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.
There are 20.
A) If a number has two digits, then the sum of its digits is less than the value of the original two-digit number.
'17' is a prime number. The sum of the digits is 1 + 7 = 8 > 7.
12
I am 87 .
The sum of all the digits of all the positive integers that are less than 100 is 4,950.
10
77
81
There are 20.
Mmm...... i think 31
112
22, 2+2= 4 and half of 22 is 11 which is odd
i am less than 100. the sum of my digits is 4 half of me is an odd number
24.
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.
31