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

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

Step-by-Step calculation
Reverse − 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/Alternative
Reversal 352; 352 − 253 = 99. Digit sum 2 + 5 + 3 = 10; middle digit 5 equals 2 + 3.

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

Final Answer
253