5050 ie 101 x 50
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.
5050, according to the program I quickly whipped up.
The sum of the all prime numbers from 1 to 100 is 1,161
101
100
5050 ie 101 x 50
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.
5050, according to the program I quickly whipped up.
1 + 1 = 2 The sum of the digits is therefore 2.
The sum of the all prime numbers from 1 to 100 is 1,161
The sum of all the the integers between 1 and 2008 (2 through 2,007) is 2,017,036.
The sum of all the odd numbers from 1 through 100 is 10,000
101
There are only 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. So the answer to the question asked is 45.The sum of all of the integers less than 100 and greater than 0 can be found in the following way:(99 + 1) + (98 + 2) + (97 + 3) + ... + (51 + 49) + 50 = 100 + 100 + 100 + ... + 100 + 50 = 49 X 100 + 50 = 4900 + 50 = 4850.
I am not at all sure about "present". If you mean percent, the answer depends on which 100 numbers. Between 900 and 999, for example, the answer is 0%. For the FIRST 100 numbers: 1 to 100, the answer is 6%.
5, 14, 23, 32, 41, 50