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# pls help

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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?

Jul 16, 2020

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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?

Modular multiplicative inverse is defined if $$\gcd(130,n)=\gcd(231,n)=1$$

$$\begin{array}{|rcll|} \hline \text{factor}~ 130 &=& 2*5*13 \\ \text{factor}~ 231 &=& 3*7*11 \\ \hline \end{array}$$

The first prim number = 2. This number is a prime factor in 130, so $$\gcd(130,2) \ne 1$$
The 2nd prim number = 3. This number is a prime factor in 231, so $$\gcd(231,3) \ne 1$$
The 3rd prim number = 5. This number is a prime factor in 130, so $$\gcd(130,5) \ne 1$$
The 4th prim number = 7. This number is a prime factor in 231, so $$\gcd(231,7) \ne 1$$
The 5th prim number = 11. This number is a prime factor in 231, so $$\gcd(231,11) \ne 1$$
The 6th prim number = 13. This number is a prime factor in 130, so $$\gcd(130,13) \ne 1$$
The 7th prim number = 17. This number is not a prime factor in 130 and not a prime factor in 231
, $$\gcd(130,17) = \gcd(213,17) = 1$$

The smallest integer n is 17

check:

$$\begin{array}{|rcll|} \hline 17^{-1} \pmod{130} &\equiv& 23 \\ 17*23 & \equiv& 1 \pmod{130}~ \checkmark \\\\ 17^{-1} \pmod{231} &\equiv& 68 \\ 17*68 & \equiv& 1 \pmod{231}~ \checkmark \\ \hline \end{array}$$ Jul 17, 2020