UPSC CSE Prelims 2024
The problem states that the sum of two-digit numbers AB and CD results in the three-digit number 1CE, where A, B, C, D, E denote distinct digits.
We can express these numbers using their place values:
The given equation is (10A + B) + (10C + D) = 100 + 10C + E.
Let's analyze the sum based on place values:
If B + D = E, then the sum of the tens digits (A + C) must generate a carry-over of 1 to the hundreds place, and the tens digit of the sum must be C. This implies that A + C = 10 + C.
A + C = 10 + C
A = 10
This result is not possible because A must be a single digit (0-9). Furthermore, as the tens digit of a two-digit number AB, A cannot be 0. Therefore, this case is invalid.
If B + D = 10 + E, then a carry-over of 1 is added to the sum of the tens digits (A + C).
For the sum to be 1CE, the sum of the tens digits (A + C + 1, including the carry-over) must result in 10 + C (where 'C' is the tens digit of 1CE and '1' is the hundreds digit).
A + C + 1 = 10 + C
A + 1 = 10
A = 9
This value of A = 9 is a valid single digit. Given that A, B, C, D, E must be distinct, a valid set of digits can be found (e.g., A=9, C=1, B=2, D=8, E=0, where 92 + 18 = 110, which matches the 1CE format). Hence, A = 9 is the uniquely determined value.
Options 2 (8), 3 (7), and 4 (Cannot be determined due to insufficient data) are incorrect.