What's the order of $2$ in $U_{143}$ Kindly read my comments, proving that $2^{60}=1 \pmod{143}$ doesn't prove that it's the order of 2 as we need to prove that 60 is the smallest number to achieve this result...
I know that in $U_{11}$, $o(2)=10$ and that in $U_{13}$, $o(2)=12$
I need to find $o(2)$ in $U_{143}$. How can I find that?
I think this has something to do with the fact that both $11$ and $13$ are prime numbers and that $143 = 11\cdot 13$
I tried to calculate $\mathrm{lcm}(10,12)$ which is $60$. I chose those 2 numbers in specific because they are orders of 2 as mentioned above.
Is there any mathematical law which is relative to the facts mentioned above?
 A: Yes.  Pretty easily $\pmod {143}$, we have $2^{20}\
\not\equiv 1, 2^{30}\not\equiv 1$.  Thus the conclusion, since no divisors of $20,30$ (hence $60$) can be the order.  (Just check $\pmod {11}$ and $\pmod {13}$.)
(For instance,  since you computed that the order of $2$ in $U_{13}$ is $12$, we have it, since neither $20$ nor $30$ is a multiple of $12$.)
A: Even though this may not answers precisely the question of the OP, it could be useful to be posted.
Let's make clear that $U_{143}$ is the group of units of the commutative ring $\Bbb Z_{143}$. In addition,
$$U_{143}=\{a\in \Bbb Z_{143}:\gcd(a,143)=1,\ 1\leq a \leq 142\}.$$
As $\gcd(2,143)=1$, it is clear that $2\in U_{143}$.
The order of this group is $|U_{143}|=\varphi(143)=120$. But using a result which yields from Lagrange's Theorem, $\mathrm{ord}(2)\mid |U_{143}|$ and therefore $$\mathrm{ord}(2)\in \{1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120\}.$$ So, it suffices to calculate the powers
$$2^{a}\bmod{143}$$
where $a=1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60$ or $120$. The smallest such $a$ which gives us $1\bmod{143}$ is the order of $2$ modulo $143$.
Hope that helps.
