How to calculate last four digits of $2^{2017}$? can u suggest me any short trick of solving these kind of problems. How to find last four digits of any number raised to some power ?
 A: Since $10^4 = 2^4 \times 5^4$, my first reaction is one should use Chinese remainder theorem.
For $2^4$, it is clear $2^{2017} \equiv 0 \pmod {2^4}$.
For $5^4$, we use following theorem

For $a,b \in \mathbb{Z}_{+}$ such that $\gcd(a,b) = 1$, we have $$a^{\varphi(b)} \equiv 1 \pmod b$$ where $\varphi(\cdot)$ is the Euler totient function.

For $a = 2$ and $b = 5^4$, we have $\varphi(5^4) = 4(5)^3 = 500$. This leads to
$$2^{2017} = 2^{4(500)+17} \equiv 2^{17} \equiv 65536\times 2 \equiv 1072\pmod {5^4}$$
It turns out we are lucky. Since $1072 \equiv 0\pmod {2^4}$ already, we don't need CRT to conclude $2^{2017} \equiv 1072 \pmod {10^4}$. 
As a result, the last $4$ digits of $2^{2017}$ is $1072$.
A: Hint: You want to find $2^{2017} \bmod 10000$. To do that, it suffices to find it $\bmod 16$ and $\bmod 625$. Finding it $\bmod 16$ is easy; to find it $\bmod 625$, use Euler's theorem that, if $\gcd(a,n)=1$, then
$$a^{\varphi(n)} \equiv 1\bmod n,$$
where $\varphi(n)$ is the Euler totient function. Can this reduce $2^{2017}$ into something more manageable?
A: Like  How to find last two digits of $2^{2016}$,
I shall find $2^{2017-4}\pmod{10^4/2^4}$
As $2^2=5-1$
$\displaystyle2^{2012}=(5-1)^{1006}\equiv1-\binom{1006}15+\binom{1006}25^2-\binom{1006}35^3\pmod{5^4}$
Now $\displaystyle1006\equiv6\pmod{5^3}\implies\binom{1006}15\equiv5\cdot6\pmod{5^4}$
$\displaystyle\binom{1006}2\equiv15\pmod{5^2}\implies\binom{1006}25^2\equiv5^2\cdot15\pmod{5^4}$
and $\displaystyle\binom{1006}3\equiv0\pmod5\implies\binom{1006}35^3\equiv0\pmod{5^4}$
$\displaystyle\implies2^{2012+1}\equiv2(1-5\cdot6+5^2\cdot15)\equiv346\cdot2\pmod{5^4}$
Multiply out by $2^4$
A: This is a particular case of computing $a^k$ modulo $n$. Here $n=10000$.
The systematic method for doing this is "repeated squaring". As
$2^{2017}=2(2^{1013})^2$, first work out $2^{1013}$ modulo $10000$,
that is its last four digits. If these last four digits form the number
$a$ then $2^{2017}\equiv 2a^2\pmod{10000}$, so find the last four digits
of $2a^2$.
Of course to find the last four digits of $2^{1013}$ one proceeds recursively: $2^{1013}=2(2^{506})^2$, so find the last four digits of $2^{506}$ etc.
A: To compute this in terms of powers of $5$ and reduce modulo $5^4$ you can compute $$2^{2017}=2\cdot (1-5)^{1008}\equiv 2\cdot\left(1-5\binom{1008}1+25\binom {1008}2-125\binom {1008}3\right)\equiv$$and because of the factors $5^r$ you can reduce the binomials modulo $125, 25, 5$ respectively so $$\equiv2\cdot\left(1-5\cdot 8+25\cdot3-125\cdot 1\right)\equiv $$(which I did in my head using the indicated short cut) $$\equiv2\cdot-89\equiv 447$$
And $447$ leaves remainder $-1$ when divided by $16$ while $625$ leaves remainder $1$, so $447+625=1072$ will match residues modulo both $16$ and $125$.
A: As others have noted, you want $2^{2017} $ mod $10000$.
Using the Carmichael function of $10000=2^45^4$, which is $5^34=500$,
we have $2^{2017}\equiv{2^{17}}=131072\equiv1072\bmod10000$.
