Find the least significant digit of $2^{3A}$, where $A=10^{100}$ Can anyone please tell me - 


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*What is the least significant digit in $2^{3A}$?

*How do we define least significant digit?

 A: The least significant digit of a positive integer, when the integer is given in decimal form, is the units digit, the digit furthest to the right. So the least significant digit of $78675$ is $5$. The least significant digit of $3341116$ is $6$. The least significant digit of $17$ is $7$. The least significant digit of $17450$ is $0$. 
Another way of putting it is that the least significant digit of the positive integer $n$ is the remainder when you divide $n$ by $10$.
Now that you know what least significant digit means, you can attack the problem. The important thing is that "last digits" cycle. For example, successive powers of $8$ end in $8,4,2,6,8,4,2,6,\dots$.  
A: In powers of $2$ you can see a pattern:
$$2^1=2$$ $$2^2=4$$ $$2^3=8$$ $$2^4=16$$ $$2^5=32$$ $$2^6=64$$
So the last digits of powers of 2 are always $2,4,8,6$ in that order. 
For example: 
$$2^{15}=....8$$ because $15 \mod 4=3$ and the third ending is $8$.
Now you have to calculate $3*10^{100} \mod 4$.
$10^{100}$ means that you have $1....00000$ a $1$ and after that hundred $0$.
So $10^{100}$ is divisible by $4$ because it is the multiply of $100$ so $3*10^{100}$ is also divisibly by $4$. 
That's why $3*10^{100} \mod 4=0$.
$0$ means that your number's last digit is the 4th in the pattern, so $6$.
Edit: In that case if your question wasn't $2^{3A}$ but ${2^3}^A$ than it's also quiet simple. ${2^3}^A=8^A$ and in the power of $8$ you can also see a pattern $(8,4,2,6)$ and you can use the method that I used previously. 
