Explanation:

- The two-digit numbers are AB and CD, meaning 10×A + B and 10×C + D.

- Their sum gives a three-digit number 1CE, or 100 + 10×C + E.

- So, (10A+B) + (10C+D) = 100 + 10C + E.

- Simplifying, 10A + B + 10C + D = 100 + 10C + E.

- Cancel 10C both sides: 10A + B + D = 100 + E.

- Since all digits are distinct and AB and CD are both two-digit numbers, A = 1.

- 10A + (B + D) = 100 + E. For two-digit AB, the maximum (when A=9) is 90 + (B + D), let's see if A=9 works.

- If A = 9, then 10A = 90, so B + D = 100 + E - 90 ? B + D = 10 + E.

- B + D is at most 18 (since digits are =9), so E can run from 0–8. All possible.

- For smaller A, the sum doesn’t work for distinct digits as neatly as with A=9.

Option 1: 9 is the correct answer.

- Option 4 is incorrect; there’s enough data.

- Option 2 and 3 do not fit as neatly, nor do they satisfy constraints as obviously as 9.