ABCDC - BEAAC = BADAD

Date: 03/25/2002 at 20:09:00
From: Arthur
Subject: Wierd math problem

ABCDC - BEAAC = BADAD 

If D=0 the point of this problem is to find out what all the letters 
stand for knowing that d is equal to zero and the problem has to equel 
BADAD.

Thanks,
Arthur R.

Date: 03/26/2002 at 08:21:02
From: Doctor Ian
Subject: Re: Wierd math problem

Hi Arthur,

Here is the original problem:

    ABCDC
  - BEACC
  -------
    BADAD
 
We're told that D is zero - although that's not much of a hint. (Do 
you see why it can't really be anything else?)

    ABC0C
  - BEACC
  -------
    BA0A0
 
If we assume that each letter has to stand for a different digit, then 
we can look at the tens place, 

    ABC0C
  - BEACC
  -------
    BA0A0
       ^
       |

and see that we're going to have to carry a 1 from the hundreds place, 
which means that 9-C = A.  If nothing else, this means that if we ever 
figure out what C is, we'll know what A has to be; or if we ever 
figure out what A is, we'll know what C has to be. So let's include 
that in what we know:

    ABC0C          9 - C = A
  - BEACC
  -------
    BA0A0

Now, when we carried that 1, we were left with C-1 on top in the 
hundreds column. If there are no other carries, then it must be true 
that

  (C-1) - A = 0

If there is a carry to that column, then it must be true that

      C - A = 0

But this would mean that A and C are equal; and we're assuming that 
each letter has to be a different digit. So we can rule this out.  
Therefore, we know that 

  (C-1) - A = 0
     
      C - 1 = A

So let's add that to what we know:

    ABC0C          9 - C = A
  - BEACC          C - 1 = A
  -------
    BA0A0

Now, this is interesting, because we have two equations with two 
variables; and that means there is only one pair of values that can 
make both equations true.  

  9 - C = A

  C - 1 = A

In fact, the two things on the left are both equal to the same thing, 
so they must be equal to each other:

  9 - C = C - 1
     
Can you solve this to get C?  That will tell you the value of A as 
well.  And that should get you most of the way towards a solution. 

Can you take it from here?

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/