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Stupid number trick V (Posted on 2013-04-09) Difficulty: 2 of 5
(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?

No Solution Yet Submitted by Jer    
Rating: 1.5000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Solution - the Fallacy | Comment 3 of 4 |
(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*[a-c] + 10*[b-b] + [c-a]
Reverse
100*[c-a] + 10*[b-b] + [a-c]
Add
        Zero.

The above fallacy ignores the fact that [c-a] 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*[a-1] +(100+ 10*[b-1]) + (10+[c])
    100*[c]       + 10*[b]    +    [a]
Subtract
  100*[a-c-1] +(100 + 10*[b-b-1]) + [10+c-a]
Reverse
  100*[10+c-a] + (100 + 10*[b-b-1]) + [a-c-1]
Add
100*[a-c-1+10+c-a] +(200 -20) + [10+c-a+a-c-1]
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 2013-04-15 04:47:02

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