2013th Fibonacci number/17?

Using WolframAlpha with the input Fib(2013) (mod 17) I get a remainder of 2, and

with the input Fib(200) (mod 5) I get remainder of 0, so the 200th Fibonacci is

divisible by 5.

I'm guessing, though, that you want an analytic approach. I'm still working on that,

and I will also star your question with the hope that someone else might notice and

help out.

Edit: Every kth Fibonacci number is a multiple of F(k). Since F(5) = 5 we have

that F(200) is a multiple of 5.

Edit #2: I appreciate your kind words. :) I'm learning some interesting results

while trying to solve this question. Every 3rd number is even, so since 2013

is divisible by 3 we know that F(2013) is even. Also, we have the property

gcd(F(m), F(n)) = F(gcd(m,n)) to work with.

Now 2013 = 3*11*61, so gcd(F(2013), F(11)) = F(gcd(2013,11)) = F(11) = 89.

Thus F(2013) is divisible by 89, which is prime.

It is also divisible by F(61) = 2,504,730,781,961 = 4513 * 555,003,497.

There has been so much research done on Fibonacci numbers that there must

be some way of solving your question analytically; it will just take some time

sifting through all the information.

Edit #3: The Pisano period for modulo 17 is 36, i.e., the residues have a

period of 36I haven't been able to find this particular sequence of residues yet,

but this is the best approach to this type of problem. There are many

elegant and fascinating properties associated with Fibonacci numbers, but

for your particular question I think we basically have to rely on tables created

by others; there doesn't appear to be a simple formula to apply in this case.

Edit #4: Since 2013 is congruent to 33 (mod 36) and since the Pisano period

is 36, we know that F(2013) will have the same residue mod 17 as F(33).

Now F(33) = 3,524,578 which is congruent to 2 (mod 17), so 2 is the remainder

when we divide F(2013) by 17. Cool. :)