A three-digit number is such that the sum of its digits is 12 and when its digits are reversed, the new number is 198 more than the original number. Find the original number.

Let the hundreds, tens and units digits be a, b, c. Original number = 100a + 10b + c Reversed number = 100c + 10b + a Given: a + b + c = 12 …(1) (100c + 10b + a) − (100a + 10b + c) = 198 Simplify: 100c + 10b + a − 100a − 10b − c = 198 99c − 99a = 198 99(c − a) = 198 ⇒ c − a = 2 …(2) From (1): a + b + c = 12 From (2): c = a + 2 Substitute: a + b + (a + 2) = 12 2a + b + 2 = 12 2a + b = 10 …(3) a and c are digits (0–9) and a ≠ 0. Try integer values: From (2), c = a + 2 ≤ 9 ⇒ a ≤ 7 Try a = 3 ⇒ c = 5, then (3): 2×3 + b = 10 ⇒ b = 4 Check sum: 3 + 4 + 5 = 12 (correct) Original number = 345 Reversed = 543 Difference = 543 − 345 = 198 (correct) Answer: The number is 345.