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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