This task involves finding all three-digit numbers such that the sum of their digits equals 23. Here’s how we can systematically solve the problem:

Explanation:

Let the number be represented as $ABC$, where:

• $A$ is the hundreds digit ($1 \leq A \leq 9$),
• $B$ is the tens digit ($0 \leq B \leq 9$),
• $C$ is the units digit ($0 \leq C \leq 9$).

The condition is: $A + B + C = 23$ Each number satisfying this condition can be written down as a valid solution.

Systematic Method:

1. Start by choosing $A$, the largest digit (since it’s the hundreds place). Then find all combinations of $B$ and $C$ that satisfy $B + C = 23 - A$.
2. Ensure all digits $A, B, C$ are within valid ranges.

Here are some examples to extend the numbers provided:

• For $A = 9$:

  • $B + C = 23 - 9 = 14$:
    • $B = 9, C = 5$: $995$
    • $B = 8, C = 6$: $986$
    • $B = 7, C = 7$: $977$
  • So, numbers include $995, 986, 977$.

• For $A = 8$:

  • $B + C = 23 - 8 = 15$:
    • $B = 9, C = 6$: $896$
    • $B = 8, C = 7$: $887$
  • So, numbers include $896, 887$.

• For $A = 7$:

  • $B + C = 23 - 7 = 16$:
    • $B = 9, C = 7$: $797$
  • So, $797$.

Full Solution:

Using this method, we find all valid three-digit numbers: $599, 689, 995, 986, 977, 896, 887, 797$

Let me know if you would like further details or if you'd like me to solve similar problems! Here are 5 follow-up questions to expand on this topic:

1. How many three-digit numbers can you form where the digits add up to a number other than 23, such as 21 or 24?
2. How would you generalize this method for $n$-digit numbers?
3. Can you write an algorithm to automatically calculate all such combinations?
4. What happens if the problem allows repetition of digits but includes additional constraints (e.g., odd or even sums)?
5. Can this problem be extended to include negative digits (e.g., in modular arithmetic)?

Tip: Organizing your calculations systematically (starting with the largest digits) helps avoid errors and omissions.