Least positive integer satisfying congruence relation How do I go about finding the solution to this problem?
Find the least positive intger, $M$, such that $M^{49}\equiv21 \pmod{209}$.
I'm thinking we need to rewrite the left hand side somehow to reduce the exponent, but I don't know how since $M$ is unknown. I'm also thinking some theorem could be applied cleverly, but don't know which.
Thanks!

HINT
Observe that $209=11 \times 19$
Thus consider the equivalent system:
$M^{49}\equiv-1 \pmod{11}$
$M^{49}\equiv2 \pmod{19}$
then
$M^{49}\equiv-1 \pmod{11} \implies M^{-1}\equiv-1 \pmod{11} \implies M\equiv-1 \pmod{11} $
$M^{49}\equiv2 \pmod{19}\implies M^{13}\equiv2 \pmod{19}$
to simplify the last consider that
$ord_{19}(M) |18 \implies ord_{19}(M)=2,3,6,9,18\implies ord_{19}(M)=18$
[infact assuming: $ord_{19}(M)=2,3,6,9$ leads to contradictions]
thus $\pmod{19}$
$M^{9}\equiv -1, \ M^{13}\equiv2 \implies M^{4}\equiv -2 \implies M^{8}\equiv 4\implies M^{24}\equiv 7\implies M^{37}\equiv 14$
$M^{13}\equiv2, \ M^{24}\equiv 7\implies\ M^{37}\equiv 14$
$M^{18}\equiv 1, \ M^{37}\equiv 14 \implies M \equiv 14 \pmod{19}$
finally the system to be solved is
$$
\left\{
\begin{array}{c} M \equiv -1 \pmod{11} \\
M \equiv 14 \pmod{19} \end{array} \right.$$
