How can we count in base eleven? We work in base 10. We know base 3 would count 1, 2, 10, 11, 12, 20. How would we count in base 11?
 A: You would have to add a symbol for $10$, such as $A$. Then it would be
$$\begin{array}{|c|c|c|c|c|c|c|}
\hline 
\text{Base 10} & 1 & 2 & 3 & \dots & 9 & 10 & 11 \\ \hline
\text{Base 11} & 1 & 2 & 3 & \dots & 9 & A & 10 \\ \hline
\end{array}$$
A: Typically one would write a special symbol that represents ten, say, $A$. Then, the first few natural numbers in base eleven would be written as
$$0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 20, 21$$
and so on. In this system, 


*

*the quantity one hundred would be written as $91$, 

*the quantity one hundred and nine would be written as $9A$, and 

*the quantity one hundred ten would be written as $A0$, 

*the quantity one hundred twenty would be written as $AA$, and 

*the quantity one hundred twenty-one would be written as $100$.
A: Use the symbol "A" to represent 10.  So you get $1,2,3,4,5,6,7,8,9,A,10,11,12,13,14,15,16,17,18,19,1A,20,\dots$
A: base $2$  we only have two digits $0,1$
base $10$ we have $10$ digits $0,1,2,3,4,5,6,7,8,9$
base $11$ we would have $11$ digits $0,1,2,3,4,5,6,7,8,9,\beta$
where $\beta$ is a new digit
A: You know how counting in Hexadecimal is done right? It could be done similarly with just the A included.Leave out the other alphabets.
