Monkeys taking bananas
Three clever monkeys have a pile of bananas in front of them.
- The first monkey takes some bananas from the pile, keeps of them for himself, and shares the rest equally amongst the other two.
- The second monkey takes some more bananas from the pile, keeps of them for himself, and shares the rest equally amongst the other two.
- The third monkey takes all the remaining bananas from the pile, keeps of them for himself, and shares the rest equally amongst the other two.
Each monkey receives an integer number of bananas whenever the bananas are divided, and the number of bananas the first, second and third monkey have at the end is in the ratio .
What is the minimum possible amount of bananas in the pile?
The answer is 408.
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1 solution
OK - Here's my attempt to explain: Say that Monkey 1 took 8A bananas, Monkey 2 took 8B bananas, and monkey 3 took 24C bananas (these are the lowest numbers to make the fractions come out to integers). Thus:
[1] Monkey 1 = 6A + 3B + 11C
[2] Monkey 2 = A + 2B + 11C
[3] Monkey 3 = A + 3B + 2C
and [1]:[2]:[3] = 3:2:1
[1] = [2] + [3] (by the ratios given); 6A+3B+11C=2A+5B+13C; therefore C = 2A-B [4]
We also know that [2] = 2*[3], So
A + 2B + 11C = 2(A + 3B + 2C); therefore 7C = A + 4B [5]
Subbing [4] into [5]: 7(2A-B)=A+4B; A=11B/13 [6]
But nobody took fractions of bananas, so let 13Y = B [7] in order to get rid of fractions
Subbing into [6]: A=11B/13 = 11Y [8]; which gives us both A and B in terms of Y ([7] and [8])
Subbing [7] and [8] into [1], [2], and [3]:
Monkey 1 = 6A + 3B + 11C [1] = 66Y + 39Y = 11C = 105Y + 11C [9]
Monkey 2 = A + 2B + 11C = 11Y + 26Y + 11C = 37Y + 11C [10]
Monkey 3 = A + 3B + 2C = 11Y + 39Y + 2 C = 50Y + 2C [11]
We still know that [2] = 2 [3], and therefore [10] = 2 [11]
37Y + 11C = 2 * (50Y + 2C) --> 37Y + 11C = 100Y + 4C --> 7C = 63Y, C = 9Y [12]
subbing [12] back into our monkeys banana count ([9], [10], [11]):
Monkey 1 [9] = 105Y + 11C = 105Y + 99Y = 204Y
Monkey 2 [10] = 37Y + 11C = 37Y + 99Y = 136Y
Monkey 3 [11] = 50Y + 2C = 50Y + 18C = 68Y
Total number of bananas = 204Y + 136Y + 68Y = 408Y
The lowest total would be when Y = 1, which gives us 408 bananas in the pile.