Posted by **Please help me understand** on Thursday, February 9, 2012 at 1:39am.

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

- Math Help -
**Reiny**, Thursday, February 9, 2012 at 8:17am
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

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