Coconut PilesDate: 10/12/98 at 20:47:02 From: Fallon Stillman Subject: Algebra Puzzle Problem Here's the problem: Gilligan and his buddies were stranded on a desert island. Gilligan, the professor, Ginger, Mary Anne, the Skipper, and Fred the monkey were gathering coconuts. One evening they all rounded up all the coconuts they could find and put them in one large pile. Being exhausted from working so hard they decided to wait and divide them up evenly in the morning. During the night Gilligan awoke and separated the nuts into five equal piles, with one left over which he gave to Fred the monkey. Gilligan took one pile, hid it, pushed the other four back together, and went back to his hut. He was followed by Ginger, Mary Ann, the Professor, and the Skipper, each dividing them up equally with one remaining nut going to Fred. The next morning the remaining nuts were divided up equally with one remaining going to Fred. What is the least number of coconuts they could have started with? Date: 10/13/98 at 13:17:58 From: Doctor Rob Subject: Re: Algebra Puzzle Problem Hello Fallon, See the following web page for a related, but not identical(!) problem: http://mathforum.org/dr.math/problems/koestler12.18.97.html The difference is that in this problem, the monkey gets his coconut before the five-way split, and in the problem on the other Web page, he gets it after. Subtle, but important. The equations for your problem are these. Let a be the number of coconuts to start with. After the first person and the monkey take their coconuts, the number left is b = (4/5)*(a-1). After the second person and the monkey take their coconuts, the number left is c = (4/5)*(b-1). The third person leaves d = (4/5)*(c-1) coconuts. The fourth person leaves e = (4/5)*(d-1) coconuts. The fifth person leaves f = (4/5)*(e-1) coconuts. At the end, f = 5*g + 1, where g is the number of coconuts each person gets in the morning. Now when you perform the chain substitution to find the equation relating a and g, you get: 4*(4*[4*(4*[4*(a-1)/5-1]/5-1)/5-1]/5-1)/5 = 5*g + 1 1024*a - 15625*g = 11529 For this you have to find the smallest positive integer solution. You can follow the model in the cited problem to figure this out. Note that g has to be odd, so g = 2*h+1, and: 512*a - 15625*h = 13577 Now h must be odd, so h = 2*i + 1, and: 256*a - 15625*i = 14601 You can continue in this way, reducing one or the other of the coefficients on the lefthand side until one of them is reduced to 1. Then that variable can be expressed as an integer times the remaining variable (say z), plus another integer. Undoing all the substitutions will express everything as an integer function of z, and the value of z which gives the smallest positive a is the one you want. Then you can compute that smallest positive value of a. There is another way, using the Extended Euclidean Algorithm. You can use that to find that: 1024*10849 - 15625*711 = 1 Now multiply by 11529, and reduce the a-value modulo 15625 to get the answer. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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