Yes it was answered previously but I am sorry that I did not follow the pattern. I was told that the answer that I submitted was incorrect. Please help if you can. Thank you so very much.
THE PROBLEM READS:
A three-digit number increases by 9 if we exchange the second and third digits. The same three-digit number increases by 90 if we exchange the first and second digits. By how much will the value increase if we exchange the first and third digits?
Thank you
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
= 99(a-c)
but if we add #1 and #2
we get
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
e.g.
672
-276
-----
396
+693
-----
1089