5050 ie 101 x 50
5,050
1+0 = 1
The answer to the smallest possible value of the sum of all the digits is 1. the number can either be 100 or 1000 - either way the sum is still one.
82
100
1 + 1 = 2 The sum of the digits is therefore 2.
5050 ie 101 x 50
The sum of all the digits of all the positive integers that are less than 100 is 4,950.
5, 14, 23, 32, 41, 50
Digits are from 1 to 9the sum of the 9 greatest digits is = 9+9+9+9+9+9+9+9+9 = 81the sum of the 9 lowest digits is = 1+1+1+1+1+1+1+1+1+1= 9and so :81 + 9 = 90
Since a dozen is 12, then the digits are 1 and 2. To get the sum, you simply add 1 + 2, which = 3.
5,050
1 + 23 + 4 + 5 + 67 = 100 is one way. 1 + 2 + 34 + 56 + 7 = 100 is another. 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 so you need to move two non-adjacent digits that sum to 8 into the tens' places. This is because: Start with a sum of 28 Lose two digits that sum to 8 from the units so that you have 28 - 8 = 20 Gain two digits that in the tens place that sum to 8 so that 20 + 80 = 100.
A dozen is equal to 12. The sum of the digits of 12 are 1 + 2 = 3.
1+0 = 1
The answer to the smallest possible value of the sum of all the digits is 1. the number can either be 100 or 1000 - either way the sum is still one.