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ABCD * 4 = DCBA


Date: 05/23/2000 at 14:54:30
From: Kerri Chaffee
Subject: Cryptarithm

Find digits A, B, C, and D that solve the following cryptarithm:

     ABCD * 4 = DCBA

I have tried it many times but I can not seem to get the right digits 
to answer the question.

If someone could answer this for me it would be great.
Thanks.


Date: 05/24/2000 at 10:59:45
From: Doctor TWE
Subject: Re: Cryptarithm

Hi Kerri - thanks for writing to Dr. Math.

Start by writing down what you know. We know:

       ABCD
     *    4
       ----
       DCBA

Let's see if we can eliminate some possibilities. A can't be 3 or 
larger (do you see why?), so it must be 1 or 2. Let's try A = 1 
first.

If A = 1, then we have:

       1BCD
     *    4
       ----
       DCB1

but 4*D can't end in a 1, and there's no carry coming from digits to 
the right. So A must be 2. Thus we have:

       2BCD
     *    4
       ----
       DCB2

Now the only digits that produce _2 when multiplied by 4 are 3 
(because 3*4 = 12) and 8 (8*4 = 32). So D must be either 3 or 8. But 
it can't be 3 can you see why?), so D = 8. Now we have:

         3    <- carried from the 8*4
       2BC8
     *    4
       ----
       8CB2

Can you take it from here? Remember to add the 3 carried from the 
units place when you multiply C*4, and notice that when multiplying 
the B*4, there can't be any carry over into the 1000's place.

I hope this helps. If you have any more questions, write back.

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