pls help

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:

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