A box of candies can be equally divided by weight without cutting pieces between three, four or seven people.
Each piece is an integral number of ounces.
What is the least number of pieces of candy the box could contain? The candies may be of different weights.

(In reply to About 9 pieces by Federico Kereki)

Umm... this logic doesn't work. With this logic, there wouldn't be any solutions above 9 at all. Cuz if you can use this logic to say there isn't a 10 piece solution, you can extend it to say there isn't a 11 piece solution. But clearly there's an 11 piece solution.

But you can use a similar proof that David used to show that there isn't a 9-piece solution. Here goes:

Let x be the # people that get a single piece of 12N in a group of 7. Of course, there are 7-x people who have at least 2 pieces. There will also be 9-x pieces left over. So the number of pieces per remaining people is:

(9-x)/(7-x)>=2. Solving that yields x>=5. So at least 5 people have a single piece of 12N. So out of the 9 pieces, 5 of them are 12N and 4 of them are unknown.

If you split these 9 pieces among 4 people, 1 person will get at least 2 of the 12N pieces, which means he will have 24N when split among 4 people. But they should each have 21N.

Therefore, there is no 9-piece solution.

Posted by np_rt on 2004-10-08 18:59:03