I did not have enough room to show this question so I'll do it here. 742 - 247 =495 next take 495 + 594 =1089... You can use all numbers other than 0.......321 - 123 =198 next take 198 + 891 =1089......Always comes up 1089

999 - 999 = 0 0 + 0 /= 1089 111 - 111 = 0 0 + 0 /= 1089 211 - 112 = 98 98 + 98 /= 1089 Your stated facts are untrue.

LOL..Gonna have to take your word for it my dear friend..LOL :)

This doesn't work always. 741-147=594 842-248=594 321-123=198 The concept is like this. When you reverse, the hundred's and ten's place alone interchange their values contributed to the total. So the difference will be nearest hundred corresponding to the unit's and hundred's place minus the difference itself. Example, 741-147, the difference between 7 and 1 is 6; correspondingly its 600-actual difference, ie, 600-6=594. Similarly works for all other numbers.

OK, here ya go: For any three digit non-repeating number under the rules you have set forth, whenever you subtract one from the other by these rules, the MIDDLE number of the result will ALWAYS be a 9. 321-123, for example: you will end up having to borrow from the '2' in the tens place of the '321', which leaves a '1', for which you must agan borrow from the '3' in the hundreds place to make '11'. Two from eleven is nine. You ALWAYS end up having to borrow from the ten's place. Which means the result for the middle number will ALWAYS be NINE. Now, when you take the final result (198, in this case) you will notice that the first digit and the last digit alre ALWAYS going to add up to total NINE. In this case, 1+8 = 9. When you flip the numbers around and add the two together, you get a curious result of the summation of a series of nines: 198+891: The ones place ALWAYS adds up to NINE. The tens place ALWAYS adds up to 18. And when you carry the "1" from "18" into the hundreds place, the result will ALWAYS be TEN. As you pointed out, this is ONLY true for a three digit number with NO REPEATING DIGITS. It is a quirk of base ten mathematics, just as even numbers are always divisible by two, numbers ending in "0" are always divisible by five and ten (except for the number "0" itself), and any number that you can add all the digits together and get a result which is divisible by 9, then the original number must also be divisible my nine as well. (For example, 134,577: Added together, these digits equal 27. Add 2 and 7 together and you get 9. Therefore the original number; 134,577 must be divisible by 9.) And furthermore, if you pick a smaller number, like 123, and from it subtract 321 by your rules, you will get -198. (-198) + (-891) = -1089. So the results are still the same. Does this help?

My digits are a,b, and c with a>b>c abc = 100a +10b +c reverse them: cba = 100c + 10b + a subtract: 100(a-c) + (c-a) = 99(a-c) = 90(a-c) + 9(a-c) Rule for multiplcation of a digit d>1 by 9: ... 9(d) = 10(d-1) + (10-d) So If a-c is greater than 1, I now have 100(a-c-1) + 10(10-a+c) + 10(a-c-1) + 10-a+c = 100(a-c-1) + 90 + (10-a+c) now reverse the number: = 100(10-a+c) + 90 + (a-c-1) and add: = 100(9) + 180 + 9 = 1089 I made an assumption that d>1 so it only works as long as the first digit "a" is at least two greater than the last "c" and it doesn't matter what the middle number is. Example: 391 -> 193 -> 198 -> 891 -> 1089 (edited to fix typos)