Prove or disprove $a^{10}=b^{10} \pmod{10\alpha}$ Today I got a question:
Find the remainder when $2^{1990}$ is divided by $1990$
I tried as follows
$199$ is a prime, so by Fermat's theorem
$$2^{199}\equiv 2 \pmod{199}$$
Now I used if $$a\equiv b \pmod{\alpha}$$
Then $$a^{10}\equiv b^{10}\pmod{ 10\alpha}$$
Then$$2^{1990}\equiv 2^{10}\pmod{1990}$$
Which is the answer.
But I am not able to prove the property that I had used. I would be pleased if someone would help me to prove or disprove the property. If I had used a wrong property then please provide other method to find the remainder.  
 A: The claimed property is false. Let's look at the exact instance you attempt to use here, namely $\, a \equiv 2\pmod{\!199}\Rightarrow\, a^{\large 10}\equiv 2^{\large 10}\pmod{\!1990}.\,$ Notice that $ $ odd $\,a:= 201 \equiv 2\,\pmod{\!199}\,$ $\, $ however we have $ $ odd $\,a^{\large 10}\!\neq 2^{\large 10}\!+1990k= $ even, so it fails. Below is one correct proof.
By Fermat $\, 2^{\large 4}\! \equiv 1\pmod{\! 5},\,\  2^{\large 198}\!\equiv 1\pmod{\!199}\ $ so $\ \color{#c00}{2^{\large 396}\equiv \bf 1}\pmod {5\cdot 199}$
Therefore $\,\ {\rm mod}\,\ \color{#0a0}{5\cdot 199\!:\,\ 2^{\large  1998}}\! \equiv  2^{\large 9+5(396)}\! \equiv  2^{\large 9} (\color{#c00}{2^{\large 396}})^{\large 5}\!\equiv 2^{\large 9}{\color{#c00}{\bf 1}}^{\large 5}\!\equiv \color{#0a0}{2^{\large 9}} $
So $\,\ 2^{\large 1990}\!\bmod 1990\, =\, 2\,(\color{#0a0}{2^{\large 1989}\!\bmod\ 5\cdot 199})\, =\, 2(\color{#0a0}{2^{\large 9}}) = 2^{\large 10} =1024$
We used  $\ ca\bmod cn =\, c\,(a\bmod n)\ $ in the prior line. See here for more on that. 
