N is a positive integer and F(N) denotes the sum of the base ten digits of N.

Derive an algorithm to determine the values of N such that F(2^N) is divisible by 10.

The digital roots of the powers of 2 follow the cycle 1,2,4,8,7,5.  (Of course multiples of 3 are not possible so 3,6,9 are not in this cycle.)
So F(2^N) cannot be a multiple of 30.

There doesn't seem to be much of a pattern to those N's that otherwise comply except that as N increases they seem to get more common.

N  F(2N)
 6 10
13 20
26 40
41 50
46 70
51 80
57 80
58 70

OEIS does not contain this sequence, but I did use https://oeis.org/A001370 to compile it.

The problem asks for an algorithm.  I don't know of a simpler way than just directly computing F(2^N).

However, if I were told that a number, X was in the sequence I could probably guess F(2^X)if X isn't too big.

For example if I were told F(2¹⁰⁰) was in the sequence (I don't know if it really is) I'd look at the cycle to see it's digital root is 7.  So F(2¹⁰⁰)would have to be 70 or 160 or 250, etc..

2¹⁰⁰ has 31 digits.

31*an average digit of 4.5=139.5 which is closest to 160.  

However, from the OEIS mentioned above:

It is believed that a(n) ~ n*9*log_10(2)/2, but this is an open problem.

So even this idea may not work for large N.

Posted by Jer on 2015-04-12 10:39:57