2022 AMC 10A Problems/Problem 19

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 $-1 \cdot 9 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \pmod{17}$

Solution in Progress

~KingRavi

Video Solution By ThePuzzlr

https://youtu.be/TGcGamPXdNc

~ MathIsChess

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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2022 AMC 10A Problems/Problem 19

Contents

Problem

Solution

Video Solution By ThePuzzlr

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