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Difference between revisions of "2022 AMC 10A Problems/Problem 19"

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==Problem==
 
==Problem==
  
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Define <math>L_n</math> as the least common multiple of all the integers from <math>1</math> to <math>n</math> inclusive. There is a unique integer <math>h</math> such that  
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Define <math>L_n</math> as the least common multiple of all the integers from <math>1</math> to <math>n</math> inclusive. There is a unique integer <math>h</math> such that
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<cmath>\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{17}=\frac{h}{L_{17}}</cmath>
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<math>\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\ldots+\frac{1}{17}=\frac{h}{L_{17}}</math>
 
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What is the remainder when <math>h</math> is divided by <math>17</math>?
 
What is the remainder when <math>h</math> is divided by <math>17</math>?
  
 
<math>\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9</math>
 
<math>\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9</math>
  
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==Solution==
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==Solution 1==
  
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Notice that <math>L_{17}</math> contains the highest power of every prime below <math>17</math>. Thus, <math>L_{17}=16\cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17</math>.
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Notice that <math>L_{17}</math> contains the highest power of every prime below <math>17</math> since higher primes cannot divide <math>L_{17}</math>. Thus, <math>L_{17}=16\cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17</math>.
  
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When writing the sum under a common fraction, we multiply the denominators by <math>L_{17}</math> divided by each denominator. However, since <math>L_{17}</math> is a multiple of <math>17</math>, all terms will be a multiple of <math>17</math> until we divide out <math>17</math>, and the only term that will do this is <math>\frac{1}{17}</math>. Thus, the remainder of all other terms when divided by <math>17</math> will be <math>0</math>, so the problem is essentially asking us what the remainder of <math>\frac{L_{17}}{17}</math> divided by <math>17</math> is. This is equivalent to finding the remainder of <math>16 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13</math> divided by <math>17</math>.
+
When writing the sum under a common fraction, we multiply the denominators by <math>L_{17}</math> divided by each denominator. However, since <math>L_{17}</math> is a multiple of <math>17</math>, all terms will be a multiple of <math>17</math> until we divide out <math>17</math>, and the only term that will do this is <math>\frac{1}{17}</math>. Thus, the remainder of all other terms when divided by <math>17</math> will be <math>0</math>, so the problem is essentially asking us what the remainder of <math>\frac{L_{17}}{17} = L_{16}</math> divided by <math>17</math> is. This is equivalent to finding the remainder of <math>16 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13</math> divided by <math>17</math>.
  
 
We use modular arithmetic to simplify our answer:
 
We use modular arithmetic to simplify our answer:
  
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This is congruent to <math>-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</math>
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This is congruent to <math>-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</math>.
  
 
Evaluating, we get:  
 
Evaluating, we get:  
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&\equiv 5\pmod{17}
 
&\equiv 5\pmod{17}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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Therefore the remainder is <math>\boxed{\textbf{(C) } 5}</math>.
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Therefore the remainder is <math>5</math> and the answer is <math>\boxed{\textbf{(C) } 5}</math>.
 
  
 
~KingRavi
 
~KingRavi
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~Scarletsyc
 
~Scarletsyc
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~wangzrpi
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== Solution 2 ==
 +
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As in solution 1, we express the LHS as a sum under one common denominator. We note that <cmath>\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{17} = \frac{\frac{17!}{1}}{17!} + \frac{\frac{17!}{2}}{17!} + \frac{\frac{17!}{3}}{17!} + \dots + \frac{\frac{17!}{17}}{17!}</cmath>
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Now, we have <math>h = L_{17}\left(\frac{\frac{17!}{1} + \frac{17!}{2} + \frac{17!}{3} + \dots + \frac{17!}{17}}{17!}\right)</math>. We'd like to find <math>h \pmod{17},</math> so we can evaluate our expression <math>\pmod{17}.</math> Since <math>\frac{\frac{17!}{1}}{17!}, \frac{\frac{17!}{2}}{17!}, \dots, \frac{\frac{17!}{16}}{17!}</math> don't have a factor of <math>17</math> in their denominators, and since <math>L_{17}</math> is a multiple of <math>17,</math> multiplying each of those terms and adding them will get a multiple of <math>17.</math> <math>\pmod{17}</math>, that result is <math>0.</math> Thus, we only need to consider <math>L_{17}\cdot \frac{\frac{17!}{17}}{17!} = \frac{L_{17}}{17} \pmod{17}.</math> Proceed with solution <math>1</math> to get <math>\boxed{\textbf{(C) }5}</math>.
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~sirswagger21
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== Solution 3 ==
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Using Wolstenholmes' Theorem, we can rewrite <math>1 + \frac{1}{2} \dots + \frac{1}{16}</math> as <math>\frac{17^2 n}{(17 - 1)!} = \frac{17^2 n}{16!}</math> (for some <math>n \in \mathbb{Z}</math>). Adding the <math>\frac{1}{17}</math> to <math>\frac{17^2 n}{16!}</math>, we get <math>\frac{17^3 n + 16!}{17!}</math>.
 +
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Now we have <math>\frac{17^3 n + 16!}{17!} = \frac{h}{L_{17}}</math> and we want <math>h \pmod{17}</math>. We find that <math>\frac{L_{17}(17^3 n + 16!)}{17!} = \frac{L_{16}(17^3 n + 16!)}{16!} = h</math>. Taking <math>\pmod{17}</math> and multiplying, we get <math>L_{16}(17^3 n + 16!) \equiv 16! \cdot h \pmod{17}</math>.
 +
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Applying Wilson's Theorem on <math>16!</math> and reducing, we simplify the congruence to <math>L_{16}(0 - 1) \equiv -L_{16} \equiv -h \pmod{17}</math>. Now we proceed with Solution 1 and find that <math>L_{16} \equiv 5 \pmod{17}</math>, so our answer is <math>\boxed{\textbf{(C) }5}</math>. 
 +
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~kn07
 +
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==Video Solution (⚡️3 min⚡️)==
 +
https://youtu.be/3g39lB6XLAE
 +
 +
~Education, the Study of Everything
  
 
== Video Solution By ThePuzzlr ==  
 
== Video Solution By ThePuzzlr ==  
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~ MathIsChess
 
~ MathIsChess
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==Video Solution by SpreadTheMathLove==
 +
https://www.youtube.com/watch?v=Wz19lcfF_m8
  
 
== See Also ==
 
== See Also ==

Latest revision as of 19:55, 31 October 2024

Problem

Define $L_n$ as the least common multiple of all the integers from $1$ to $n$ inclusive. There is a unique integer $h$ such that \[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{17}=\frac{h}{L_{17}}\] What is the remainder when $h$ is divided by $17$?

$\textbf{(A) } 1 \qquad \textbf{(B) } 3 \qquad \textbf{(C) } 5 \qquad \textbf{(D) } 7 \qquad \textbf{(E) } 9$

Solution 1

Notice that $L_{17}$ contains the highest power of every prime below $17$ since higher primes cannot divide $L_{17}$. Thus, $L_{17}=16\cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17$.

When writing the sum under a common fraction, we multiply the denominators by $L_{17}$ divided by each denominator. However, since $L_{17}$ is a multiple of $17$, all terms will be a multiple of $17$ until we divide out $17$, and the only term that will do this is $\frac{1}{17}$. Thus, the remainder of all other terms when divided by $17$ will be $0$, so the problem is essentially asking us what the remainder of $\frac{L_{17}}{17} = L_{16}$ divided by $17$ is. This is equivalent to finding the remainder of $16 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13$ divided by $17$.

We use modular arithmetic to simplify our answer:

This is congruent to $-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}$.

Evaluating, we get: \begin{align*} (-1) \cdot 9 \cdot 35 \cdot 11 \cdot 13 &\equiv (-1) \cdot 9 \cdot 1 \cdot 11 \cdot 13 \pmod{17} \\ &\equiv 9 \cdot 11 \cdot (-13) \pmod{17} \\ &\equiv 9 \cdot 11 \cdot 4\pmod{17} \\ &\equiv 2 \cdot 11 \pmod{17} \\ &\equiv 5\pmod{17} \end{align*} Therefore the remainder is $\boxed{\textbf{(C) } 5}$.

~KingRavi

~mathboy282

~Scarletsyc

~wangzrpi

Solution 2

As in solution 1, we express the LHS as a sum under one common denominator. We note that \[\frac{1}{1} + \frac{1}{2} + \dots + \frac{1}{17} = \frac{\frac{17!}{1}}{17!} + \frac{\frac{17!}{2}}{17!} + \frac{\frac{17!}{3}}{17!} + \dots + \frac{\frac{17!}{17}}{17!}\]

Now, we have $h = L_{17}\left(\frac{\frac{17!}{1} + \frac{17!}{2} + \frac{17!}{3} + \dots + \frac{17!}{17}}{17!}\right)$. We'd like to find $h \pmod{17},$ so we can evaluate our expression $\pmod{17}.$ Since $\frac{\frac{17!}{1}}{17!}, \frac{\frac{17!}{2}}{17!}, \dots, \frac{\frac{17!}{16}}{17!}$ don't have a factor of $17$ in their denominators, and since $L_{17}$ is a multiple of $17,$ multiplying each of those terms and adding them will get a multiple of $17.$ $\pmod{17}$, that result is $0.$ Thus, we only need to consider $L_{17}\cdot \frac{\frac{17!}{17}}{17!} = \frac{L_{17}}{17} \pmod{17}.$ Proceed with solution $1$ to get $\boxed{\textbf{(C) }5}$.

~sirswagger21

Solution 3

Using Wolstenholmes' Theorem, we can rewrite $1 + \frac{1}{2} \dots + \frac{1}{16}$ as $\frac{17^2 n}{(17 - 1)!} = \frac{17^2 n}{16!}$ (for some $n \in \mathbb{Z}$). Adding the $\frac{1}{17}$ to $\frac{17^2 n}{16!}$, we get $\frac{17^3 n + 16!}{17!}$.

Now we have $\frac{17^3 n + 16!}{17!} = \frac{h}{L_{17}}$ and we want $h \pmod{17}$. We find that $\frac{L_{17}(17^3 n + 16!)}{17!} = \frac{L_{16}(17^3 n + 16!)}{16!} = h$. Taking $\pmod{17}$ and multiplying, we get $L_{16}(17^3 n + 16!) \equiv 16! \cdot h \pmod{17}$.

Applying Wilson's Theorem on $16!$ and reducing, we simplify the congruence to $L_{16}(0 - 1) \equiv -L_{16} \equiv -h \pmod{17}$. Now we proceed with Solution 1 and find that $L_{16} \equiv 5 \pmod{17}$, so our answer is $\boxed{\textbf{(C) }5}$.

~kn07


Video Solution (⚡️3 min⚡️)

https://youtu.be/3g39lB6XLAE

~Education, the Study of Everything

Video Solution By ThePuzzlr

https://youtu.be/TGcGamPXdNc

~ MathIsChess

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=Wz19lcfF_m8

See Also

2022 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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All AMC 10 Problems and Solutions

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