# Q. 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?

(a) 1

(b) 2

(c) 3

(d) 4

Correct Answer: (c) 3

#### Question from UPSC Prelims 2020 CSAT Paper

**Explanation : **

## A3BC + DE2F = 15902

As per the given condition in the question, each letter represents a different digit greater than 3.

So we can replace the letters with 4, 5, 6, 7, 8, or 9.

A3BC + DE2F = 15902**Step 1: Unit digit**

If we add C & F, then we should get 12. Only then can we get 2 at the unit place in the sum (15902).

So, C, F can be (4, 8) or (5, 7)**Step 2: Tens digit**

We got a carry of 1 from 12. Now, we know that the tens digit of the sum, 15902 is 0.

So, B + 2 = 9

Or B = 7

Hence, C, F cannot be (5, 7). They must be (4, 8).**Step 3: Hundreds digit**

We got a carry of 1 from 10. Now, we know that the hundreds digit of the sum, 15902 is 9.

So, E + 3 = 8

Or E = 8 – 3 = 5

Hence, we found that B =7, C = 4/8, E = 5 and F = 4/8

So, A/D = 6/9**So, difference between A and D = 9 – 6 = 3**