# Puzzle always equals 1089

by William Behringer
(Opelousas, Louisiana USA)

1089 Puzzle

As one who is not and never has been "good" in math there is a question that has ever puzzled me and I have never found anyone who can explain it. Here's hoping that you can explain it in a way I can "get".

Why can one take any three numbers from 1 to 9, ie 357 825 or 912, reverse there sequence, ie 357 would become753. Then subtract the smaller from the larger. When that answer is obtained, reverse that number and add ie 753 minus 357 = 396. Reverse 396 ie 693 and add will always equal 1089. This is true regardless of what the starting numbers were. I can't for the life of me understand this. How can this process result in the same answer "1089" whether you start with 123 or 987? Please explain this to me.

### Comments for Puzzle always equals 1089

 Dec 20, 2012 equals 1089 by: Staff Answer Restatement of the rule presented in the problem statement: 1. Select any three digits from 1 to 9. 2. Form any 3-digit number from the digits you selected. (The first and last digits must have a difference of at least 2.) 3. Reverse the three digits to form a second 3-digit number. 4. Find the difference between the original 3-digit number (in paragraph 2) and the 3-digit number which has been reversed (in paragraph 3). 5. Add to this result (in paragraph 4) the number produced by reversing its digits (in paragraph 3). 6. The sum will always be 1089. This 1089 rule can be shown to be true using ordinary algebra. The same rule can also be applied to 2-digit numbers. For 2-digit number the final sum will always be 99 (rather than 1089). This is a property of 2-digit numbers. For example: Beginning with the number 57 │57 - 75│ = 18 18 + 81 = 99 Beginning with the number 82 │82 - 28│ = 54 54 + 45 = 99 Beginning with the number 91 │91 - 19│ = 72 72 + 27 = 99 The proof for the 2-digit sum of 99 is less involved than the proof for the 3-digit sum of 1089, so that is what I am going to show you. N2-digit number = 2-digit number adigit in 10s place = first digit in a 2-digit number (the digit in the 10’s place) bdigit in 1s place = second digit in a 2-digit number (the digit in the 1’s place) N2-digit number = 10 * adigit in 10s place + bdigit in 1s place N = 10 * a + b N = 10a + b Preversed 2-digit number = 10 * b + a P = 10 * b + a P = 10b + a N - P = (10a + b) - (10b + a) N - P = 10a + b - 10b - a N - P = 10a - 10b + b - a N - P = 10a - 10b + b - a -10 + 10 = 0, so we can write N - P = 10a - 10b + b - a + 10 - 10 N - P = 10a - 10b - 10 + b - a + 10 N - P = 10(a - b - 1) + (10 + b - a) Qreversed 2-digit number of N - P = 10 * (10 + b – a) + (a - b - 1) Q = 10 * (10 + b - a) + (a - b - 1) Q + (N - P) = 10 * (10 + b - a) + (a - b - 1) + 10(a - b - 1) + (10 + b - a) Q + (N - P) = 100 + 10b - 10a) + a - b - 1 + 10a - 10b - 10 + 10 + b - a Q + (N - P) = 100 - 1 - 10 + 10 + 10b - 10b - 10a + 10a + a - a - b + b Q + (N - P) = (100 - 1) + (- 10 + 10) + (10b - 10b) + (- 10a + 10a) + (a - a) + (- b + b) Q + (N - P) = 99 + 0 + 0 + 0 + 0 + 0 Q + (N - P) = 99 Thanks for writing. Staff www.solving-math-problems.com