Four Digit Number II (Posted on 2003-07-06)

	Submitted by Gamer
Rating: 3.0000 (4 votes)
Solution:	(Hide)
So far, the equation is: _abcd +dcba ----- abcd0 It must end in 0 in order to come out with no decimal. a must equal 0 or 1. If a = 0, d = 0, and d = a, which is not allowed. So a must equal 1. _1bcd +dcb1 ----- 1bcd0 Since d+1=0, d must equal 9. _1bc9 +9cb1 ----- 1bc90 Now, b must equal 1 or 0. If b = 1, then a = b, which is not allowed. So b = 0. _10c9 +9c01 ----- 10c90 Noting that 9+1 = 0 carries a 1, 1+c+0 = 9, c = 8, so: _1089 +9801 ----- 10890 This shows that 1089 is the only possible solution.

	Subject	Author	Date
	Puzzle Resolution	K Sengupta	2007-07-11 09:20:43
	Visual approach	Steve Herman	2004-10-23 14:04:19
	solution	ben young	2003-07-14 03:03:59
	re: Non-algorithmic solution	Gamer	2003-07-12 08:28:02
	answer	kevin	2003-07-08 10:06:03
	re: Non-algorithmic solution	Jim C	2003-07-07 08:22:05
	Thank God I have all of MY digits!	Jim C	2003-07-07 08:14:27
	Solution (Algebraic)	ryan smith	2003-07-06 20:04:10
	Non-algorithmic solution	TomM	2003-07-06 13:02:15
	re(2): Solution	Charlie	2003-07-06 06:09:26
	re: Solution	Charlie	2003-07-06 06:07:14
	Solution	Lewis	2003-07-06 04:45:24