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        riddles >> hard >> Calendar Cubes I
        
(Message started by: suid on Aug 4^th, 2002, 2:04pm)

Title: Calendar Cubes I
Post by suid on Aug 4^th, 2002, 2:04pm

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.

Title: Re: Calendar Cubes I
Post by Snack on Aug 21^st, 2002, 2:46pm

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.

Title: Re: Calendar Cubes I
Post by TonyMo on Sep 13^th, 2002, 5:23am

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.

Title: Re: Calendar Cubes I
Post by RazMeister on Nov 12^th, 2002, 3:41pm

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....