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.

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?

Three boxes are all labeled incorrectly, and you must get the labels right. The labels on the boxes read as follows:

[box 1] nails
[box 2] screws
[box 3] nails and screws

To gain the information you need to move the labels to the correct boxes, you may remove a single item from one of the boxes. You may not look into the boxes, nor pick them up and shake them, etc.

Yeah its possible just pick out one item from all the three boxes. you will find two common items and one different item. For example:- assume you got one nail and two screws.Go with the different one i.e with nail and label it as nails. the remaining two are already labeled incorrectly so correct them

Pick an item from box3 labelled nails and screws and see what it is. It must be either a nail or screw
1. If it is a nail, box3 has nails, box1 has screws and box2 has nails and screws.
2. If it is a screw, box3 has screws, box1 has nails and screws and box2 has nails.

1st one is labelled Nails, 2nd one Screws and 3rd both Nails and Screws.

we know that all the labels are wrongly placed. So... the third one CANNOT have BOTH nails and screws.
Taking out one item from the third box, if we get a Nail, then the remaining two boxes should have "Nails and Screws" and "Screws"...
Since we know that the second box has "Screws" labelled so it cannot have ONLY screws. Thus, it has Nails and Screws.

same procedure if we get screw instead of a nail...
Hope I'm correct

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.[/QUOTE]