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(2^{N})
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(2^{100}) 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(2^{100})would have to be 70 or 160 or 250, etc..
2^{100} 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 20150412 10:39:57 