Monkeys Dividing the Coconut PileDate: 07/28/98 at 22:44:37 From: Thomas Subject: Monkey Problem This is a problem I could not solve. Please help me out. Three monkeys spend a day gathering coconuts together. When they have finished, they are very tired and fall asleep. The following morning the first monkey wakes up. Not wishing to disturb his friends, he decides to divide the coconuts into 3 equal piles, but there is one left over, so he throws this odd one away, helps himself to his share, and goes home. A few minutes later the second monkey awakes. Not realizing that the first has already gone, he too divides the coconuts into 3 equal heaps, finds one left over, throws the odd one away, helps himself to his fair share, and goes home. Then the third monkey does exactly the same. What is the smallest possible number of coconuts left?" Many thanks, Thomas Date: 07/29/98 at 11:10:16 From: Doctor Anthony Subject: Re: Monkey Problem If N = original number of coconuts, then the first monkey takes (N-1)/3 as his share, leaving 2(N-1)/3 behind. Then the second monkey takes: 2(N-1)/3 - 1 2(N-1) 1 ------------ = ------ - --- 3 9 3 leaving behind: 4(N-1) 2 ------ - --- 9 3 The third monkey takes: 4(N-1)/9 - 2/3 - 1 4(N-1) 5 ------------------- = ------ - --- 3 27 9 leaving: 8(N-1) 10 8(N-1) - 30 ------ - --- = ------------- (Equation 1) 27 9 27 Here, we assume that the third monkey does not just grab everything left after the other two have gone. We require Equation 1 to be an integer, say t, and so: 8(N-1) - 30 = 27t 8N = 38 + 27t where both N and t are integers. 8N - 27t = 38 Then divide through by 8, the smaller coefficient: N - 3t - 3t/8 = 4 + 6/8 This equation must be satisfied in integers. This means that the fractions in the above equation must reduce to an integer. Thus: 3t 6 3t+6 --- + - = ---- = integer 8 8 8 To find the smallest value, we can start by testing whether t=1 results in an integer and then increasing t by 1 until we get to the first integer. For example, if t = 1, then (3(1) + 6)/8 = 9/8, not an integer. If t = 2, then (3(2) + 6)/8 = 12/8 = 3/2, not an integer. We can continue this process until we get to the first integer. This may take a while, or it may not. Alternatively, we could use the following method. First, multiply the top line by 3. This is because we want the coefficient of t in the numerator to be 1. So: 9t + 18 ------- = integer 8 Thus: t + (t + 18)/8 = integer So we want: (t + 18)/8 = integer = p t+18 = 8p t = 8p - 18 Since t must be positive, we take p = 3 and get t = 24-18 = 6. So the minimum number of coconuts left by the third monkey is 6. - Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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