We're not quite sure what the "2 digit numbers of 100" are.
Do you mean all the "2-digit numbers you say as you count to 100" ?
If that's it, then here's how to sum them:
There are 100 of them, minus (1 through 9, and 100) = 100 - 10 numbers = 90 numbers.
Make pairs of them . . .
10 + 99 = 109
11 + 98 = 109
12 + 97 = 109
13 + 96 = 109
14 + 95 = 109
.
.
etc.
There are 45 pairs, and every pair adds to 109.
So the sum of all 90 numbers is (45 x 109) = 4,905.
105 is the sum
46
Seven of them.
The sum of all palindromic numbers from 1001 to 9999 is 495000.
There are nine of them. They are:108117126135144153162171180
111 and 201
105 is the sum
105 is the sum
Seven of them.
46
The sum of all the integers between 100 and 999 is equal to 494450 - which also happens to be the answer to the sum 899 x 500 (which refers to the number of numbers multiplied by the mean number + 0.5).
There are 8 numbers between 100 and 200 (inclusive) that have a digit sum of 12, namely: 129, 138, 147, 156, 165, 174, 183, 192.
There are nine of them. They are:108117126135144153162171180
The sum of all palindromic numbers from 1001 to 9999 is 495000.
what is the least possible sum of two 4-digit numbers?what is the least possible sum of two 4-digit numbers?
The sum of the all prime numbers from 1 to 100 is 1,161
There are no 3-digit whole numbers whose digits sum to 3. The smallest 3-digit number is 100, and the largest is 999, but in neither case is the sum of the digits equal to 3.