Calendar Cubes (Posted on 2002-09-09)

	Submitted by levik
Rating: 3.3750 (8 votes)
Solution:	(Hide)
The trick to this problem is realising that by turning a "6" upside down, we can make a "9". Since there are only a maximum of 31 days in a month, we will never need a 6 and a 9 simultaneously. Because each month as an 11th and a 22nd, both the cubes will need to have the numbers 1 and 2 on them. Also, because each of the digits needs to appear next to a zero (for the first 9 days of the month), zero should also be on both cubes. This leaves us with three free spots on each cube, where we need to put the numbers 3, 4, 5, 6, 7, and 8. So, C1 : 0, 1, 2, 3, 4, 5 C2 : 0, 1, 2, 6, 7, 8

	Subject	Author	Date
	Puzzle Thoughts	K Sengupta	2023-03-21 23:10:28
	re(2): My Solution	Dej Mar	2008-09-16 02:46:06
	re: My Solution	humfoozneydannle	2005-01-02 01:56:36
	solution	humfoozneydannle	2005-01-02 01:52:01
	My Solution	Wendy	2004-02-24 17:40:00
	Neat	Lawrence	2003-08-27 13:51:09
	Reality	Nick Reed	2002-09-13 07:07:40
	re(3): Huh?	levik	2002-09-10 10:12:31
	re(2): Huh?	friedlinguini	2002-09-10 05:09:43
	re: Huh?	levik	2002-09-10 04:52:21
	upside down trick	Dulanjana	2002-09-10 03:40:53
	Huh?	TomM	2002-09-10 03:05:02
	13 digits	TomM	2002-09-09 20:12:27