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 correct answer is 9.

Let us represent the addition of the two-digit numbers in column form:

A B
+ C D
------
1 C E

Here, A, B, C, D, and E represent distinct single digits from 0 to 9. Since AB and CD are two-digit numbers, their leading digits A and C cannot be 0.

Let us analyze the addition column by column, starting from the units (ones) place:

In the units column, we add B and D to get a units digit of E. Since the sum of two single digits can be at most 9 + 8 = 17, the addition can produce a carry-over to the tens place. Let this carry-over be represented by $c$, where $c$ can be either 0 or 1.

Now, let us examine the tens column. In the tens column, we add the digits A and C along with the carry-over $c$ from the units column. The result of this addition gives a tens digit of C and a hundreds digit of 1. Mathematically, this can be written as:

$A+C+c=10+C$

We can simplify this equation by subtracting C from both sides:

$A+c=10$

Since A is a single-digit number, its maximum possible value is 9. For the sum of A and $c$ to equal 10, the carry-over $c$ cannot be 0. Thus, the carry-over $c$ must be 1.

Substituting $c=1$ into the equation:

$A+1=10$

Solving for A gives:

$A=9$

Therefore, the value of A is 9.