Find the smallest natural number using Chinese Remainder Theorem Question is find the smallest natural number $x$ such that,
\begin{align}
x &\equiv 1 \pmod 2 \\
x &\equiv 2 \pmod 3 \\
x &\equiv 3 \pmod 4\\
x &\equiv 4 \pmod 5\\
x &\equiv 5 \pmod 6\\
x &\equiv 6 \pmod 7\\
x &\equiv 7 \pmod 8\\
x &\equiv 8 \pmod 9\\
x &\equiv 9 \pmod {10}\\
x &\equiv 10 \pmod {11}\\
x &\equiv 11 \pmod {12}\\
x &\equiv 0 \pmod {13}\\
\end{align}
However when I use the proof of Chinese remainder theorem, I cannot even get over the first step where I must find inverse modulo: $$b_1 \frac{6227020800}{2}\equiv 1 \pmod 2$$ which I presume is $$0\cdot \frac{6227020800}{2} + 2\cdot 1 \equiv 1$$ hence 0? 
I am somewhat confused. A friend did it using another method, but I would like to learn how to use the Chinese remainder theorem.
$$x\equiv a_1 b_1 \frac{M}{m_1} + \cdots + a_k b_k \frac{M}{m_k} \pmod M$$

$2,3,\ldots,12\mid x+1\iff  27720\mid x+1$ since $27720$ is the lcm of the divisors.
So $\, 13n = x  = 27720\,\color{#c00} k - 1\ $ so $\bmod 13\!:\,\ \color{#c00} k\equiv \dfrac{1}{27720}\equiv\dfrac{ -12}4\equiv -3\equiv \color{#c00}{10}$ 
hence $\, x = 22720(\color{#c00}{10}+13m)-1 = 277199 + 13\cdot 27720\,m$ 
