There are 45 three-digit numbers where the sum of the digits equals 9.
Understanding the Problem
We are looking for three-digit numbers (numbers between 100 and 999) where the individual digits add up to 9. This is a combinatorics problem that requires finding all valid combinations of three digits summing to 9, given that the first digit cannot be zero.
Method to Find the Solution
Here's how we can determine the 45 three-digit numbers:
- First Digit: The first digit can range from 1 to 9. It cannot be 0 because then it would not be a three digit number.
- Remaining Sum: The remaining two digits need to sum up to (9 - first digit).
- Counting Combinations: We need to list the combinations of two digits that add up to the remaining sum.
Example Combinations
Let's explore a few examples:
- If the first digit is 1, then the other two digits must sum to 8: (0,8), (1,7), (2,6), (3,5), (4,4), (5,3), (6,2), (7,1), (8,0) - 9 possibilities
- If the first digit is 2, then the other two digits must sum to 7: (0,7), (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), (7,0) - 8 possibilities
- If the first digit is 3, then the other two digits must sum to 6: (0,6), (1,5), (2,4), (3,3), (4,2), (5,1), (6,0) - 7 possibilities
- And so on...
We can continue this pattern and we will notice the number of possibilities for the last two digits decreases by one for every increment of the first digit.
The Solution
If you count all such combinations, you will find that there are exactly 45 three-digit numbers where the sum of the digits is 9.
- The possibilities are:
- 108, 117, 126, 135, 144, 153, 162, 171, 180
- 207, 216, 225, 234, 243, 252, 261, 270
- 306, 315, 324, 333, 342, 351, 360
- 405, 414, 423, 432, 441, 450
- 504, 513, 522, 531, 540
- 603, 612, 621, 630
- 702, 711, 720
- 801, 810
- 900
Conclusion
The reference confirms that 45 three-digit numbers have a sum of 9.
