Let A3BC and DE2F be four-digit numbers where each letter represents a different digit greater than 3, If the sum of the numbers is 15902, then what is the difference between the values of A and D?

The correct option is 3.

Step-by-step explanation:

We are given two four-digit numbers $A3BC$ and $DE2F$.

Each letter represents a distinct digit strictly greater than 3. Therefore, the six variables $\{A,B,C,D,E,F\}$ must be a permutation of the six digits:

$\{4,5,6,7,8,9\}$

The addition problem is set up as follows:

$A3BC+DE2F=15902$

Let us analyze the addition column-by-column, starting from the units place:

1. Units Column:
We have $C+F$ ending in 2.
Since every digit is at least 4, the minimum possible sum of two distinct digits is $4+5=9$.
Therefore, $C+F$ must equal 12:

$C+F=12$

This gives a carry of 1 to the tens column.

2. Tens Column:
Including the carry of 1 from the units place, the sum in the tens column ends in 0:
$1+B+2=10$

$B+3=10\Rightarrow B=7$

This gives a carry of 1 to the hundreds column.

3. Hundreds Column:
Including the carry of 1 from the tens place, the sum in the hundreds column is 9:
$1+3+E=9$

$4+E=9\Rightarrow E=5$

Since the sum is 9, there is no carry (carry of 0) to the thousands column.

4. Thousands Column:
With no carry from the hundreds column, the sum in the thousands column gives 15:
$A+D=15$

5. Determining the digits for A and D:
We have already determined that $B=7$ and $E=5$.
The remaining available digits from $\{4,5,6,7,8,9\}$ are $\{4,6,8,9\}$.
From Step 1, $C+F=12$. The only pair from $\{4,6,8,9\}$ that adds up to 12 is $\{4,8\}$.
Thus, the remaining digits for $A$ and $D$ must be $\{6,9\}$, which satisfy:

$6+9=15$

6. Calculating the difference:
The difference between the values of A and D is:

$|A-D|=|9-6|=3$

Thus, the difference between the values of A and D is 3.