math>2</math>, which is not a multiple of 2.
 +
 +
When p=3, <math>n=6,7,8,18,19,20,21,22</math> will all work, and this can be easily tested through taking the numerator mod 3. p=5 is only satisfied when the <math>L_n</math> contains <math>(5,10,15,20)</math>, so when <math>n\ge 20</math>. However, this is just an overlap with the previous values, so there are <math>\boxed{\textbf{B) }8}</math>.
 +
 +
~skibbysiggy
 +
 +
== Video Solution ==
 
We will use the following lemma to solve this problem.
 
We will use the following lemma to solve this problem.
  
 
---------------------------
 
---------------------------
 
Denote by <math>p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m}</math> the prime factorization of <math>L_n</math>.
 
Denote by <math>p_1^{\alpha_1} p_2^{\alpha_2} \cdots p_m^{\alpha_m}</math> the prime factorization of <math>L_n</math>.
−
For any <math>i \in \left\{ 1, 2, \cdots, m \right\}</math>, denote <math>\sum_{j = 1}^{\left\lfloor \frac{n}{p_i^{\alpha_i}} \right\rfloor} \frac{1}{j} = \frac{a_i}{b_i}</math>, where <math>a_i</math> and <math>b_i</math> are relatively prime.
+
For any <math>i \in \left\{ 1, 2, \ldots, m \right\}</math>, denote <math>\sum_{j = 1}^{\left\lfloor \frac{n}{p_i^{\alpha_i}} \right\rfloor} \frac{1}{j} = \frac{a_i}{b_i}</math>, where <math>a_i</math> and <math>b_i</math> are relatively prime.
 
Then
 
Then
−
<math>k_n = L_n</math> if and only if for any <math>i \in \left\{ 1, 2, \cdots, m \right\}</math>, <math>a_i</math> is not a multiple of <math>p_i</math>.
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<math>k_n = L_n</math> if and only if for any <math>i \in \left\{ 1, 2, \ldots, m \right\}</math>, <math>a_i</math> is not a multiple of <math>p_i</math>.
 
---------------------------
 
---------------------------
  
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Therefore, the answer is <math>\boxed{\textbf{(D) }8}</math>.
 
Therefore, the answer is <math>\boxed{\textbf{(D) }8}</math>.
  
−
<math>\textbf{NOTE: Detailed analysis of this problem}</math> (particularly the motivation and the proof of the lemma above) <math>\textbf{can be found in my video solution:}</math>
+
<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>
  
 +
==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 21:10, 4 November 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 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 \[\sum_{i=1}^{n}\frac{L_n}{i} \equiv \sum_{i=1}^{\left\lfloor\tfrac{n}{p^e}\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p^e) \equiv \sum_{i=1}^{\left\lfloor\tfrac{n}{p^e}\right\rfloor}\frac{L_n}{ip^e} \ (\operatorname{mod} \ p).\] 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

Solution 2

Notice that $\sum_{i=1}^{n}\frac{1}{i} = \frac{\sum_{i=1}^{n}\frac{L_n}{i}}{L_n}$. Thus, in order for $L_n$ to reduce to a smaller value, the numerator and denominator must share a common factor.

We can start by testing some values of $n$, and quickly observe that $n=6$ is the first number that satisfies the desired conditions. In particular, $\sum_{i=1}^{6}\frac{1}{i} = \frac{\sum_{i=1}^{6}\frac{L_6}{i}}{L_6} = \frac{147}{60} = \frac{49}{20}$. Since a factor of three is shared, we are motivated to observe factors of three in the numerator $\sum_{i=1}^{6}\frac{60}{i}$. Notice that $v_3(L_6) = 1$, since $3^1$ is the greatest power of three that occurs in $1,2,3,4,5,6$. Consider the sum in the numerator in mod 3. We have $(\frac{60}{1}, \frac{60}{2}, \frac{60}{3}, \frac{60}{4}, \frac{60}{5}, \frac{60}{6} )\rightarrow (0,0,2,0,0,1)$. Adding 0+0+2+0+0+1 we get 0 mod 3, which is why we are able to simplify $\frac{147}{60}$.

Thus, we should be taking the numerator mod p, where p is a prime.

Quickly, we wish to show that the numerator and denominator can never share a factor of 2. Say we have least common multiple $L_n$. Let $v_2(L_n) = e$. In other words, let $2^e$ be the largest power of 2 such that $2^e$ divides $L_n$. Notice that among all numbers $1$ through $n$, only one of them can be a multiple of $2^e$, being $2^e$ itself. The next multiple of $2^e$ would be $2 \cdot 2^e = 2^{e+1}$, contradicting the fact that $v_2(L_n) = e$. This means that in the sum $\sum_{i=1}^{n}\frac{L_n}{i}$, there will be $1$ value that is $1$ mod $2$, resulting from $\frac{L_n}{2^e}$, and the rest are all $0$s. Clearly, the numerator will then be $1$ mod $2$, which is not a multiple of 2.

When p=3, $n=6,7,8,18,19,20,21,22$ will all work, and this can be easily tested through taking the numerator mod 3. p=5 is only satisfied when the $L_n$ contains $(5,10,15,20)$, so when $n\ge 20$. However, this is just an overlap with the previous values, so there are $\boxed{\textbf{B) }8}$.

~skibbysiggy

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)
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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