Find a number that satisfies two congruences I have an exercise of finding a number that satisfies two congruences. 
$$
\left\{
\begin{array}{cc}
x &\equiv 10 \mod 30 \\
x &\equiv 5 \mod 101
\end{array}
\right.
$$ 
The exercise suggests 4 steps to solve:
Step 1: find integers $s$ and $t$ such as $30s + 101t = 1$
Step 2: use s and t to find a number satisfying: 
$$
\left\{
\begin{array}{cc}
a &\equiv 10 \mod 30 \\
a &\equiv 0 \mod 101
\end{array}
\right.
$$ 
Step 3: use s and t to find a number satisfying:
$$
\left\{
\begin{array}{ccc}
b &\equiv& 0 \mod 30 \\
b &\equiv& 5 \mod 101
\end{array}
\right.
$$ 
Step 4: use the result from step 2 or 3 to find x. ($x \equiv 10 \mod 30 $,$x \equiv 5 \mod 101$)
I went through the first 3 steps without problems, but I dont't know how to do step 4 since I could also use s and t to find x (without going to step 2 or 3). 
Thank you
 A: writing $$x=10+30m$$ and $$x=5+101n$$ where $m,n$ are integers. From here we get
$$5=101n-30m$$ this is an linear Diophantine equation in $m,n$.Solving this equation we get
$$n=84+101C,n=25+30C$$ where $C$ is an integer
A: Since $\gcd(30,101)=1$ by CRT we know that an unique solution exists $\mod{30\cdot101}$, notably


*

*$x \equiv 10 \mod 30\implies x=10+30k$

*$x \equiv 5 \implies10+30k\equiv5\implies30k\equiv96\mod 101$


thus we need to find the inverse of $30 \mod 101$ by Euclidean algorithm that is $64$ indeed
$$64\cdot30-101\cdot19=1$$
then


*

*$30k\equiv96 \mod 101\implies64\cdot30k\equiv 64\cdot96\mod 101 \implies k\equiv 84 \mod 101$


and


*

*$x=10+30k=10+30\cdot(84+101s)=2530+30\cdot101s$



$$\implies x\equiv 2530 \mod 30\cdot 101$$

A: Well, I solved it using a logical approach.
We can see that 
$x - 10 = 30a$
and 
$x - 5 = 101b$
($a$ and $b$ are random whole numbers)
Now if we observe, it's very clear that $x$ is a multiple of $10$. Also $b$ in the second equation will have to be an odd multiple of $5$. So putting some values of $b$, we get,
$x = 510, 1520, 2530,...$ and so on. Out of these values only the ones who have  $b$ of the format $5(6k - 1)$ fulfill both the conditions of equation $1$ and $2$. Putting $b = 5(6k - 1)$ in equation $2$ we get, the following,
$x = 10(303k - 50)$ where $k$ is any natural number.
Feel free to correct me if I am wrong anywhere and improve the answer. 
A: The second congruence implies there exists $m$ such that $$x=101m+5.$$ Substituting this into the first congruence gives $$101m+5\equiv 10\bmod 30.$$ Now, reducing this gives $$11m+5\equiv 10\bmod 30.$$ This simplifies to $$11m\equiv 5\bmod 30,$$ $$m\equiv 25\bmod 30.$$ Then, this means there exists an $n$ such that $$m=30n+25.$$ Substituting this in the first equation involving $x$ gives $$x=101(30n+25)+5,$$ that is $$x=3030n+2530,$$ implying that $x\equiv 2530\bmod 3030.$
