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 $\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{17}=\frac{h}{L_{17}}$ 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 $17$ . 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

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 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
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