A three-digit number is such that the sum of its digits is 11. If you reverse its digits, the new number is 297 more than the original number. Find the original number.

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.