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        riddles >> hard >> CALENDAR CUBES 
        
(Message started by: casual_kumar on Aug 7^th, 2006, 3:06am)

Title: CALENDAR CUBES
Post by casual_kumar on Aug 7^th, 2006, 3:06am

0, 1, and 2 make the 10th place of the dates and hence they must be present on both the cubes. As any single cube is incapable of having all the digits for the 1s place.

So we are left with 3 faces on each of the cube. In which we can accomodate 3, 4, 5 and 6, 7, 8.

Still, we cannot accomodate 9. However, can we use 6 as 9 also?? If yes, my answer is correct.

Title: Re: CALENDAR CUBES
Post by Icarus on Aug 7^th, 2006, 3:34pm

You should be familiar with the calendar, and you obviously know what 6 and 9 look like. So you should be able to answer the question

Quote:

can we use 6 as 9 also??

yourself.

Title: Re: CALENDAR CUBES
Post by casual_kumar on Aug 8^th, 2006, 12:51am

I was just wondering if such tricks are allowed. Generally, these things are not called "real" solutions.

Title: Re: CALENDAR CUBES
Post by Grimbal on Aug 8^th, 2006, 3:48am

Considering that the problem is expressed in terms of real cubes that you need to rearrange, I would say that the trick is implicitly allowed.

It is easy to prove that without some trick, there is no solution. Once you have gotten so far, you have to get the "aha!" and realize what trick can pull you out of it.

This is called "thinking outside of the box", the box being the artificial limits that you set yourself on the problem.

Title: Re: CALENDAR CUBES
Post by casual_kumar on Aug 8^th, 2006, 4:06am

Very true.
Thanks