The sum of the digits of all two-digit numbers is 810. Here's how we arrive at that answer.
Breaking Down the Problem
We need to find the sum of the digits of each two-digit number (10 through 99) and then add those sums together.
Calculation
Let's analyze the tens and units digits separately:
- Tens Digits: The tens digits are 1, 1, 1, 1... , 2, 2, 2... up to 9, 9, 9,... Each tens digit (1 to 9) appears 10 times. So the sum of the tens digits is (1+2+3+4+5+6+7+8+9) 10 = 45 10 = 450.
- Units Digits: The units digits cycle through 0 to 9 for each tens digit. So, the sum of the units digits is (0+1+2+3+4+5+6+7+8+9) 9 = 45 9 = 405.
Now, let's add the sum of the tens digits and the sum of the units digits: 450 + 405 = 855.
Example Table
| Two-Digit Number | Sum of Digits |
|---|---|
| 10 | 1 + 0 = 1 |
| 11 | 1 + 1 = 2 |
| 12 | 1 + 2 = 3 |
| ... | ... |
| 98 | 9 + 8 = 17 |
| 99 | 9 + 9 = 18 |
Calculating by hand each of these numbers adds to 855.
Considering the Sum of All Two-Digit Numbers
The reference provided the sum of all two-digit numbers, calculated as: (10+99)/2 90 = 4905. This is the sum of the numbers themselves*, not the sum of their individual digits.
Final Answer
The sum of the digits of all two-digit numbers is 855.
