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Topic: Calendar Cubes I (Read 1972 times) |
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suid
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Calendar Cubes I
« on: Aug 4th, 2002, 2:04pm » |
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The solution to this problem is as follows:
Cube 1: 0 1 3 5 7 2
Cube 2: 0 2 4 6 8 1
By turing the '6' upside down, we create a '9'.
Using this, we can create the numbers needed.
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Snack
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Posts: 3
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Re: Calendar Cubes I
« Reply #1 on: Aug 21st, 2002, 2:46pm » |
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This can also be solved using a base 6 notation.
Cube 1: 0 1 2 3 4 5
Cube 2: 0 1 2 3 4 5
Here you can make numbers 0 - 35.
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TonyMo
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Posts: 5
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Re: Calendar Cubes I
« Reply #2 on: Sep 13th, 2002, 5:23am » |
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In fact, so long as both cubes have 0, 1 and 2, then the digits 3, 4, 5, 6, 7 and 8 can be distributed arbitrarily between the cubes.
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RazMeister
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Re: Calendar Cubes I
« Reply #3 on: Nov 12th, 2002, 3:41pm » |
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Yeah guys, so basically this is also possible:
CUBE 1: 012345
CUBE 2: 012678
yeah, again the 6 can b rotated to look like 9....
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