The strange clock (Posted on 2004-02-17)

	Submitted by Tristan
Rating: 2.6667 (6 votes)
Solution:	(Hide)
I will call the hands A, B, and C, from fastest to slowest. So A goes around in 3 minutes, B in 4, and C in 6. This puzzle only deals with integer minutes, because all hands point directly at numbers only every minute. Hand B has only three possibilities. It could have moved from 1 to 10, 2 to 11, or 3 to 6. The number of minutes it takes mod 4 is 1 or 3. Hand C, in that time could have moved 2,6, or 10 numbers over in that time. The only possibility left considering this is from 1 to 11. Hand B is left only with the possibility of 3 to 6, and hand A goes from 2 to 10. So this is what happened: Hand A went from 2 to 10, Hand B went from 3 to 6, Hand C went from 1 to 11, The time this took mod 12 is 5.

	Subject	Author	Date
	Yeppers	John	2004-02-24 23:59:07
	Solution	sherif	2004-02-23 18:51:19
	I think me knows it!	Mitch Mullings	2004-02-22 19:59:31
	i think i got it :)	evan	2004-02-21 03:08:57
	unique solution	pierre	2004-02-19 14:44:58
	re(3): Solution	Charlie	2004-02-18 15:31:10
	re(2): Solution	Mital	2004-02-18 13:12:58
	re: Solution	SilverKnight	2004-02-18 12:59:10
	Solution	Mital	2004-02-18 12:57:35
	my solution -no calculators	Ady TZIDON	2004-02-18 03:49:37
	Solution	Gabe	2004-02-17 13:03:22
	solution	Charlie	2004-02-17 11:46:54
	Solution	Wengel	2004-02-17 10:54:36