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## #1 2013-08-21 01:46:52

SPACKlick
Member
Registered: 2013-08-21
Posts: 3

### Absolutely Christmas Crackers Puzzle

3 x 4 x 4 = 48. Sadly at least two crackers must be exact duplicates of ones already made.

BUT if you could handle the sad look on people's faces, you could consider a missing item as a possible combination. This gives 4 x 5 x 5 = 125 different crackers. Many crackers would be missing one item, some missing two, and one cracker with nothing in at all! (This solution by Eric Chen.)

4x5x5=100 not 125

Also rather than Many and Some why not say 40 crackers will be missing 1 item and 14 will be missing 2?

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## #2 2013-08-24 09:25:58

Nehushtan
Member
Registered: 2013-03-09
Posts: 897
Website

### Re: Absolutely Christmas Crackers Puzzle

SPACKlick wrote:

40 crackers will be missing 1 item and 14 will be missing 2?

Correction: 11 will be missing exactly two items.

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## #3 2013-08-24 17:15:29

MathsIsFun
Registered: 2005-01-21
Posts: 7,659

### Re: Absolutely Christmas Crackers Puzzle

Thanks both of you, I have amended the puzzle (use refresh): Absolutely Christmas Crackers Puzzle - Solution

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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