Math Is Fun Forum

SPACKlick

Member

Registered: 2013-08-21

Posts: 3

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?

Nehushtan

Member

Registered: 2013-03-09

Posts: 957

Re: Absolutely Christmas Crackers Puzzle

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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MathsIsFun

Administrator

Registered: 2005-01-21

Posts: 7,713

Re: Absolutely Christmas Crackers Puzzle

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

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