Question: This problem is giving me the hardest time:

> 
> Prove or disprove that there is a Fibonacci number that ends with 2014 zeros.
> 
> 
> 

I tried mathematical induction (for stronger statement that claims that there is a Fibonacci number that ends in any number of zeroes), but no luck so far.

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Related question: [Fibonacci modular results](https://math.stackexchange.com/questions/60340/fibonacci-modular-results?rq=1)

Answer: 
Just observation:
$$F\_{750}\equiv 0 \pmod{1000}$$
$$F\_{7500}\equiv 0 \pmod{10000}$$
$$F\_{75000}\equiv 0 \pmod{100000}$$
$$F\_{750000}\equiv 0 \pmod{1000000}$$
$$F\_{7500000}\equiv 0 \pmod{10000000}$$

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[Here (Pisano Period)](http://mathworld.wolfram.com/PisanoPeriod.html) is said, that
the sequence of last $d$ digits of Fibonacci numbers has a period of $15·10^{d-1}$.   

Our $75\cdot 10^{d-2}$ is a half-period (knowing that $F\_0=0$). 

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I think must be some property of Fibonacci numbers:
$$
\large{
k|F\_n \quad \Rightarrow \quad
k^d|F\_{k^{d-1}n}.
}$$
Given any n (even one with 2014 trailing zeros) there has to be values b and k so that $F\_b\equiv F\_{b+k} \pmod{n}$ AND $F\_{b+1}\equiv F\_{b+k+1} \pmod{n}$ (which is equivalent to the claim the mod n residues have to eventually form a repeating cycle, and proven in other answers).

Once such b and k values are found, by the "two hop" identity using hops of 1 and k starting from index b we get
$F\_bF\_{b+k+1} = F\_{b+1}F\_{b+k} - (-1)^{b}F\_kF\_1$.

Now, notice that $F\_bF\_{b+k+1}\equiv F\_{b+1}F\_{b+k} \pmod{n}$ by construction, so it must be $F\_k\equiv 0 \pmod{n}$, that is, $F\_k$ has 2014 trailing zeroes.

edit: corrected typo on exponent to (-1)