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http://digicc.com/fido/I know we have a lot of math wizards on here, so anybody who has the time to figure out how this works, please share the secret.

Ok, that's cool. Anyone figure it out?

[spoiler?]not sure this is proper explanationSuppose you take the three digit number, each of the digits represented by a letter, as in "abc" Then you rearrange the digits to get the second number, lets say it's rearranged to be "bca" and if bca is less than abc, you subtract bca from abc and get a new numberin the number abc, it is (a*100) + (b*10) + (c), bca is (b*100) + (c*10) + aso abc - bca = 100a +10b +c - 100b - 10c -awhich is 99a -90b - 9cwhatever a, b, and c are, 99a - 90b - 9c = 9(11a - 10b - c) or a multiple of nine.For any number that is a multiple of nine, the individual digits add up to a multiple of nine ( 72-> 7+2=9, 108-> 1+8=9)Since you pick one of the digits of a multiple of nine and tell the thing the other digits, all it has to do is figure out what non-zero value it needs to add to the sum of the digits to get the digits to add up to a multiple of nine

Cool puzzle. I only had time to surf the 'Net for the solution:http://britton.disted.camosun.bc.ca/jbfido.htm

Cool, thanks Zuben!

what about my answer? I was first and didn't have to look it up was the background colored text too subtle?

Quote, originally posted by joatmon »what about my answer? I was first and didn't have to look it up was the background colored text too subtle?Yep. I just thought you were whoring vertical space. Didn't even see text there until I went back and highlighted it. But yes, you explained the math (algebraic) much better than anywhere else I've seen.