I can't find out, how to solve this. Will you give me some advice what to do in 4th step? Lot of thanks.
This is my example: $7^{30}\equiv x\pmod{ 100}$ I want to compute it this way.
These are my steps:
For small CRT problems there are often many optimizations we can use to simplify computation. Here, since one modulus $= 4\,$ is very small, we can reduce to checking $\,4\,$ candidate solutions, viz. $\,x\equiv 24\pmod{25}\!\iff\! x = 24+25k\!\iff\!$ $x\equiv 24,49,74,99\pmod{\!100},\,$ so we reduce to checking which of these candidates $\,x\,$ also satisfies $\,x\equiv 1\pmod{\!4},\,$ which is easy to test [also easy to compute: $\bmod 4\!:\ \color{#c00}1\equiv x\equiv 24+25k\equiv \color{#c00}k\,$ so $\,x \equiv 24+25(\color{#c00}1) \equiv 49\pmod{\!100}$].
The above method will generally be the quickest when one of the moduli $\,m\,$ is very small, since it involves checking only $\,m\,$ candidate solutions. Another optimization also applies to this system: calculating the Bezout identity for the moduli $\,m,n = 25,4\,$ is trivial since $\,25\ {\rm mod}\ 4 = 1,\, $ so $\ \ \color{#c00}{25}+\color{#0a0}{(-6)4} = 1,\,$ and we can easily transform a Bezout identity to a CRT solution as follows
$ \!\!\qquad\color{#c00}{j m} + \color{#0a0}{k n} = 1\ \ \ \Rightarrow\ \ \begin{eqnarray}&&x\equiv a\!\!\!\pmod{\! m}\\ &&x\equiv b\!\!\!\pmod{\! n}\end{eqnarray}$ $\!\iff\!$ $\begin{eqnarray} x&\equiv&\ a\,\color{#0a0}{kn}\, +\, b\,\color{#c00}{jm}&&({\rm mod}\,\ {mn})\\ &\equiv& a+(b\!-\!a)\color{#c00}{jm}&&({\rm mod}\,\ mn)\end{eqnarray}$
$\ \ \color{#c00}{25} + \color{#0a0}{(-6)4} = 1\ \ \ \Rightarrow\ \ \begin{eqnarray}&&x\equiv a\!\!\!\pmod{\!25}\\ &&x\equiv b\!\!\!\pmod {\!4}\end{eqnarray}$ $\!\iff\!$ $\begin{eqnarray} x&\equiv&\ a\,\color{#0a0}{(-24)} + b\,\color{#c00}{(25)}&&({\rm mod}\,\ {mn})\\ &\equiv&\ a\ +\ (b\!-\!a)\,\color{#c00}{(25)}&&({\rm mod}\,\ mn)\end{eqnarray}$
For the special case $\ a,b\, =\,-1,1\ $ we obtain $\ x\equiv (-1)\color{#0a0}{(-24)}+(1)\color{#c00}{(25)} \equiv 49\pmod{\!100}.\,$ With practice, we can do the above in under a minute of mental arithmetic in most small cases.
The same optimization works generally if one of the moduli is $\pm1$ modulo the other, since that immediately yields the Bezout identity for their gcd $= 1.\,$ This is essentially an optimization of the extended Euclidean algorithm when it takes only a single step. Compare the second form of the solutions above to Easy CRT.
$$7^2=49=50-1\implies7^4=(50-1)^2=50^2-2\cdot50+1^2\equiv1\pmod{100}$$
As $\displaystyle30\equiv2\pmod4,7^{30}\equiv7^2\pmod{100}$
$\displaystyle7\equiv-1\pmod4\implies7^{30}\equiv(-1)^{30}\implies7^{30}\equiv1\pmod4\ \ \ \ (1)$
$\displaystyle7^2=49\equiv-1\pmod{25}\implies7^{30}=(-1)^{15}\implies7^{30}\equiv-1\pmod{25}\ \ \ \ (2)$
Apply Chinese Remainder Theorem on $(1),(2)$
The problem is to reduce 7^30 to a manageable form. Start by reducing powers of 7 to mod 100. It turns out that 7^10 = 49 mod 100.
Therefore (7^10)^3 = 49^3 mod 100. 49^3 mod 100 turns out to be 49 mod 100.
So 7^30 = 49 mod 100.
