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Re: Finding A, B, and C
Posted:
Jan 21, 2005 3:10 PM


On Sun, 02 Jan 2005 21:39:58 GMT, N. Silver wrote: >Sylvain Croussette answered on alt.math.undergrad: >> st9468s4 (EJC)@drexel.edu wrote: > >> ABC I have to find the values of A, B and C where A>B>C >> CBA >>_________ >>= CAB > >Assuming A,B and C are single digits: >Rewrite as: > CAB >+CBA > > ABC > >>From the 1st column: B+A=C, or B+A=C+10, if there is a carry. > From the 2nd column also A+B=B; this is the same as the 1st column >so it's impossible, unless there is a carry from the 1st; so it's >1+A+B=B+10, because A+B produced a carry in the 1st column. > From the third column 1+C+C=A. >Rewrite these equations: >1) A+BC=10 >2) A+BB=9 >3) A2C=1 >Clearly from equation 2, A=9, so from equation 3, C=4 so B=5. > > > I was wondering where did 10 come from? Is it just some kind of random number you pick because I have been working on a very similar equation
ABC ACB + CAB
the answer is alike, A=4 B=5 C=9. Could you also tell me what kind of method are you using, I've tried with substitution, elimination, division, etc
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