Probate PuzzleDate: 12/15/2003 at 08:01:24 From: Shruti Subject: Brain Teaser When my uncle in Mathura died he left a will instructing his lawyers to divide his estate of Rs.1920000 in this manner: Every son should receive three times as much as a daughter and every daughter should get two times as much as their mother. What is my Aunty's share? This not a sum in my homework but a sum from Shakuntala Devi's puzzle book. What is confusing about this sum that it is not stated how many sons or daughters the uncle has. Date: 12/15/2003 at 10:24:50 From: Doctor Ian Subject: Re: Brain Teaser Hi Shruti, We know that there is one mother, so we can say that her share is 'x'. We don't know how many daughters there are, although we can say that this number is an integer, d. Similarly, we don't know how many sons there are, but we can say that the number is an integer, s. This means the equation describing the situation is 1,920,000 = 6sx + 2dx + x Can we assume that this all has to work out nicely, with no remainders? Since it's a puzzle (and not a real-world problem), maybe we can. We can factor an x out on the right side, 1,910,000 = x(6s + 2d + 1) Now, we have a product; and if the product is to work out exactly, x must be a factor of the number on the left. What are the factors of the number on the left? To find that out, we need to find its prime factors: x(6s + 2d + 1) = 2^10 * 3 * 5^4 How does this help us? Now we can consider the possible numbers of sons and daughters. For example, could there be 1 son and 0 daughters? If so, then 2^10 * 3 * 5^4 = x(6*1 + 2*0 + 1) = 13x But 13 isn't a factor of the total amount. So this is _not_ possible. How about 0 sons and 1 daughter? If this is the case, then 2^10 * 3 * 5^4 = x(6*0 + 2*1 + 1) = 3x 2^10 * 5^4 = x So this is one possible solution. If we can choose values of s and d such that (6x + 2d + 1) adds up to the product of any of the prime factors, that will be a solution, too. For example, 0 sons and 2 daughters will work, since 6*0 + 2*2 + 1 = 5 and 5 is one of the prime factors. How can we find more solutions? One way is to systematically consider possible pairs of (s,d) values: . . . . . . . . (0,2) (1,2) (2,2) (3,2) . . (0,1) (1,1) (2,1) (3,1) . . (0,0) (1,0) (2,0) (3,0) . . and so on--basically, all the points on an s-d plane that have integer coordinates--and see which ones work out. The problem with this is that there are infinitely many pairs to be tried! However, if you're willing to make assumptions about the maximum numbers of sons and daughters, you can try them all. Another way to look for solutions is to try to use reasoning to see where solutions might lie. For example, I know that 15 is a possible factor, since 3*5 = 15. And I know that 12 + 2 + 1 = 15. That tells me that (s=2,d=1) is a solution. The problem with this approach is that it's not very systematic, which means you're likely to miss some possible solutions. A third way to proceed would be to make a list of possible factors of the total bequest, 2 = 6s + 2d + 1 3 = 6s + 2d + 1 4 = 6s + 2d + 1 5 = 6s + 2d + 1 6 = 6s + 2d + 1 8 = 6s + 2d + 1 etc. and then solve for one of the variables in terms of the other, e.g., 8 = 6s + 2d + 1 8 - 2d - 1 = 6s (8 - 2d - 1)/6 = s (7 - 2d)/6 = s This equation tells us that d can't be greater than 4, since that would give us a negative value for s. So we just have to try assigning the values 0, 1, 2, and 3 to d. None of those work, so we know that there are no solutions corresponding to this particular factor. Does that make sense? This is nice, because it means we can try all the different factors, and try all the possible solutions for those factors. It's a lot of work, but we can be confident that if a solution exists, we'll find it. By the way, an equation like this has a name: Diophantine. To see another example of a situation that leads to a Diophantine equation, look at 100 Animals, 100 Dollars, No Algebra http://mathforum.org/library/drmath/view/63981.html Or you can look for other information about Diophantine equations, on the Internet or in your math books. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ |
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