$ 7^{50} \cdot 4^{102} ≡ x \pmod {110} $ The way I would solve this would be: $$ (7^3)^{15} \cdot 7^5 \cdot (4^4)^{25} \cdot 4^2 $$ and take it from there, but I know that this is most likely in an inefficient way.  Does anyone have more efficient methods?
 A: Do prime factorization on $110$. It's $110 = 5 \cdot 2 \cdot 11$.
Now work with the prime factors as moduli.
$$7^{50} \cdot 4^{102} \equiv 0 \pmod 2 \text{ trivial, as 2 \mid 4}$$ 
Then apply Fermat's Little Theorem so we have:
$$7^4 \equiv 1 \pmod 5 \implies 7^{50} \equiv 49 \equiv 4 \pmod 5$$
$$4^4 \equiv 1 \pmod 5 \implies 4^{102} \equiv 4^2 \equiv 1 \pmod 5$$
Now multiply them and we have:
$$7^{50} \cdot 4^{102} \equiv 4 \cdot 1 \equiv 4 \pmod 5$$
Now repeat the method with the last factor:
$$7^{10} \equiv 1 \pmod {11} \implies 7^{50} \equiv 1 \pmod {11}$$
$$4^{10} \equiv 1 \pmod {11} \implies 4^{102} \equiv 4^2 \equiv 16 \equiv 5 \pmod {11}$$
Multiply them and we have:
$$7^{50} \cdot 4^{102} \equiv 1 \cdot 5 \equiv 5 \pmod {11}$$
Now just apply CRT to the three congruence relation to get the final answer.
A: $4^{102}$ is just $2^{204}$ therefore $7^{50}*4^{102}\equiv 7^{50}*2^{204}\bmod 110$. So if we can find a reduced number to which $2^{204}$ is congruent to mod 55 then we can know its congruence mod 110 since we know its even.
using euler's theorem we get that $2^{\phi (55)}\equiv 1 \bmod55$
since $\phi(55)=40$
$2^{204}\equiv 2^4 \bmod 55$ so $2^{204}\equiv 16 \bmod110$.
using the fact that $\phi(110)$ is $40$
we get $7^{50} \equiv 7^{10}$ make a list:
7,49,13,91,87,59,83,31,107,89.
Therefore :$7^{50}*4^{102}\equiv 89*16\equiv 104 \bmod 110$
