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:
$$
\large{
k|F\_n \quad \Rightarrow \quad
k^d|F\_{k^{d-1}n}.
}$$
Let $\ell\_n$ be the last $2014$ digits of of the $n$th of the Fibonacci sequence. Note if we know $\ell\_{k+1}$ and $\ell\_k$, then we know $\ell\_{k-1}$. So if for two natural numbers $n,k$ we find that $\ell\_n=\ell\_{n+k}$ and $\ell\_{k+1}=\ell\_{n+k+1}$, then it follows that 
$$\begin{align\*}\ell\_{k-1}&=\ell\_{n+k-1}\\\ell\_{k-2}&=\ell\_{n+k-2} \\ &\dots \\\ell\_1&=\ell\_{n+1} \end{align\*}$$And since $\ell\_1=0$, it follows that $\ell\_{n+1}=0$ and hence the $(n+1)$st Fibonacci number will end in $2014$ zeroes.

Now let us prove that such a number exists. Consider the following pairs: $$\begin{align\*} \ell\_1&,\ell\_2 \\ \ell\_{2}&,\ell\_3 \\ &\vdots \\ \ell\_{10^{48}}&, \ell\_{10^{48}+1} \\ \ell\_{10^{48}+1}&, \ell\_{10^{48}+2} \end{align\*}$$Since there are $\left({10^{24}}\right)^2$ different values of the pairs, by the Pigeonhole Principle, there exists $i,j$ where $1<i<j\leq 10^{48}+1$ such $\ell\_{i}=\ell\_{j}$ and $\ell\_{i+1}=\ell\_{j+1}$. It follows that $(\ell\_1,\ell\_2)=(\ell\_{j-i+1},\ell\_{j-i})$ hence the $j-i+1$st Fibonacci number ends in $2014$ zeroes.