More Monkeys and Nuts
Date: 06/07/99 at 20:29:17 From: Michael Holt Subject: Further monkey puzzling - point raised Dear Dr(s) Math, I have looked through your archive and found a couple of variants on the monkeys sharing nuts theme. None is specifically the same as the problem I have (I don't think) although it is very similar to one that is answered at some length with formulae and substitution, etc. The person who set me this puzzle suggested that it was possible to deduce the answer without going to this trouble, and gave me a clue which I will add at the end of the puzzle. The puzzle: Five monkeys collect nuts, which they leave in a pile. During the night one wakes up and divides the pile into 5. There is one left over. He eats the remainder nut and his 1/5 share, and mixes the nuts together. The second monkey comes down, divides the pile into 5 with one left over, and so it goes on; each left pile divides into 5 with one left over and each monkey eats its share plus one. The question is, what is the least number of nuts that they had amassed in order for this to work? The clue: Apparently, by asking what would have happened if there were 4 more nuts collected, the answer can be found without lengthy calculations - in fact I am told of somebody who solved it in his head over lunch given this advice. I have played around with the idea at length but have grown frustrated, as I have not yet solved the problem. What do you think? Thanks for your help; I look forward to hearing from you. Yours sincerely, Michael Holt
Date: 06/10/99 at 12:42:20 From: Doctor Peterson Subject: Re: Further monkey puzzling - point raised Hi, Michael. I've been playing with this off and on for a few days, and getting confused because of the variations among versions of the puzzles! My biggest frustration has been trying to make my answer agree with one of the others. Yes, this clue does help, at least with your version of the puzzle. Let's suppose that the initial number of nuts is X_0. I divide the pile by five and have one left over, so each pile is (X_0-1)/5, and the number I leave is X_1 = (X_0 - 1) * 4/5 The same process is repeated five times until I have a fairly complicated expression for X_5, the number left at the end. Suppose we set 4 nuts next to the pile and don't touch them through the whole process (so they AREN'T part of the dividing by five - although that's not quite how you described the hint). The total number of nuts is Y_0 = X_0 + 4 Y_1 = X_1 + 4 = (X_0 - 1) * 4/5 + 4 = (Y_0 - 4 - 1) * 4/5 + 4 = (Y_0 - 5) * 4/5 + 4 = Y_0 * 4/5 - 4 + 4 = Y_0 * 4/5 So at each step, the total number of nuts is just multiplied by 4/5, and we can easily see that Y_5 = Y_0 * (4/5)^5 For this to be a whole number, Y_0 must be a multiple of 5^5, and the smallest possible value would be exactly that, or 3125. Then X_0 is four less than this, or 3121. Now let's try it out: X_0 = 3121 X_1 = (3121 - 1) * 4/5 = 2496 X_2 = (2496 - 1) * 4/5 = 1996 X_3 = (1996 - 1) * 4/5 = 1596 X_4 = (1596 - 1) * 4/5 = 1276 X_5 = (1276 - 1) * 4/5 = 1020 This final number, of course, is 4^5 - 4. This problem is NOT the same as http://mathforum.org/dr.math/problems/koestler12.18.97.html where the recursion equation is X_1 = X_0 * 4/5 - 1 and there is an extra round at the end in which the pile is divided into five openly. I don't know whether there's a similar trick here. Your problem IS, however, the same as http://mathforum.org/dr.math/problems/stillman10.12.98.html except that there we have the extra round, so the answer using our method should be 5^6 - 4 = 15621, and the number each one gets at the end should be 4^6 - 4 = 4092. The equation found there is 1024*a - 15625*g = 11529 Plugging in our values for a = X_0 and g = X_6/4 (the number each gets at the end), we see 1024*15621 - 15625*1023 = 15995904 - 15984375 = 11529 and we have it. The trick amounts to a change of variables that simplifies the equation. To find such a change, you could try Y = X + a, and see what value of a will help: X_1 = (X_0 - 1) * 4/5 Y_1 - a = (Y_0 - a - 1) * 4/5 Y_1 = Y_0 * 4/5 + a - (a + 1)*4/5 To make the equation simpler, we want a - (a + 1)*4/5 = 0 5a = 4(a + 1) a = 4 Or, you can just have a flash of insight and try adding four nuts for no reason at all. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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