[seqfan] Re: What are the possible digit-sums for Fibonacci numbers?

[Joseph Myers]

On Tue, 26 Dec 2016, israel at math.ubc.ca wrote:

> Well yes, Fibonacci(n) mod 10^k will be periodic, but it will always take 
> values of 0 mod 10^k. It's likely that (for n > 0) these will have lots of 
> other digits making the digit sums large, but I don't see any prospect of 
> controlling those.

Mod 10^k - 1 may be more useful.  If Fibonacci(n) has digit sum 6, so does 
Fibonacci(n) mod 9999 - and no value of Fibonacci(n) mod 9999 has digit 
sum 6.

Eliminating sums larger than 9 may be possible that way, but harder (you 
get more possible sums mod 10^k - 1).  And such a modulus can't eliminate 
11, because one possibility is that the digits in places with values 
10^(0 mod k) add up to 1 and all other nonzero digits are in places with 
values 10^(k-1 mod k) (total of such digits being 10), which results in 
value 2 mod 10^k - 1, which occurs infinitely many times.  (There may 
still be possibilities from combining information mod 10^a with 
information mod 10^b - 1.)

-- 
Joseph S. Myers
jsm at polyomino.org.uk