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I need some help on how to begin a problem. The question is:

Solve $[17][x]=[27]$ in $Z_{350}$. Show how to get the answer without guessing. Express your answer as a class $[x]$ where $x$ is greater than or equal to $0$ and less than $350$.

This is what I tried: $a-b=nk$ where $k$ is some integer. $17x=27$, so $a=17,be=350,c=27$. I then tried to take the $\gcd$ of $a$ and $b$ but the answer to that is $1$. I figured I did something wrong.

Viktor Vaughn
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Jack
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  • This is what I tried. a-b=nk where k is some integer. 17x=27, so a=17,be=350,c=27. I then tried to take the GCD of a and b but the answer to that is 1. I figured I did something wrong. – Jack Sep 22 '17 at 01:12
  • You didn't do anything wrong and you seem to be on the right track. Since $\gcd(17, 350) = 1$ then you can write $17r + 350s = 1$ for some integers $r, s$. This $r$ will be the multiplicative inverse of $17$ in $Z_{350}$, so multiplying both sides of $17x = 27$ by $r$ will cancel the factor of $17$ on the LHS and give you the answer for $x$. So you just have to find the $r$ and $s$ as above, which you can do by reversing the steps of the Euclidean algorithm. – Viktor Vaughn Sep 22 '17 at 01:19

3 Answers3

1

$\!\bmod 350\!:\ \dfrac{27}{17}\equiv\dfrac{17+(10-350)}{17}\equiv 1-20\equiv 331.\ $ The idea is $\,\overbrace{27\equiv 350}^{\Large \color{#c00}{a\ \equiv\ n}\,}\pmod{\!17}\ $ and

$\!\bmod\ n\!:\quad \dfrac{a}{b}\ \equiv\ \dfrac{a-j\,n}b\ $ is an exact quotient if $\ j\equiv \color{#c00}{\dfrac{a}n}\pmod{b}\ \ \,$[so $\, \color{#c00}{a\equiv n} \Rightarrow j\equiv 1$]

Bill Dubuque
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the beginning is to solve $$ 17 t \equiv 1 \pmod {350}. $$ This is often called the Extanded Euclidean Algorithm, I prefer the continued fraction way of writing this:

$$ \gcd( 350, 17 ) = ??? $$

$$ \frac{ 350 }{ 17 } = 20 + \frac{ 10 }{ 17 } $$ $$ \frac{ 17 }{ 10 } = 1 + \frac{ 7 }{ 10 } $$ $$ \frac{ 10 }{ 7 } = 1 + \frac{ 3 }{ 7 } $$ $$ \frac{ 7 }{ 3 } = 2 + \frac{ 1 }{ 3 } $$ $$ \frac{ 3 }{ 1 } = 3 + \frac{ 0 }{ 1 } $$ Simple continued fraction tableau:
$$ \begin{array}{cccccccccccc} & & 20 & & 1 & & 1 & & 2 & & 3 & \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 20 }{ 1 } & & \frac{ 21 }{ 1 } & & \frac{ 41 }{ 2 } & & \frac{ 103 }{ 5 } & & \frac{ 350 }{ 17 } \end{array} $$ $$ $$ $$ \begin{array}{ccc} \frac{ 1 }{ 0 } & \mbox{digit} & 20 \\ \frac{ 20 }{ 1 } & \mbox{digit} & 1 \\ \frac{ 21 }{ 1 } & \mbox{digit} & 1 \\ \frac{ 41 }{ 2 } & \mbox{digit} & 2 \\ \frac{ 103 }{ 5 } & \mbox{digit} & 3 \\ \frac{ 350 }{ 17 } & \mbox{digit} & 0 \\ \end{array} $$

$$ 350 \cdot 5 - 17 \cdot 103 = -1 $$ $$ -350 \cdot 5 + 17 \cdot 103 = 1 $$

$$ 103 \cdot 17 \equiv 1 \pmod {350}, $$ $$ \frac{1}{17} \equiv 103 \pmod {350}. $$

$$ (27 \cdot 103) \cdot 17 \equiv 27 \pmod {350}. $$

Will Jagy
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Given $[17][x]=[27]$ in $Z_{350}$, we need to find $k=[17]^{-1}$ such that $[17]k=[1]$ and thus $[x]=[27]k$.

Since $17$ is coprime to $350$, we can be sure that the inverse exists and can find it through the extended Euclidean algorithm:

$\begin{array}{c|c} n & s & t & q \\ \hline 350 & 1 & 0 & \\ 17 & 0 & 1 & 20 \\ 10 & 1 & -20 & 1 \\ 7 & -1 & 21 & 1 \\ 3 & 2 & -41 & 2 \\ 1 & -5 & 103 & 3 \\ \end{array}$

This gives $-5\cdot 350 + 103\cdot 17 =1$ and thus $17^{-1}\equiv 103\bmod 350$.

Then $103\cdot 27 = 2781 = 2800-19\equiv 331 \bmod 350$.

Joffan
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