Pick two numbers! (Posted on 2005-11-22)

	Submitted by e.g.
Rating: 3.8889 (9 votes)
Solution:	(Hide)
Suppose I answered that one of the tickets' digits summed 13. The tickets could have been numbered ABCD and ABC(D+1) with D<9 and you couldn't work out the numbers, since all you'd have is that A+B+C+D=12. So I answered "NO" to the second question, and the numbers aren't ABCD and ABC(D+1). If the tickets had been numbered ABC9 and AB(C+1)0, with C<9, then the sum of the digits would have been even, so this case is impossible. Finally, as the numbers cannot be A999 and (A+1)000 --for the sum exceeds 25-- the tickets must have been AB99 and A(B+1)00, so A+B=3, allowing 0399-0400, 1299-1300, 2199-2200, and 3099-3100. In three of these I would have answered "YES" to the question about a number appearing more than twice; since my daughter could work out the numbers, I must have answered "NO" and the tickets were 1299-1300.

	Subject	Author	Date
	the DAUGHTER= DEUS EX MACHINA	Ady TZIDON	2020-09-02 06:08:27
	Answer	K Sengupta	2008-11-16 14:11:30
	ticket trivia	cassie	2006-01-05 08:32:35
	No Subject	James	2005-11-25 15:53:23
	Answer to the Two Ticket Problem	Gordon Steel	2005-11-24 16:58:55
	*sizzle*	FastJazzy	2005-11-23 22:52:43
	my answer	Scott	2005-11-23 04:11:14
	re: I disagree	James	2005-11-23 00:32:18
	re: I disagree	Mindy Rodriguez	2005-11-23 00:31:38
	I disagree	James	2005-11-23 00:25:25
	re: soln	Mindy Rodriguez	2005-11-22 22:55:36
	my answer	Vernon Lewis	2005-11-22 12:46:55
	Solution	Neil	2005-11-22 12:42:14
	Solution	Roscoe	2005-11-22 12:18:04
	My solution	pcbouhid	2005-11-22 11:48:19
	soln	doubledos	2005-11-22 11:00:52