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

Solution

ATQ, Let the number be 100a + 10b + c, where a is hundreds digit, c is units digit. Given: a + b + c = 11 …(1) Reversed number = 100c + 10b + a Given: (100c + 10b + a) − (100a + 10b + c) = 297 ⇒ 100c + 10b + a − 100a − 10b − c = 297 ⇒ 99c − 99a = 297 ⇒ 99(c − a) = 297 ⇒ c − a = 3 …(2) From (1): a + b + c = 11 From (2): c = a + 3 Substitute: a + b + (a + 3) = 11 2a + b + 3 = 11 2a + b = 8 …(3) Digits 0–9, and a ≠ 0. Try integer values: From (2), c = a + 3 ≤ 9 ⇒ a ≤ 6 Test a = 2 ⇒ c = 5, then (3): 2×2 + b = 8 ⇒ b = 4 Number = 245 Check reversed 542 − 245 = 297 (correct) Sum of digits = 2 + 4 + 5 = 11 (correct). Answer: 245.