A 3-digit number has digit sum 10. The middle digit equals the sum of the other two, and reversing the digits increases the number by 99. What is the number?

Problem restatementLet the number be 100a + 10b + c with a + b + c = 10, b = a + c, and reverse − original = 99.

Concept/ApproachTranslate the conditions into equations and solve for digits a, b, c.

Step-by-Step calculationReverse − original = (100c + 10b + a) − (100a + 10b + c) = 99(c − a) = 99Therefore, c − a = 1 ⇒ c = a + 1b = a + c = a + (a + 1) = 2a + 1Sum: a + b + c = a + (2a + 1) + (a + 1) = 4a + 2 = 10 ⇒ a = 2Then c = 3, b = 5 ⇒ Number = 253

Verification/AlternativeReversal 352; 352 − 253 = 99. Digit sum 2 + 5 + 3 = 10; middle digit 5 equals 2 + 3.

Common pitfallsSetting reverse − original = 0 or 990 by misplacing place values; remember only hundreds and units exchange roles.

Final Answer253