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 a) is easy, b) is not (Posted on 2011-02-18)
1, 2, 4, 8, 1, 3, 6, 1 ... is a non-cyclic series where a(n) represents the leading digit of 2^n.

a) How many ones are there within the first 3000 members?
b) Same question for a digit d, other than digit one.

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 Why a) is easy... | Comment 3 of 8 |
Every time the power of two passes a power of 10 it must give a number starting with 1 exactly once.  It must be at least once since 2/.9999 > 2 so you cant skip over.  It must be at most once since 1*2 = 2 so you cant hit 1 twice.

2^3000 is a 904 digit number (about 1.23*10^903) so there are 904 ones.

Part b) cannot really be done so easily because the other digits can be skipped over.  The best I can do at this point is to give an approximation based on the probability of a digit being skipped.  As the previous poster points out this involves Benfords law.

For the solution Steve gave, he only failed to realize he would need to round his decimal up.  The others (I am pretty sure) will all need to be rounded down.   I do not have time to confirm this.

 Posted by Jer on 2011-02-18 16:51:57

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