If you're finding the sum of the numbers, all you need is the first and last numbers and the number of numbers.
There are 90 numbers in between 10 and 99, so n = 90.
The first number is 10, so a = 10
The last number is 99, so z = 99.
Then just put your a, n and z values into the formula and simplify.
But if you're finding the sum of all the digits, that's another story entirely...
OK, well each group of 10 numbers will end in 0, 1, 2, 3, ..., 9.
There's 9 groups of these, so there's 9 x 1, 9 x 2, 9 x 3, ..., 9 x 9.
There's also 10 more 1's (because the first digit of the first 10 numbers is 1), 10 more 2's, 10 more 3's ... 10 more 9's.
So that makes 19 x 1 + 19 x 2 + 19 x 3 + ... + 19 x 9, or if you like,
19 x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)
Hope that helped
Hello, not happy jan!
Instead of asking "What's the formula?", we can use a primitive approach.
Imagine adding up the 90 two-digit numbers . . .What is the sum of the digits of all 2-digit numbers from 10 and 99?
. . . . . . . . . . . . .
In the units-column, we have the sum of the digits 0 to 9 (45) ... nine times.
Sum of the units digits: .
In the tens-column, we have: .
. .
Sum of the tens digits: .
Therefore, the sum of the digits is: .