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 Candy Box (Posted on 2004-10-08)
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.

 See The Solution Submitted by Brian Smith Rating: 3.5000 (4 votes)

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 Another division | Comment 2 of 17 |

We know that it certainly can't be done with only 7 pieces, as this would require each to be equal in order for 7 people to share equally. But then the other numbers of people could not share equally.

The total number of ounces must be divisible by 3, 4 and 7, and therefore by 84.

If one takes a row of ounces that's 84 long, and puts dividing markers at intervals equal to 1/3, 1/4 and 1/7 of the way along it, the dividers will be after ounces numbered 28, 56, 21, 42, 63, 12, 24, 36, 48, 60 and 72.  Sorted, that's 12, 21, 24, 28, 36, 42, 48, 56, 60, 63, 72.  Combining the ounces between markers, we get piece sizes of 12, 9, 3, 4, 8, 6, 6, 8, 4, 3, 9, 12.  As it results from a symmetrical diagram and there are an even number of pieces, there are pairs of each size: 3, 4, 6, 8, 9 and 12-ounce sizes.  That's 12 pieces in all.

I don't know if it can be done with fewer pieces than 12, but it certainly requires at least 8, as indicated in the first paragraph.

 Posted by Charlie on 2004-10-08 12:55:10

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