find the last 4 dgits of $2^{2016}$

Question

I am trying to find the last 4 digits of $2^{2016}$.

Here is where I am up to:

2^2016 = x mod 10000

2^2016 = x mod 625 - I used Eulers function to find that 2^16 = 1 mod 625 = 536

2^2016 = x mod 16

Now I am stuck. The second equation can not use Euler so how do I continue?

Thanks

Thanks for that but how does that help me? Don't I need to somehow use the chinese remaindor theorem to solve the 2 equations? I am just not sure as to how. — qua, Dec 31 '20 at 16:49
You know the remainders modulo $625$ and $16$. They are $536$ (aassuming you did it correctly) and $0$. What's stopping you? — Jyrki Lahtonen, Dec 31 '20 at 17:00
probably brain freeze. Does it make sense to continue as follows? x=0(mod16) x=536(mod625) and then solve for x? — qua, Dec 31 '20 at 17:03
Note $\bmod 4!:\ 2^{\large 2}\equiv 0\Rightarrow 2^N\equiv 2^{N\large \bmod!2},$ by modular order reduction. Now you can apply CRT, but it's easier to use the mod distributive law, e.g. see here, and here and here. $\ \ $ — Bill Dubuque, Dec 31 '20 at 17:49

score -1 · Answer 1 · answered Dec 31 '20 at 16:46

-1

$2^{2016} = 16 \times 2^{2012} \equiv 0 \mod{16}$.

answered Dec 31 '20 at 16:46

Karan Elangovan

1 Answers1