Teacher2Teacher 
Q&A #403 
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From: Loyd <loydlin#aol.com> To: Teacher2Teacher Public Discussion Date: 2003122111:07:53 Subject: Re: magik123 problem On 2003122005:44:59, miltos karam wrote: > >let X be a possitive integer,A the number of even digits B the number >of odd digits and C the number of digits ..... >we create the number ABC and we continue like that for the new >number,if we do that we will end up at number 123!!! >but how can we prove that? >for example :X=43526,A=3,B=2,C=5 then ABC=X1=325..A1=1,B1=2,C1=3 and >ABC=123. > First, the problem only works if there are some odd and even digits. That is, 22222 wouldn't work. The last digit will always be odd since adding even to odd always results in odd. You can only end up with three digits because the last number is the sum of the even and odd and numbers like 325 above shows that there is 1 even digit and one odd digit, and since the last digit is always odd and will always be the sum of the other two, then you have to have 3 for last digit. You will always have one even and two odd. If you have a number like 777778 then you have 15,43. Here you have to treat 43 as a single odd digit. The final number is 1 even and two odd and the sum of 1 and 2 is three. Thus, 123.
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