Bases (Maths) question?

This is tricky because you run out of numbers at 9, so you have to substitute letters.

0,1,2,3,4,5,6,7,8,9,A, B, 10, 11 and so on

23 consists of one decimal 12 and a decimal 11, so its 1B

Like in all bases, you have a one's column. In base 12, the column is the twelve's column.

....| 12^3 | 12^2 | 12^1 | 1 |

....| 0 | 0 | 1 | 11 |

From 23 base 10, you can remove one 12, so you put it into the twelve's column. Now you have 11 left over. These go in the one's column.

Since it is base 12, you can possibly have a 1..11 as a value in any column.

Sometimes in base twelve, and eleven is written as a 'E'.

1E or 1 E or | 1 | E | would be the answer. The verticle lines are column delimiters.

23 = 12 + 11

= (1*(12^1))+(11*(12^0))

= 1B (becaz in 12 base 0 to 9 is same thn 10 = A and 11 = B)

so, ans = 1B

use 12 as the base for the exponents

5^1=5

5^2=25

5^3=125

here

12^1=12

12^2=144

6*12^1+1*12^0

61

imagine that 23 represent months.

convert that into year(12)+months.=1 year +11months.

in decimal system first number is 1 and the last is 10.

in duodecimal system numbers consists of 1,2,3 ....8,9,A,B,10,

that means 11 decimal is equal to B duodecimal.

therefore the answer is 1B