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Problem statement

Five sailors were cast away in an island, inhabited only by monkeys. To provide food, they collected all the coconuts that they could find. During the night, one of the sailors woke up and decided to take his share of the coconuts. He divided them into 5 equal piles and noticed that one coconut was left over, so he threw it to the monkeys; he then hid his pile and went to sleep. A little later a second sailor awoke and had the same idea as the first one. He divided the remainder of the coconuts into 5 equal piles, discovered also that one coconut was left over, and threw it to the monkeys. He then hid his share and went back to sleep. One by one the other three sailors did the same thing, each throwing one coconuts to the money. The next morning, all sailors looking sharp and ready for breakfast, divided the remaining pile of coconuts into five equal parts, and no coconuts was left over this time. Find the smallest number of coconuts in the original pile.

My solution was, each time a sailer take his share, I recalculate the number of coconut:

  1. $n = 5\cdot q_0 + 1 $ $\to$ # left = $\frac{4}{5}\cdot q_0 = \frac{4}{5}\cdot \frac{n - 1}{5}$

  2. $\frac{4}{5}\cdot q_0 = 5 \cdot q_1 + 1 \to$ # left $= \frac{4}{5}\cdot q_1 ....$

Continuing this process, I ended up with a very strange fraction:

$\frac{(256\cdot n - 464981)}{1953125}$

Then I set this fraction to $5\cdot k$, since the last time # coconuts is divisible by $5$, to solve for $n$.
Am I in the right direction? Any hint would greatly appreciated.

Thanks,
Chan

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    Another monkey question! Are we going to be seeing a "monkeys" tag soon? :)2011-02-07

2 Answers 2

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You wrote that $n = 5q_0 + 1$, so then the first sailor threw away a coconut, and kept $q_0$ coconuts. The remaining number of coconuts is $4q_0$!

So then you write:

$n = 5q_0 + 1$ $4 q_0 = 5q_1 + 1$ $4 q_1 = 5q_2 + 1$ $4 q_2 = 5q_3 + 1$ $4 q_3 = 5q_4 + 1$ $4 q_4 = 5q_5$

Now you know that $q_5 > 0$, so $n$ will be smallest when $q_5 = 1$. (Of course if they had nothing for breakfast, do the following with $q_0 = 0$).

You must also consider that when the fifth sailor was taking his share, after throwing away a coconut, the remaining number of coconuts was divisible by $5$. So

$ q_4 = \frac{5q_5}{4}$

must be an integer. Let's write $q_5 = 4k_0$. Then writing the equations backwards, and expressing $q_3$ with $q_{4}$ yields:

$4 q_4 = 5q_5 = 20k_0 \text{ so } q_4 = 5k_0$ $4 q_3 = 5q_4 + 1 = 5^2k_0 +1 $

Again you must make sure that $\frac{5^2k_0 +1}{4}$ is an integer. Continue on until you reach $n$. You will be reexpressing the $k_0$ from here on (that's why I added the subscript).

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    @micak: Tha$n$k for your ki$n$d$n$ess ;) I got $n$ = 18746. Does it seem right to you? I use 4k1 - 1. Could you explai$n$ how both `4k1 - 1` a$n$d `4k1 + 3` work? I think `4k1 - 1` is less than `4k1 + 3` right?2011-02-07
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You asked for a hint, so I won't give the full answer:

Step (1): Realize that if K is an answer, so is K + n*5^(5+1) for any n. Step (2): The hint for this step will probably give the whole thing away so it is withheld Step (3):

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    Thanks! But I'm not sure I understand your hint. You said: `K + n*5^(5+1) for any n`, this is not an equation, so how could I find K?2011-02-07