Fun with modulo #1

Sure sir, its here.

Since, we have to calculate the remainder of $\large\ \frac { { 2 }^{ 1990 } }{ 1990 }$ , it is equivalent to calculate the remainder of $\large \frac { { 2 }^{ 1989 } }{ 995 }$ .

Now observe that $995=199 \times 5$ .

First let us calculate $\large\ { 2 }^{ 1989 }\pmod{ 5 }$ .

As ${ 2 }^{ 4 }\equiv1 \pmod{5}$ , ${ 2 }^{ 1989 }\equiv2 \pmod{5}$ .

Again , observe that ${ 2 }^{ 198 }\equiv1 \pmod{199}$ ,

So, ${ 2 }^{ 1989 }\equiv114 \pmod{199}$ .

Now apply C.R.T to get

$199x + 114 = 5k + 2$ for some $x$ and $k$ .

Interospecting we get $512$ as a least common value of these equations.

Now, as we eliminated $2$ in the beginning , we multiply $512$ with $2$ to get $\boxed{1024}$ , the desired answer.