Proving Something is Always So

Date: 04/01/98 at 02:48:12
From: Emma
Subject: Prove result is always 1089

I have a question about proving something is always so.

John said, "Write down any 3 digit integer and write down the number 
obtained by reversing the digits. Find the difference between these 
two numbers. Repeat this process with your answer but this time add 
the two numbers. The result is always 1089."

EXAMPLE: 734 - 437 = 297     and     297 + 792 = 1089.

My first question is how do I go about writing a proof? What is 
necessary for me to prove that the statement is always correct?

My next question is a problem that I have come upon while 
experimenting with different numbers. I have found that my result is 
a negative number. Is this still correct?

EXAMPLE: 123 - 321 = -198    and   -198 + -891 = -1089. 

How do I explain my findings? Help would be much appreciated. 

Thanks,
Emma

Date: 04/01/98 at 05:04:45
From: Doctor Pete
Subject: Re: Prove result is always 1089

Hi,

Usually the hard part about proving these kinds of problems is 
figuring out what to start with. The idea we need is "how do we 
express 3-digit numbers?" Well, a 3-digit number, say, 426, is 
actually a representation of:

     426 = 400 + 20 + 6 = 4(100) + 2(10) + 6(1)

So we represent any 3-digit number N in terms of its hundreds, tens, 
and units digits:

     N = 100h + 10t + u

where h may be an integer between 1 and 9 (if h is 0, then N is only 
2 digits), and t and u may be integers from 0 to 9.

That said, the process of reversing the digits gives you a number:

     M = 100u + 10t + h

Now, suppose that in reversing the digits of N to get M, we find that 
M < N. Then we want the difference N-M, which is

     N - M = 100(h-u) + 10(t-t) + (u-h)
           = 100(h-u) + (u-h) > 0

since M < N implies N-M > 0.  It follows that h-u > 0, or u-h < 0.  
(What happens when M = N?) Well, you can't have a negative units 
digit, which is how we have so far represented N-M, so we "borrow" 100 
from the hundreds digit (we have no tens to borrow but we carry this 
through):

     N - M = 100(h-u-1) + 100 + (u-h)
           = 100(h-u-1) + 90 + 10 + (u-h)
           = 100(h-u-1) + 10(9) + (10+(u-h))
           = 100(h-u-1) + 10(9) + (10-(h-u))

Note that the units digit is now written 10-(h-u), which is clearly 
between 0 and 9 since h-u > 0. So now we take this number, call it P.  
Reverse the digits to get Q:

     Q = 100(10-(h-u)) + 10(9) + (h-u-1)

Add the two:

     P + Q = 100(10-(h-u) + h-u-1) + 10(9+9) + (10-(h-u) + h-u-1)
           = 100(9) + 10(18) + 9
           = 1089

So we're done. The only thing left is to consider the case when
N < M, which gives you negative numbers if you take the difference 
N-M. (Note that it would be the same if you took the absolute value 
difference, |N-M| = |M-N|), but you would use similar reasoning.) 
The only case in which the process fails, which I hinted at earlier, 
is when N = M. For example, 464, or 121, reversing these numbers and 
subtracting gives 0, and 0 + 0 = 0. Can you see where the proof 
"breaks down" in this case?

Also, observe that a crucial step in the proof is to obtain N-M = P in 
its hundreds-tens-units form; i.e., its digits. If we did not do this, 
we could not reverse them to find Q.

-Doctor Pete,  The Math Forum
Check out our web site! http://mathforum.org/dr.math/