Students will see the 1089 trick. Then ask students if they can prove the trick works for all non-palindromic three-digit numbers. If they get stuck, offer the hint to represent their first three-digit number as \(100a + 10b + c.\) Here's the proof: The first three-digit number can be represented as \(100a + 10b + c.\) Its reversal is \(100c + 10b + a.\) The difference between the two is \(99a - 99c.\) This is a multiple of \(99.\) The difference can't be zero, because our first number was non-palindromic. Our possible differences are thus the non-zero three-digit multiples of \(99:\) \(099,\) \(198,\) \(297,\) ... \(981.\) Taking any of these, and adding them to their reverse, we'll always get \(9\) units, \(18\) tens, and \(9\) hundreds. More concisely, \(9 + 180 + 900 = 1089.\)

Conclude by giving your students this challenge:

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