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

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This is congruent to <math>-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</math>
 
This is congruent to <math>-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</math>
  
−
Evaluating, we get: <cmath>-45 \cdot 7 \cdot 11 \cdot 13 \pmod{17} \cong 6 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</cmath>
+
Evaluating, we get: <cmath>-45 \cdot 7 \cdot 11 \cdot 13 \pmod{17} \equiv 6 \cdot 7 \cdot 11 \cdot 13 \pmod{17}</cmath>
−
<cmath> \cong 42 \cdot 11 \cdot 13 \pmod{17} \cong 8 \cdot 11 \cdot 13 \pmod{17}</cmath>
+
<cmath> \equiv 42 \cdot 11 \cdot 13 \pmod{17} \equiv 8 \cdot 11 \cdot 13 \pmod{17}</cmath>
−
<cmath> \cong 88 \cdot 13 \pmod{17} \cong 3 \cdot 13 \pmod{17}</cmath>
+
<cmath> \equiv 88 \cdot 13 \pmod{17} \equiv 3 \cdot 13 \pmod{17}</cmath>
−
<cmath>\cong 39 \pmod{17} \cong 5\pmod{17}</cmath>
+
<cmath>\equiv 39 \pmod{17} \equiv 5\pmod{17}</cmath>
  
 
Therefore the remainder is <math>5</math> and the answer is <math>\boxed{C}</math>
 
Therefore the remainder is <math>5</math> and the answer is <math>\boxed{C}</math>

Revision as of 13:32, 12 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: \[-45 \cdot 7 \cdot 11 \cdot 13 \pmod{17} \equiv 6 \cdot 7 \cdot 11 \cdot 13 \pmod{17}\] \[\equiv 42 \cdot 11 \cdot 13 \pmod{17} \equiv 8 \cdot 11 \cdot 13 \pmod{17}\] \[\equiv 88 \cdot 13 \pmod{17} \equiv 3 \cdot 13 \pmod{17}\] \[\equiv 39 \pmod{17} \equiv 5\pmod{17}\]

Therefore the remainder is $5$ and the answer is $\boxed{C}$


~KingRavi

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

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