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Re: Finding A, B, and C
Posted:
Jan 21, 2005 6:12 PM


Laura Haliman writes:
> ABC I have to find the values of A, B and C where A>B>C > CBA >_________ >= CAB
The subtraction ABC  CBA = CAB can be rewritten in an equivalent form using addition: CAB + CBA = ABC.
CAB + CBA  ABC
Let's add. We add the right column of numbers first: A+B = C or if A+B were larger than 9, we would have a carry of 1. (Remember, the 1 is really a ten. It is in the tens' column.) In that case, we would know that A+B = 10 + C.
Let's add the middle column of numbers. It's A+B again. If there were no carry, it would have to be A+B = C, but A+B = B. So, there was a carry afterall. And, actually, A+B = 10 + C.
Because we now know that A+B is larger than 9, and because we also know there is a carry of 1, from the middle column, we get: 1+A+B = 10+B, which reduces to: A = 9.
We still know that A+B is larger than 9. So a 1 was carried into the left column from the middle column. We have 1+C+C = A. But A = 9. Then, 2C+1 = 9 ==> C = 4.
Backtracking to A+B = 10+C, and substituting gives us 9+B = 10+4 ==> B = 5.
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