How to solve $7^n\equiv 1 \pmod {180}$? I've used Euler's theorem: $a^{\phi(m)}\equiv 1 \pmod m$, where $a$ and $m$ are relatively prime.
and I got: $7^{48}\equiv 1 \pmod{180}$.
My question is: Is there a smaller $n$ satisfies the equation: $7^n\equiv 1 \pmod{180}$?
For example: By using Euler's theorem to solve $2^n\equiv 1 \pmod{15}$
             we get: $2^8\equiv 1 \pmod{15}$, where $n=8$.
             However, $2^4\equiv 1 \pmod{15}$ also satisfies the equation where $n=4$.
How to find the smallest $n$?
 A: To find $n$, first try $24=\frac{48}{2}$. We are succesful, so we try $12=\frac{24}{2}$. Success again. The prime divisors of $12$ are $2$ and $3$, so you only need to check $6$ and $4$, which are not successful.
A: Try working  modulo the prime powers that make up $180$. Note that $180$ is $2^2 \times 3^2 \times 5$. 


*

*Modulo $2^2$ you get that $7^n$ is $1$ if $n$ is a multiple of $2$ (and it is easy to see it is an only if). 

*Modulo $5$ you get that $7^n$ is $1$ if $n$ is a multiple of $4$, and it is not hard to see it is an only if by checking $7^2$ is not $1$. 

*Modulo $3^2$ you get that $7^n$ is $1$ if $n$ is a multiple of $6$. Yet in fact $7^3$ is already $1$ too. But $7^2$ is not $1$. 
Putting this together you get that modulo $180$ the $n$ for which $7^n$ is $1$ are  multiples of the LCM of $2,4,3$ that is $12$.
Note, if your know it you could also use Carmichael's $\lambda$ function  instead of $\varphi$ to get the upper bound of $12$ right away. 
A: Since $a^{\phi(n)}\equiv 1 \bmod n$ for $(a,n)=1$ by Euler's theorem, we have $7^{48}\equiv 1\bmod 180$, and we can check the divisors $d$ of $48$, whether or not $7^d\equiv 1\bmod 180$ holds. This follows from the fact, that the order of $a$ divides an exponent $m$ with $a^m\equiv 1$. Now $24$ and $12$ do work, but smaller ones do not work.
