A three-digit number is such that the sum of its digits

Solution

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.