To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$.

Question

To Find $a$ such that $2^{1990} \equiv a\pmod {1990}$.

$1990 = 2 \times 5 \times 199$. Now $a \equiv 0 \pmod {2}$, $a \equiv 4 \pmod{5}$ and $a \equiv 29 \pmod{199}$. Taking first two together we get $a \equiv 4 \pmod {10}$.

$199x \equiv 1\pmod {10}$ has a solution $x =-1$ and $10x \equiv 1\pmod{199}$ has a solution $x = 20$.

Thus using Chinese Remainder Theorem we have,

$a \equiv 4(-1)199 + 29(20)10 \equiv 5004 \equiv 1024\pmod {1990}$.

Thus our $a = 1024$.

Is the solution correct? Is there any better proof?

Answer 1

Checking your answer of $a=1024=2^{10}$: $$\begin{align}2^{1990}&\equiv 0\equiv 2^{10}&\pmod 2\\ 2^{1990}&\equiv \left(2^{4}\right)^{495}2^{10}\equiv 2^{10}&\pmod{5}\\ 2^{1990}&\equiv \left(2^{198}\right)^{10}2^{10}\equiv 2^{10}&\pmod{199} \end{align}$$

After your edits, your answer is correct.

Answer 2

Yes, it is correct. More simply, by little Fermat $\,2^4\equiv 1\pmod 5\,$ and $\,2^{198}\equiv 1\pmod{199}\,$ so $\,4,198\mid 1980\,\Rightarrow\,2^{1980}\equiv 1\,\Rightarrow\,2^{1990}\equiv 2^{10}\,$ mod $5$ and mod $199$. Finally, combining these we conclude that $\,2^{1990}-2^{10}\,$ is divisible by $\,2,5,199,\,$ so also by their lcm = product $=1990$.