Hi Friends,

Ten people land on a deserted island. There they find lots of coconuts and a monkey.

During their first day they gather coconuts and put them all in a community pile.

After working all day they decide to sleep and divide them into ten equal piles the next morning.

That night one castaway wakes up hungry and decides to take his share early.

After dividing up the coconuts he finds he is one coconut short of ten equal piles.

He also notices the monkey holding one more coconut. So he tries to take the monkey's coconut to have a total evenly divisible by 10. However when he tries to take it the monkey conks him on the head with it and kills him.

Later another castaway wakes up hungry and decides to take his share early.

On the way to the coconuts he finds the body of the first castaway, which pleases him because he will now be entitled to 1/9 of the total pile. After dividing them up into nine piles he is again one coconut short and tries to take the monkey's slightly bloodied coconut.

The monkey conks the second man on the head and kills him.

One by one each of the remaining castaways goes through the same process, until the 10th person to wake up gets the entire pile for himself.

What is the smallest number of possible coconuts in the pile, not counting the monkeys?

Regards,
Rijus.
--------
Real Inspirational Journey.........Unanimously & Sincerely.

5039

By my calculations, if each time someone gets up to divide the piles and is one short, then the total number INCLUDING the monkey's must be divisible by 10, 9, 8, 7, 6, 5, 4, 3 and 2. In other words, the number of coconuts must have those numbers as factors. The easiest way to find a number with those factors is to multiple them together (10 * 9 * 8 * ... 2, or 10!) which is 3628800. But this is not the smallest number.

Since 3 is a factor of 9, any number divisible by 9 is hence divisible by 3. Similarly, 2 is a factor of 4, 4 is a factor of 8 and 5 is a factor of 10. Numbers with 9 as a factor have a 2/3 chance of also being a factor of 6.

Taking this into account, we can reduce the factors down to 10, 9, 8 and 7 (and possibly 6). Multiplying these together gives us 5040. This is divisible by 6, so we have found the smallest number. Since the monkey has one of these coconuts, the answer is 5039 coconuts in the pile.

That's my reasoning...anyone got a better solution?

i am not sure but is the answer 7559???

well as to how i went about doing it ......

for ease of calculation lets assume x is the total no of coconuts (including the one with the monkey...)

and this value of x must be LCM of all numbers from 1 to 10

and that value is 7560 so the number of coconuts excluding the one from the monkey is 7559

correct me if i am wrong

cheers,
cooltwins