Q48. If the sum of the two-digit numbers AB and CD is the three-digit number 1CE, where the letters A, B, C, D, E denote distinct digits, then what is the value of A?

The problem states that (10A + B) + (10C + D) = 100 + 10C + E, where A, B, C, D, E are distinct digits. From the units column, B + D = E or B + D = 10 + E (if there's a carry-over of 1 to the tens column).

From the tens column, A + C + (carry-over from units) = 10 + C (since the sum results in a three-digit number 1CE, implying a carry-over of 1 to the hundreds column). Let's analyze the tens column: A + C + (carry-over from units) = 10 + C.

If there is no carry-over from the units column (B+D=E), then A + C = 10 + C, which simplifies to A = 10. This is impossible as A must be a single digit (0-9). Therefore, there must be a carry-over of 1 from the units column (B+D = 10+E).

Now, with a carry-over of 1 from the units column, the tens column equation becomes: A + C + 1 = 10 + C. Subtracting C from both sides: A + 1 = 10. Solving for A: A = 9. Since A, B, C, D, E must be distinct digits, and A=9, we can find a valid combination (e.g., 98 + 17 = 115, where A=9, B=8, C=1, D=7, E=5 are all distinct).

Thus, A is uniquely determined as 9. This question tests logical reasoning and number properties.