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Lepton
1:41+ Arse Scratcher

 Posted: Fri Jul 16, 2010 9:45 am    Post subject: 1 California license plates have the form X-YYY-ZZZ where X is a digit [1-9], Y is a letter [A-Z], and Z is a digit [0-9]. A fun game to play, while waiting in Los Angeles traffic, is to see if the digit X is a factor of the number ZZZ. For example, in the plate 4ABC108 this is true, and in the plate 3DEF400 it is not true. What percentage of all possible license plates of this format satisfy the factorization criteria?
Quailman
His Postmajesty

 Posted: Fri Jul 16, 2010 10:20 am    Post subject: 2 Well, there are two possible outcomes - either it is or it isn't. So the answer is 1/2, or 50%. That was quick.
ralphmerridew
Daedalian Member

 Posted: Fri Jul 16, 2010 10:34 am    Post subject: 3 If the first digit is 0, there are no ways to fill the other 3 digits to get such a plate. If the first digit is d > 0, there are floor (1000/d) ways. Total probability is (0 + floor (1000/1) + floor (1000/2) + ... + floor (1000/9)) / 10000 == (0 + 1000 + 500 + 333 + 250 + 200 + 166 + 142 + 128 + 111) / 1000 == 2827/10000 == 28.27%
Jack_Ian
Big Endian

 Posted: Fri Jul 16, 2010 2:35 pm    Post subject: 4 Assuming ZZZ cannot be 000… 100 x (999/1 + 998/2 + 999/3 + 996/4 + 995/5 + + 996/6 +994/7 + 992/8 + 999/9)/(9 x 999) = 31.387%
Jack_Ian
Big Endian

 Posted: Thu Jul 22, 2010 10:59 pm    Post subject: 5 So who's right?
ralphmerridew
Daedalian Member

 Posted: Fri Jul 23, 2010 1:12 am    Post subject: 6 Well, I'm wrong because I overlooked the part that said X is 1-9.
Lepton*
Guest

 Posted: Fri Jul 23, 2010 5:58 am    Post subject: 7 Jack_Ian is right, I think. I overlooked the XYYY000 case (shameful). Oh, there's an emoticon for that.
Zag
Tired of his old title

 Posted: Fri Jul 23, 2010 6:12 am    Post subject: 8 But why would we assume ZZZ can't be 000?
Jack_Ian
Big Endian

 Posted: Fri Jul 23, 2010 6:33 am    Post subject: 9 Because it's a number-plate that exists in the real world and not just in Puzzlemania.
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