<s>[INST] Find all solutions using the Chinese Remainder Theorem Find all solutions using the Chinese Remainder Theorem.
$$
\begin{cases}x \equiv 3 \pmod{4}\\
x \equiv 5 \pmod{21}\\
x \equiv 7 \pmod{25}
\end{cases}$$
I can see that $4$,$21$, and $25$ are all pairwise relatively prime. I then proceeded to use$ Z=(C_1)\times(X_1)\times(B_1) +(C_2)\times(X_2)\times(B_2) +(C_3)\times(X_3)\times(B_3)$.
Let $C_i $ be the remainder and$ b_i$ be the modulus.
$3=C_1,5=C_2,7=C_3$. 
$4=b_1,21=b_2,25=b_3$.
Let $B=b_1\times b_2\times b_3$
Let $B_i=B/b_i$
Then, $525=B_1,100=B_2,84=B_3$
Now I try and solve $(B_i)(X_i)≡ 1 \mod b_i$
I can see that $(B_1)(X_1)≡ 1 \mod b_1 $yields $525(X_1)≡ 1 \mod 4 $with$ X_1 $being $1$.
For $(B_2)(X_2)≡ 1 \mod b_2$, I get $100(X_2)≡ 1 \mod 21$. $X_2$ turns out to be $-1$.
For $(B_3)(X_3)≡ 1 \mod b_3 = 84(X_3)≡ 1 \mod 25$, I needed to use the extended euclidean algorithm to find the gcd and $X_3$, which turned out to be $-11$.
Work: 
$$m   \;n \;  q \; r$$
      $$\\84\;25  \;3 \;9$$
      $$\\25 \;9  \;2 \;7$$
      $$\\9   \;7  \;1\;  2$$
      $$\\7 \;  2   \;3 \;1$$
$1=7-(3\times2)
\\=7-(3(9-7))=(4\times7)-(9\times3)
\\=(4(25-(9\times2)))-(9\times3)=(4\times25)-(9\times-11)
\\=(4\times25)-(11(84-(25\times3)))=(37\times25)-(11\times84)$
Thus, $X_3=-11$
It turns out that I computed the wrong $(X_2)$, it is instead $4$.
With this, I can compute the correct $Z$ such that it will satisfy the three congruence equations.
$Z=(3\times525\times1)+(100\times5\times4)+(84\times7\times-11)=-2893$
Using a calculator, I found that 
$$Z \equiv 3 \pmod{4}$$
$$Z \equiv 5 \pmod{21}$$
$$Z \equiv 7 \pmod{25}$$
 [/INST] I should at least mention that there is another way to solve
$\begin{cases}
   x \equiv 3 \pmod{4}\\
   x \equiv 5 \pmod{21}\\
   x \equiv 7 \pmod{25}
\end{cases}$
We need to find integers $\alpha,\beta, \gamma \in \mathbb Z_{2100}$ (where $2100 = 4 \cdot 21 \cdot 25$) such that
$\begin{cases}
   \alpha \equiv 1 \pmod{4}\\
   \alpha \equiv 0 \pmod{21}\\
   \alpha \equiv 0 \pmod{25}
\end{cases}, \qquad 
\begin{cases}
   \beta \equiv 0 \pmod{4}\\
   \beta \equiv 1 \pmod{21}\\
   \beta \equiv 0 \pmod{25}
\end{cases},\; \text{and} \qquad
\begin{cases}
   \gamma \equiv 0 \pmod{4}\\
   \gamma \equiv 0 \pmod{21}\\
   \gamma \equiv 1 \pmod{25}
\end{cases}$
It will follow that $x \equiv 3\alpha + 5 \beta + 7 \gamma \pmod{2100}$
where $2100 = 4 \cdot 21 \cdot 25$.
To solve $\begin{cases}
   \alpha \equiv 1 \pmod{4}\\
   \alpha \equiv 0 \pmod{21}\\
   \alpha \equiv 0 \pmod{25}
\end{cases}$
we first note that $\alpha \equiv 0 \pmod{21}$ and 
$\alpha \equiv 0 \pmod{25}$ implies $\alpha \equiv 0 \pmod{525}$.
So $\alpha = 525t$ for some integer $t$. 
Thus
\begin{align}
   \alpha \equiv 1 \pmod 4
   &\implies 525t \equiv 1 \pmod 4 \\
   &\implies t \equiv 1 \pmod 4 \\
   &\implies \alpha \equiv 525 \pmod{2100}
\end{align}.
Similarly
\begin{align}
   \beta\equiv 1 \pmod{21} 
   &\implies 100t \equiv 1 \pmod{21} \\
   &\implies -5t \equiv 1 \pmod{21} \\
   &\implies t \equiv 4 \pmod{21} \\
   &\implies \beta \equiv -20 \pmod{2100}
\end{align}
and
\begin{align}
   \gamma \equiv 1 \pmod{25} 
   &\implies 84t\equiv 1 \pmod{25} \\
   &\implies 9t \equiv 1 \pmod{25} \\
   &\implies t \equiv 14 \pmod{25} & (9 \cdot 14 = 126 \equiv 1 \pmod{25})\\
   &\implies \gamma \equiv 1176 \pmod{2100}
\end{align}.
So
\begin{align}
    x 
    &\equiv 3\alpha + 5 \beta + 7 \gamma \pmod{2100} \\
    &\equiv 3(525) + 5(-20) + 7(1176) \pmod{2100} \\
   &\equiv 1307 \pmod{2100} \\
\end{align}
</s>