Using Chinese Remainder Theorem find the remainder of $a$ $\pmod{ 90 }$ given that: $a\equiv 6\pmod {10}$ and $a\equiv 1\pmod 9$ 
Using Chinese Remainder Theorem find the remainder of $a$ $\pmod{ 90 }$ given that: $a\equiv 6\pmod {10}$ and $a\equiv 1\pmod 9$

I studied CRT many times but don't get it so as to solve the above simple problem! I don't know how to apply CRT for it...
Please help
 A: The first congruence says $a=6+10k$ for some $k$.  Put it in the second congruence to get $6+10k \equiv 1 (\bmod{9}).$  Solve for $k$.  Plug in for $a$.  You should get $46$.
A: Not a direct answer to your question. there is an easier solution if we can observe the following.
$a\equiv 6\equiv 46\pmod {10}$
The above step was written after going through the numbers $6,16,26,36,46$ because $46$ can be expressed as $9k +1$ (where k is an integer) and therefore, the next expression can be also made in the same format as provided below.
$a\equiv 1\equiv 46\pmod {9}$
Hence,
$a\equiv 46\pmod {LCM(10,9)}$
$a\equiv 46\pmod {90}$
A: To apply the CRT in a generalisable way we need to find:
$x\equiv 1 \bmod 9;\equiv 0\bmod 10$ which is easy as $x=10$ will do.
$y\equiv 0\bmod 9; \equiv 1\bmod 10$ and $y=81$ will do (or $y=-9$)
Now notice that if we take $px+qy$ modulo $9$ we get $x$, and modulo $10$ we get $y$.
So the answer would be $1\times 10+6\times 81\bmod 90$ which comes out at $46$.
A: The solution follows  from Bézout's identity, from which results an explicit formula  for the inverse isomorphism in the Chinese remainder theorem:
\begin{align}
\varphi\colon\mathbf Z/mn\mathbf Z&\longrightarrow \mathbf Z/m\mathbf Z\times\mathbf Z/n\mathbf Z, \\
x\bmod mn&\longmapsto(x\bmod m,x\bmod n).
\end{align}
for $m,n$ coprime. If $um + vn=1$ is a Bézout's relation between $m$ and $n$, this inverse isomorphism is given by
\begin{align}
\varphi^{-1}\colon \mathbf Z/m\mathbf Z\times\mathbf Z/n\mathbf Z&\longrightarrow \mathbf Z/mn\mathbf Z,\\
(x\bmod m,y\bmod n)&\longmapsto yum+xvn\bmod mn.
\end{align}
A: $\underbrace{{\rm mod}\,\ \color{#c00}9\!:\ 1 \equiv\! \phantom{a}}_{\large a\ \equiv\ 1\pmod{\! 9}}\!\!\!\!\!\!\overbrace{a\equiv 6\!+\!10k}^{\large a\ \equiv\ 6\pmod{\!10}}\!\!\!\! \equiv -3+k$ $\iff \color{#c00}{k\equiv 4} $ $ \iff a=6+10(\overbrace{\color{#c00}{4+9j}}^{\large\color{#c00} k}) = 46+90j$

Remark $\ $ The same elimination method used above works generally. It leads to a simple form of CRT = Chinese Remainder Theorem  that I call Easy CRT. Below is one simple way to present it. You can find many example applications in prior posts here.
Compare the above to the formula below where $\rm {\rm mod}\ 9\!:\, \dfrac{b-a}{m} = \dfrac{1-6}{10}\equiv \color{#c00}{\dfrac{4}1\equiv k}$
Theorem $ $  (Easy CRT) $\rm\ \ $ If $\rm\ m,\:n\:$ are coprime integers then 
$$\rm \begin{eqnarray}\rm x&\equiv&\rm a\,\ (mod\ m) \\
\rm x&\equiv&\rm b\,\ (mod\ n)\end{eqnarray} \ \iff\ \ x\, \equiv\, a + m\ \bigg[\frac{b-a}{\color{#c00}m}\ mod\ n\:\bigg]\ \ (mod\ m\:n)$$
Proof $\rm\,\ m,n\,$ coprime $\:\rm\Rightarrow\, \color{#c00}{{\large\frac{1}m} = m^{-1}}\!\pmod n\, $ exists, by Bezout or Euler's $\phi$ Theorem. 
$\rm\ (\Leftarrow)\ \ \ mod\ m:\,\ x \equiv a + m\left[\cdots\right] \equiv a,\ $ and $\rm\ mod\ n\!\!:\ x \equiv a + m\,\color{#c00}{\large\frac{1}m}\,(b-a) \equiv b$
$\rm (\Rightarrow)\ \ $ The solution is unique $\rm\, (mod\ m\,n)\, $ since if $\rm\ x',\,x\ $ are solutions then $\rm\ x'\equiv x\ $ mod $\rm\:m,n\:$ hence $\rm\ m,n\mid  x'-x\ \Rightarrow\ m\,n\mid x'-x\  $ since $\rm\ m,\,n\:$ coprime $\rm\:\Rightarrow\ lcm(m,n) = m\,n$.
