<s>[INST] Smallest such $n \in \mathbb{N}$ that $2^{n} \equiv 1 \pmod{5\cdot 7\cdot 9\cdot 11\cdot 13}$ Can anybody give me a hint about how to find smallest such $n \in \mathbb{N}$ that $2^{n} \equiv 1  \pmod{5\cdot 7\cdot 9\cdot 11\cdot 13}$? 
I thought that I will find it piece by piece with help from my friend Fermat's Little Theorem, so :


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*$2^{n_{1}} \equiv 1 \mod 5$, so $n_{1}=4$.

*$2^{n_{2}} \equiv 1 \mod 7$, so $n_{2}=6$.

*$2^{n_{3}} \equiv 1 \mod 3$, so $n_{3}=2$.

*$2^{n_{4}} \equiv 1 \mod 11$, so $n_{4}=10$.

*$2^{n_{5}} \equiv 1 \mod 13$, so $n_{5}=12$.
So, since I know that $x \equiv y \mod mz \Leftrightarrow x \equiv y \mod m$ when $z \neq 0$, so my lucky was to take the lowest common multiple of $4,6,2,10,12$, which is $60$ and lo and behold it fits the bill. But is it the smallest such $n$? If so, how to explain it?
 [/INST] First of all, $p-1$  is not necessarily the Multiplicative Order ord$_pa$  for prime $p$ and $(a,p)=1$
For example, ord$_72=3\ne 7-1$
Again, $9=3^2$ is not prime, we need Euler's Totient Theorem here
Now as ord$_{11}2=10$ and ord$_{13}2=12,$ there can be no number $<$lcm
$(10,12)=60$ to satisfy $2^n\equiv1\pmod {5\cdot7\cdot9\cdot11\cdot13}$
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