-2
6000.
The number is 23.
3201
the sum of my digits is 6? answer=60
Add the digits together. The sum of the digits of 23 is 5.
A number is divisible by 9 if the sum of its digits is divisible by 9. Thus, the number 99999 is divisible by 9 (the sum of its digits is 45, and it is clearly 11111 * 9), and because 99999 is the largest 5 digit number, it must be the largest 5 digit multiple of 9. 99999 is the greatest no divisible by 9.if u divide it u`ll get 11111.
ten ones with an infinite number of zeroes.
The number is 23.
-9
256
3201
47 + 74 = 121
41
the sum of my digits is 6? answer=60
Add the digits together. The sum of the digits of 23 is 5.
The number is 45. The sum of its digits i.e. 4+5=9 Five times the sum of its digits i.e. 5 times 9 which is 45
Nine hundred thirty (930) is the highest three digit number whose digits have a sum of 12. 9+3+0=12. Any numbers higher than that will give you an sum greater than 12.
(9,999) + (9,999) = 19,998 So it looks like the sum of two 4-digit numbers can't have more than 5 digits.