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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

5 Answers 5

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$\underbrace{{\rm mod}\,\ \color{#c00}9\!:\ 1 \equiv\! \phantom{a}}_{\large a\ \equiv\ 1\pmod{\! 9}}\!\!\!\!\!\!\overbrace{a\equiv 6\!+\!10k}^{\large a\ \equiv\ 6\pmod{\!10}}\!\!\!\! \equiv -3+k$ $\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 $\rm\:\Rightarrow\ lcm(m,n) = m\,n$.

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    Thank you so much for your kind and clever guides and answers :) – 2017-01-28
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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 $46$.

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    But how to solve the equation you mentioned?! – 2017-01-27
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    So your real problem is that you are working in Chapter 3 without having mastered Chapter 2..... – 2017-01-27
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    @Hamid mod $9$ we have $\color{#0a0}{10\equiv{\bf 1}}\,$ so $\,1\equiv 6+\color{#0a0}{10k}\equiv 6+\color{#0a0}k\,$ so $\,k\equiv 1-6\equiv -5\equiv \color{#c00}4,\,$ same as in my answer. – 2017-01-28
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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, $a\equiv 46\pmod {LCM(10,9)}$

$a\equiv 46\pmod {90}$

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    For this to be useful you should say how you found the solution $46$. Otherwise it looks as if it was pulled out of a hat - like magic. – 2017-01-27
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    in this case, i went through numbers of the form $1+9m$ and $6+10n$ and it was easy to find. – 2017-01-27
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    Yes, I figured that, but you should explain that to your readers in the answer, i.e. show the steps that you took, the specific numbers tested, etc. – 2017-01-27
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    @BillDubuque, please see my edit. – 2017-01-27
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    That's clearer. Such brute-force checking of all possible values is often useful when computing with small moduli. But I would not recommend that one use such methods when one seeks to learn how to do CRT *in general,* since brute-force checking will be impractical except for very small moduli. – 2017-01-28
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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 $46$.

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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}