What digit appears in unit place when $2^{320}$ is multiplied out  Is there a way to answer the following preferably without a calculator 


What digit appears in unit place when $2^{320}$ is multiplied out ? a)$0$ b)$2$ c)$4$ d)$6$ e)$8$
     ---- Ans(d)


 A: Quickly look at the last digit of $2^n$, for $n=1$, $2$, $3$, and so on. So we keep multiplying by $2$.  For determining the last digit of $2^{n+1}$, only the last digit of $2^n$ matters.
We get $2$, $4$, $8$, $6$, $2$ and the pattern starts all over again. The pattern is periodic with period $4$. So at $n$ a multiple of $4$, we get a $6$, and $320$ is a multiple of $4$.
Remark: If we wanted the last digit of $2^{999}$, note that $996$ is a multiple of $4$. So at $996$ we get a $6$. Now count forward: $2$, $4$, $8$: the answer is $8$. Or else $1000$ is a multiple of $4$. Go backwards one step from $6$: the last digit is $8$.
A: $\rm mod\ 5\!:\, \color{#0A0}{2^4}\equiv \color{#C00}1\Rightarrow2^{320}\equiv (\color{#0A0}{2^4})^{80}\equiv \color{#C00}1^{80}\!\equiv \color{#C00}1.\,$ The only choice $\:\equiv \color{#C00}1\!\pmod 5\:$ is $6,\: $ in d).
A: As the ending digits for powers of two are
$${2}^{0} \mapsto 1$$
$${2}^{1} \mapsto 2$$
$${2}^{2} \mapsto 4$$
$${2}^{3} \mapsto8$$
$${2}^{4} \mapsto6$$
$${2}^{5} \mapsto2$$
You only have to do ${2}^{320 \mod(4)}$ to get the ending digit.
A: Since $2^{320}$ is a square $2^{320}=(2^{150})^2$ and squares last digit is always $0,1,4,9,6,5$, two of them, $2$ and $8$, are ruled out. $0$ as well, since you need a $10$ for that. Write it as $(2^{75})^4$, and check that $4$th powers end with $0,1,6,1,6,5$ to get the remaining as result.
