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## #1 2013-08-21 23:46:52

SPACKlick
Novice

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### Absolutely Christmas Crackers Puzzle

There is an error on this page

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?

## #2 2013-08-25 07:25:58

Nehushtan
Power Member

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### 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.

## #3 2013-08-25 15:15:29

MathsIsFun

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### 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