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The following problem is from the very beginning of the book and it even has a solution with explanation. However, I keep banging my head against the wall unable to understand the reasoning behind it. I would be grateful if you could explain his reasoning in a very detailed way since I cannot understand the reasoning behind every of his assumptions.

I have uploaded the problem here http://tinypic.com/r/o018id/6.

blalala
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    What part of the solution are you having trouble with? – Jemmy Apr 13 '13 at 05:46
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    The problem is the cryptarithm $AAA+BBB=AAAC$. He says $A$ has to be $1$ because when you add two 3-digit numbers and get a 4-digit number that 4-digit number must be less than $2000$ --- do you understand that part? – Gerry Myerson Apr 13 '13 at 05:47
  • Why no other digit than 1 can appear as a carry in the thousands position of the result? As well as, the following statements. – blalala Apr 13 '13 at 05:48
  • Thank you, Gerry. That was my problem. – blalala Apr 13 '13 at 05:49

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When you add two numbers, the most you can carry is $1$. This is because the maximum of each is $9$ (and here we know they are different, so one is no more than $8$) plus you might carry $1$ in from the place to the right. $9+9+1=19 \lt 20$

Ross Millikan
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