<s>[INST] Specific steps in applying the Chinese Remainder Theorem to solve modular problem splitting modulus I am trying to get an idea of how the Chinese Remainder Theorem (CRT) can be used to finish up this problem, in which the problem
$$7^{30}\equiv x\pmod{ 100}$$
is attempted by splitting the modulus into relatively prime factors $25$ and $4,$ arriving at
$$\begin{align}
7^{30}&\equiv1\pmod4\\
7^{30}&\equiv-1\pmod{25}
\end{align}$$ 
I understand that the CRT may be called upon because $m=\prod m_i,$ and we have the same $7^{30}$ value on the LHS, but I don't know how to carry it out.
The question was touched upon in this post as the second entry:

How do I efficiently compute  $a^b \pmod c$ when $b$ is less than $c.$ For instance $5^{69}\,\bmod 101.$

However, I don't see this particular point clearly worked out, perhaps because it is a multi-pronged question.

Following this presentation online, this seems to be the verbatim application of the CRT without any added concepts or shortcuts:
From @gimusi's answer (upvoted):
$$\begin{cases}
x \equiv 7^{30} \pmod4\\
x\equiv 7^{30} \pmod{25}
\end{cases}$$
rearranged into
\begin{cases}
x \equiv 1 \pmod4\\
x\equiv -1 \pmod{25}
\end{cases}
Given the general form of the equations above as $x\equiv a_i \pmod {m_i},$ the CRT states $x\equiv a_1 b_1 \frac{M}{m_1}+a_2 b_2 \frac{M}{m_2}\pmod M$ with $M=\prod m_i,$ and with 
$$b_i =\left(\frac{M}{m_i}\right)^{-1}\pmod {m_i}.$$
The inverse of $\frac{M}{m_i}$ is such that $\frac{M}{m_i}\left(\frac{M}{m_i}\right)^{-1}\pmod {m_i}\equiv 1.$ 
Calculating the components:
$$\begin{align}
a_1&=1\\
a_2&=-1\\
M&=4\times 25 =100\\
\frac{M}{m_1} &= \frac{100}{4}=25\\
\frac{M}{m_2} &= \frac{100}{25}=4\\
b_1 &= \left(\frac{M}{m_1}\right)^{-1} \pmod 4 = (25)^{-1}\pmod 4 =1\\
b_2 &= \left(\frac{M}{m_2}\right)^{-1} \pmod {25}= (4)^{-1} \pmod{25}=19
\end{align}$$
Hence,
$$x=1\cdot 25 \cdot 1 + (-1)\cdot 4 \cdot 19 = -51 \pmod{100}\equiv 49.$$
 [/INST] Welcome to Math SX! You have to use Euler's theorem as $\varphi(4)=2$, $\;\varphi(25)=20$ we have
$$ 7^{30}\equiv7^{30\bmod2}=1\mod 4,\qquad 7^{30}\equiv7^{30\bmod20}=7^{10}\mod 25$$
To find the latter power, you can use the modular fast exponentiation algorithm, but here, it will be simpler: modulo $25$,
$$7^2\equiv -1\enspace\text{so}\enspace 7^4=1,\enspace\text{hence } \;7^{30}\equiv 7^{30\bmod 4}=7^2\equiv -1.$$
Finally, since $\;25-6\cdot 4=1$ (Bézout's identity),
$$7\equiv \begin{cases}\phantom{-}1\mod4\\-1\mod 25\end{cases}\iff 7\equiv 1\cdot 25-(-1)\cdot 6\cdot 4=49\mod 100.$$
</s>