Math Quickie

[Lepton]

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]

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]

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]

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]

So who's right?

[ralphmerridew]

Well, I'm wrong because I overlooked the part that said X is 1-9.

[Lepton*]

Jack_Ian is right, I think. I overlooked the XYYY000 case (shameful).

Oh, there's an emoticon for that.

[Zag]

But why would we assume ZZZ can't be 000?

[Jack_Ian]

Because it's a number-plate that exists in the real world and not just in Puzzlemania.