Another Chain in Payment (Posted on 2003-11-17)

	Submitted by Gamer
Rating: 3.0000 (1 votes)
Solution:	(Hide)
The first chain could be split so that the fourth link from one end and sixth link from the other end was cut. This would make chains of 3, 1, 12, 1, 6. 1: Pay with the 1 chain 2: Pay with the 1 chain 3: Take back both 1 chains, pay with the 3 chain 4: Pay with the 1 chain 5: Pay with the 1 chain 6: Take back 1 and 3 chains, pay with the 6 chain 7: Pay with the 1 chain 8: Pay with the 1 chain 9: Take back both 1 chains , pay with the 3 chain 10: Pay with the 1 chain 11: Pay with the 1 chain 12: Take back 1, 3 and 6 chains, pay with the 12 chain 13: Pay with the 1 chain 14: Pay with the 1 chain 15: Take back both 1 chains, pay with the 3 chain 16: Pay with the 1 chain 17: Pay with the 1 chain 18: Take back 1 and 3 chains, pay with the 6 chain 19: Pay with the 1 chain 20: Pay with the 1 chain 21: Take back both 1 chains, pay with the 3 chain 22: Pay with the 1 chain 23: Pay with the 1 chain 4, 1, 16, 1, 32, 1, 8 is the next set. 1: Pay with a 1 link 2: Pay with a 1 link 3: Pay with a 1 link 4: Take back all 1 links, pay with the 4 link 5: Pay with a 1 link 6: Pay with a 1 link 7: Pay with a 1 link 8: Take back the 4 link and the 1 links, pay with the 8 link... Paying by continuing on with powers of two would work for this situation. Fwaff figured that (2^(x+1))(x+1) - 1 would work as a formula.

	Subject	Author	Date
	re: Partial Solution	AlexSeen	2024-08-23 09:50:24
	not sufficient information	donnmike	2003-12-06 02:22:32
	Comment to Levik	S	2003-11-17 10:45:11
	The last little bit...	fwaff	2003-11-17 10:10:52
	Most of the rest of the solution	fwaff	2003-11-17 09:57:00
	Partial Solution	Sam	2003-11-17 09:07:31