QUESTION

CSAT

Hard

Maths

Prelims 2020

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 $D$ ?

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Explanation

As per the given condition in the question, each letter represents a different digit greater than 3. Therefore, the digits can be from 4 to 9.

We have the equation: $A3BC + DE2F = 15902$

Step 1: Unit digit When we add C and F, the sum should be 12 to get a 2 in the unit place of the sum (15902). Thus, the possible pairs for C and F are $(4, 8)$ or $(5, 7)$ .

Step 2: Tens digit We have a carry of 1 from the previous step. The tens digit of the sum is 0, so: $B + 2 = 9$ , which gives $B = 7$ . This eliminates the pair $(5, 7)$ for C and F. Therefore, $C = 4$ and $F = 8$ .

Step 3: Hundreds digit We have a carry of 1 from the tens place. The hundreds digit of the sum is 9, so: $E + 3 = 8$ , which gives $E = 5$ .

Thus, we have found: $B = 7$ , $C = 4$ , $E = 5$ , $F = 8$ . Now, for A and D, the possible pair is $(6, 9)$ . Therefore, the difference between A and D is: $9 - 6 = 3$ .