N is a positive integer and F(N) denotes the sum of the base ten digits of N.
Find F(F(F(2^{2016}))) and F(F(F(F(2^{2016}))))
*** As an extra challenge, solve the puzzle without using a computer program assisted method.
The process of repeatedly applying F eventually gives the digital root of 2^2016.
Powers of 2 have digital roots in a cycle of length 6. 2016 is divisible by 6 so the digital root of 2^2016 = the digital root of 2^0 = 1.
The question is then whether F has been applied enough times.
F(2^2016) should have be about 4.5*2016*log(2) = 2731.
F(F(2^2016)) will be a two digit number [unless F(2^2016) happens to be a power of 10]
F(F(F(2^2016))) will be 10
F(F(F(F(2^2016)))) will be 1.
Checking
https://oeis.org/A001370/b001370.txt
gives
F(2^2016) = 2656
F(F(2^2016)) = 19
F(F(F(2^2016))) = 10
F(F(F(F(2^2016)))) = 1

Posted by Jer
on 20150331 12:50:47 