How many two digit numbers are there in base 5. I really don't understand how to figure this out. I know it's been asked already, but I still don't understand it.
The digits of a base 10 number goes from 0 to 9. The first digit of a two digit number (any base) cannot be zero, otherwise the number will become a 1-digit number.
So in base 10, there are 9 choices for the first digit, and 10 choices for the second, for a total of 9*10 numbers.
For a number in base 5, we have up to 5 different digits, namely 0,1,2,3,4. The digit 5, 6, ... do not exist in a number to base 5.
So the number of choices for the first digit, excluding zero, is 5-1=4. The number of choices for the second digit is 5.
So there is a total of 4*5 =20 two digit base 5 numbers, namely
10, 11, 12, 13, 14, 20, 21, 22, 23, 24, 30, 31.....34, 40, 41, 42, 43, 44.
(Remember digits 5 and above do not exist in base 5, just like we have no digit for 10 and above for base 10).
If this is still not clear, please post.
Mathmate-
Thank you very much, I completely understand your explanation now.
To determine how many two-digit numbers there are in base 5, we need to consider the possible values for each digit. In base 5, we have 5 different digits: 0, 1, 2, 3, and 4.
To find the number of choices for the first digit, we can exclude zero since it would not be the first digit of a two-digit number. Therefore, there are 4 choices for the first digit (1, 2, 3, or 4).
For the second digit, we can choose any digit from 0 to 4 since all digits are allowed in the second position. Again, there are 5 choices for the second digit.
To find the total number of two-digit numbers in base 5, we need to multiply the number of choices for the first digit by the number of choices for the second digit. Therefore, the total number of two-digit numbers in base 5 is 4 (choices for the first digit) multiplied by 5 (choices for the second digit), which equals 20.
So, there are 20 two-digit numbers in base 5.