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2022 AMC 10A Problems/Problem 19: Difference between revisions

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==Problem==
==Problem==


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  
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
 
<cmath>\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\ldots+\frac{1}{17}=\frac{h}{L_{17}}</cmath>
<math>\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\ldots+\frac{1}{17}=\frac{h}{L_{17}}</math>
 
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>?


Revision as of 05:44, 16 November 2022

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 \[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\ldots+\frac{1}{17}=\frac{h}{L_{17}}\] 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

Notice that $L_{17}$ contains the highest power of every prime below $17$. 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 $\frac{L_{17}}{17}$ 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

Video Solution By ThePuzzlr

https://youtu.be/TGcGamPXdNc

~ MathIsChess

See Also

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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