in the original number,
let the unit digit be a,
the tens digit be b
and the hundred digit be c
the appearance of our number is cba
then the value of our number is 100c + 10b + a
case1: interchange 2nd and 3rd digit
number looks like cab
value of our new number is 100c + 10a + b
so 100c + 10a + b - (100c + 10b + a) = 9
9a -9b = 9
a-b = 1 , (#1)
case2: interchange the 1st and 2nd digit
appearance of new number is bca
value of new number is 100b + 10c + a
so 100b+ 10c + a - (100c + 10b + a) = 90
90b -90c = 90
b - c = 1 , (#2)
case3: interchange 1st and 3rd
appearance of new number is abc
value of new number is 100a + 10b + c
change in value = 100a + 10b + c - (100c + 10b + a)
= 99a - 99c
but if we add #1 and #2
a - c = 2
so 99(a-c) = 99(2) = 198
A little know feature of the above is the following "math trick"
1. Pick any 3 digit number, all different and the hundreds digit greater than the unit digit
2. reverse the digits and subtract the numbers. If you get a 2 digit result, insert a 0 in the hundred place
3. reverse your subtraction answer and add the last two results,
4. You will always get 1089
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