(1) Take a 3 digit number whose first digit is greater than its last.
(2) Reverse the digits of (1) to create a smaller number.
(3) Subtract the number in (2) from (1) (Add a leading zero if this number is not 3 digits)
(4) Reverse the digits of (3)
(5) Add the numbers from steps (3) and (4)
What is the result, and why?
(In reply to
re: Solution by Ady TZIDON)
I missed seeing this posted.
Kenny M is "baiting".
In what appears below anything enclosed as such [....] refers to a digit.
The process described is:
100*[a] + 10*[b] + [c]
100*[c] + 10*[b] + [a]
Subtract
100*[ac] + 10*[bb] + [ca]
Reverse
100*[ca] + 10*[bb] + [ac]
Add
Zero.
The above fallacy ignores the fact that [ca] is negative and does not take into account any process (depending upon one's schooling) as to how the minuend is treated  I refer to "borrowing", equal addition and decomposition/renaming to mention a few.
Begin
100*[a] + 10*[b] + [c]
100*[c] + 10*[b] + [a]
Decompose/Rename the minuend before continuing:
100*[a1] +(100+ 10*[b1]) + (10+[c])
100*[c] + 10*[b] + [a]
Subtract
100*[ac1] +(100 + 10*[bb1]) + [10+ca]
Reverse
100*[10+ca] + (100 + 10*[bb1]) + [ac1]
Add
100*[ac1+10+ca] +(200 20) + [10+ca+ac1]
900 + 180 + 9
Umm? I didn't deliberately set out to answer Jer's poser, rather to show where the fallacy failed. I'm afraid my notation looks a bit untidy.

Posted by brianjn
on 20130415 04:47:02 