2

We work in base 10. We know base 3 would count 1, 2, 10, 11, 12, 20. How would we count in base 11?

pjs36
  • 17,979
LXI-I-XXI3
  • 29

5 Answers5

2

You would have to add a symbol for $10$, such as $A$. Then it would be $$\begin{array}{|c|c|c|c|c|c|c|} \hline \text{Base 10} & 1 & 2 & 3 & \dots & 9 & 10 & 11 \\ \hline \text{Base 11} & 1 & 2 & 3 & \dots & 9 & A & 10 \\ \hline \end{array}$$

Eff
  • 12,989
2

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 $100$.

Ken
  • 3,751
1

Use the symbol "A" to represent 10. So you get $1,2,3,4,5,6,7,8,9,A,10,11,12,13,14,15,16,17,18,19,1A,20,\dots$

Gregory Grant
  • 14,874
1

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

alkabary
  • 6,214
-1

You know how counting in Hexadecimal is done right? It could be done similarly with just the A included.Leave out the other alphabets.

user2277550
  • 2,194