Question

The sum of a two-digit number and the number obtained by reversing its digits is 99. If the digits differ by 3, what is the number?

• A. 36
• B. 63
• C. 54
• D. 45

Hint:

Use digit sum and difference to form equations for two-digit number problems.

Answer 1

The Correct Option is A

- Step 1: Let the number be $10a + b$, reversed number = $10b + a$. Given: $(10a + b) + (10b + a) = 99 \implies 11a + 11b = 99 \implies a + b = 9$.
- Step 2: Given $|a - b| = 3$.
- Step 3: Solve: $a + b = 9$, $a - b = 3$ or $b - a = 3$. Case 1: $a - b = 3 \implies a = b + 3 \implies b + 3 + b = 9 \implies 2b = 6 \implies b = 3, a = 6$. Number = 63.
- Step 4: Case 2: $b - a = 3 \implies b = a + 3 \implies a + a + 3 = 9 \implies 2a = 6 \implies a = 3, b = 6$. Number = 36.
- Step 5: Verify: For 36, sum = $36 + 63 = 99$, digits differ by $|3 - 6| = 3$. For 63, same. Option (1) is 36.
- Step 6: Option (1) is correct.