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

m (Solution)
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Evaluating, we get:  
 
Evaluating, we get:  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
-45 \cdot 7 \cdot 11 \cdot 13 \pmod{17} &\equiv 6 \cdot 7 \cdot 11 \cdot 13 \pmod{17} \
+
-1 \cdot 35 \cdot 9 \cdot 11 \cdot 13 \pmod{17} &\equiv 9 \cdot 11 \cdot -13 \pmod{17} \
&\equiv 42 \cdot 11 \cdot 13 \pmod{17} \
+
&\equiv 9 \cdot 11 \cdot 4 \pmod{17} \
&\equiv 8 \cdot 11 \cdot 13 \pmod{17} \
+
&\equiv 36 \cdot 11 \pmod{17} \
&\equiv 88 \cdot 13 \pmod{17} \
+
&\equiv 2 \cdot 11 \pmod{17} \
&\equiv 3 \cdot 13 \pmod{17} \
+
&\equiv 22 \pmod{17} \
&\equiv 39 \pmod{17} \
 
 
&\equiv 5\pmod{17}
 
&\equiv 5\pmod{17}
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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~KingRavi
 
~KingRavi
 +
 +
~Scarletsyc
  
 
~mathboy282(tex)
 
~mathboy282(tex)

Revision as of 21:10, 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: \begin{align*} -1 \cdot 35 \cdot 9 \cdot 11 \cdot 13 \pmod{17} &\equiv 9 \cdot 11 \cdot -13 \pmod{17} \\ &\equiv 9 \cdot 11 \cdot 4 \pmod{17} \\ &\equiv 36 \cdot 11 \pmod{17} \\ &\equiv 2 \cdot 11 \pmod{17} \\ &\equiv 22 \pmod{17} \\ &\equiv 5\pmod{17} \end{align*}

Therefore the remainder is $5$ and the answer is $\boxed{\textbf{(C) } 5}$.

~KingRavi

~Scarletsyc

~mathboy282(tex)

Video Solution By ThePuzzlr

https://youtu.be/TGcGamPXdNc

~ MathIsChess

See Also

2022 AMC 10A (ProblemsAnswer KeyResources)
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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