Date: 12/07/97 at 17:08:59 From: Jeff MacDougall Subject: Advanced Algebra Honors I can't figure this out, it is a problem for my Algebra 2 class that I and my friends can't figure out: A boy selling fruits has only three weights, but with them he can weigh any whole number of pounds from 1 pound to 13 pounds inclusive. What three weights does he have?
Date: 12/17/97 at 12:00:25 From: Doctor Mark Subject: Re: Advanced Algebra Honors Hi Jeff, Problems like this have intrigued people for a long time, and the problem you are looking at was originally solved by a man named Claude Bachet (1587-1638). Here's the idea. First of all, we have to realize that if he puts the fruit in one pan of the balance (you have to imagine that this is how the boy is going to weigh the fruit), then he can put a particular weight in either the pan the fruit is in, or the other pan, or not put the weight in any pan at all. Let the three weights be w, W, and K, where w is the lightest weight, W is the middle weight, and K is the heaviest weight. Now clearly the heaviest weight we can weigh is, by assumption, 13, and this must be when the fruit is in one pan, and all three weights are in the other pan, so: w + W + K = 13 Now the second heaviest weight we can weigh is 12, and this must be balanced when we have the two heaviest weights, W and K, in the other pan. Thus W + K = 12. Hence, w = 1. We could weigh a fruit of weight 11 by putting w in the pan with the fruit, and W+K in the other pan. Then there would be 12 pounds in each pan. How could we weigh a fruit of weight 10? If you think about it, this must involve putting the fruit in one pan, and w and K in the other pan. That means that w + K = 10. Since w = 1, this means that K is 9, and hence that W must be 3. So the weights are 1, 3, and 9. That's an interesting pattern, don't you think? The original problem that Bachet considered was how many weights, and of what values, were necessary to weigh all weights up to 40. What answer do you think he got? (Hint: what number do you get if you double 13 and add 1? And how is that related to the weights you needed? What if you do a similar thing for 40? Maybe this tells you what the pattern is. You might also think about how numbers are written in base 3...) As a historical note, Fermat's Last Theorem, which was just solved 4 years ago after defying the best mathematicians for more than 350 years, was originally scribbled in the margins of Fermat's copy of Bachet's translation of Diophantus' Arithmetica (boy, that's a lot of possessives, eh?). -Doctor Mark, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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