finding remainder by using fermat's little theorem or euler's totient I am trying to carry on for the following equation but I stuck with a big number that still needs to be smaller to calculate without calculator.
$\ 331^{51}\mod 49$
Since the $\phi(49)=42$, I carried on the problem as follows:
$=\ 331^{42}.331^{9}\mod 49$ 
which $331^{42}=1$, so;
$=\ 331^{9}\mod 49$
Since $331/49$ remains $37$ problem becomes;
$=\ 37^{9}\mod 49$
$=\ (-12)^{9}\mod 49$
$=\ (-12)^{8}.(-12)\mod 49$
Since 8 is even;
$=\ (12)^{8}.(-12)\mod 49$
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.(a bunch of conversions)...
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Finally I have found this (which is checked via wolframalpha and it is correct);
$=\ (2)^{29}.(23)\mod 49$
But it is still too big to calculate manually. So I wonder maybe one of you guys have another idea? Especially, I want to know whether I can use $49$ as $7*7$ and create two different modular equations and solve them in a way? maybe with CRT maybe with another method?

Hint:
Using binomial theorem,
$$(-12)^7=(2-14)^7\equiv2^7\pmod{49}$$
and $$12^2\equiv-3\pmod{49}$$
