Think of a 3-digit number such that the first and the last digit differ by 2 or more.

E.g. 246

Reverse the digits in the 3-digit number Subtract the smaller 3-digit number from

the larger one.

Reverse the digit in the answer and add it to the original answer.

Keep the answer to yourself and don’t let others get to know it.

Don’t tell me your answer because I can guess yours.

Why?

The number is 123. Reverse it becomes 321. Then 321 – 123 = 198 Now, reverse 198 to become 891. Then 891 + 198 = 1089.

This is only true to a particular number… How to say this is true in general?

We need the help from ALGEBRA!

In general, the 3-digit number is 100a + 10b + c.

Reverse it to become 100c + 10b + a. Subtract one number from the other

(100a + 10b + c) – (100c +10b + a)= 100a – a + 10b – 10b + c – 100c= 99a – 99c= 99 (a – c)

Since we need to keep the 99 (a – c) to be 3-digit, a – c ≥ 2.

Possible values are 2, 3, … ,9 Possible values for 99 (a – c) are 198,

297, 396, 495, 594, 693, 792, 891.

Check yourself if you have one of those numbers before the addition of its reverse…

Addition of those numbers and its reverse will always be equal to 1089.

198 891 1089

297 792 1089

396 693 1089

495 594 1089

594 495 1089

693 396 1089

792 297 1089

891 198 1089