Calculating of least significant digit of an expression I want to calculate te least significant digit (1s place) of following: 
$ 1+2^{1393} + 3^{1393}+4^{1393} $


How we can calculate this? It's very hard for me!


 A: For every number $n\in\mathbb{N}$ we have
$$n^5\equiv n\mod10$$
Using this theorem we have
$$2^{1393}\equiv2\mod10\\
3^{1393}\equiv3\mod10\\
4^{1393}\equiv4\mod10$$
Now we have
$$1+2^{1393}+3^{1393}+4^{1393}\equiv1+2+3+4\equiv10\equiv0\mod10$$
So, the least significant digit is $0$.
A: $$1+2^{1393} + 3^{1393}+4^{1393}$$ is even and $$(1+4^{1393})|5,\,\,\,\,\,(2^{1393} + 3^{1393})|5$$ Hence the last digit is $\color{red}{0}.$
A: $2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32, 2^6 = 64 ...$
So the least significant digit is a periodic functions depending on degree ($2,4,8,6$)
You should find such periods for $3$ and $4$ and subtract integer number of periods from every degree correspondently. Hope this is enough for you to solve the problem.
A: If you would calculate $k^m$ on paper you could start with $k$ and then multiply it by $k$ and so on, $m$ times in total.
Using the decimal system for the last digit $d_0$ this means actually calculating
\begin{align}
d_0^{(1)} &= k \\
d_0^{(i+1)} &= k \cdot d_0^{(i)} \mod 10
\end{align}
For $2^k$ this gives a cycle $2, 4, 8, 6$. 
For $3^k$ this gives a cycle $3, 9, 7, 1$.
For $4^k$ this gives $4, 6, 4, 6$
And 1 stays $1, 1, 1, 1$.
Adding those four terms above modulo 10 gives $0, 0, 0, 4$.
As $m = 1393 \mbox{ mod } 4 = 1$ it is the first of the above: $0$. 
For $m = 1396$ it would be a $4$.
