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What is the smallest integer n, greater than 1, such that n^{-1} mod{130} and n^{-1} mod{231} are both defined?

 Dec 20, 2018

Best Answer 

 #1
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What is the smallest integer n, greater than 1, such that
\(n^{-1} \pmod{130}\)
and
\(n^{-1} \pmod{231} \)
are both defined?

 

\(\text{$n$ must be coprime to $130$ and $231$ } \\ \text{respectively $\gcd(n,130) = \gcd(n,231) = 1$ }\)

 

Factorisation:

\(\begin{array}{|rcll|} \hline 130 &=& 2\times 5 \times 13 \\ 231 &=& 3\times 7 \times 11 \\ \hline \end{array}\)

\(\text{The next prime number after $2,3,5,7,11,13$ is ${\color{red}17}$ }\)

 

So 17 is the smallest integer, greater than 1.

\(17^{-1} \pmod {130} = 23 \\ 17^{-1} \pmod {231} = 68 \)

 

laugh

 Dec 20, 2018
 #1
avatar+21848 
+9
Best Answer

What is the smallest integer n, greater than 1, such that
\(n^{-1} \pmod{130}\)
and
\(n^{-1} \pmod{231} \)
are both defined?

 

\(\text{$n$ must be coprime to $130$ and $231$ } \\ \text{respectively $\gcd(n,130) = \gcd(n,231) = 1$ }\)

 

Factorisation:

\(\begin{array}{|rcll|} \hline 130 &=& 2\times 5 \times 13 \\ 231 &=& 3\times 7 \times 11 \\ \hline \end{array}\)

\(\text{The next prime number after $2,3,5,7,11,13$ is ${\color{red}17}$ }\)

 

So 17 is the smallest integer, greater than 1.

\(17^{-1} \pmod {130} = 23 \\ 17^{-1} \pmod {231} = 68 \)

 

laugh

heureka Dec 20, 2018

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