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# Second one (Urgent)

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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

#1
+20805
+4

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.

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

Dec 20, 2018

#1
+20805
+4

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.

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

heureka Dec 20, 2018