156953

Alison picks out a random three-digit number, which could be written as xyz. xyz can be expressed as 100x+10y+z. Then she gets the reverse of xyz, which can be written as 100z+10y+x and takes that away from 100x+10y+z. This left her with 99x-99z now automatically we know that the middle number is 9 since x>z (or else she wouldn’t have been able to subtract 100x+10y+z from 100z+10y+x) and the middle number from xyz would be the same when reversed. We also know that 99x-99z is divisible by 9 because it could be written as 9(11x-11z) so we know that the last digit and the first digit add up to 9 because the rule for all numbers divisible by nine the sum of the digits are nine. Now this proves that when our integer 99x-99z is added to its inverse the last digit is 9. Now the second last digit is 8 because 9+9=18. Now the first and last digits are 1 and 0 because of the same proof we have for the last digit but the add one because you have to carry on the 1 from the tens column and we know that 9+1=10 so the first two digits are 1 and 0. This proves that only if x≠z then when you subtract a three-digit number’s inverse from itself and the add the inverse of that to the answer when you subtract the original three-digit number’s inverse from itself, your answer will be 1089.