Question: This problem is giving me the hardest time:

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> Prove or disprove that there is a Fibonacci number that ends with 2014 zeros.
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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: 
This is a classic. The Fibonacci sequence $\pmod{m}$ is periodic for any $m$, since there are only a finite number of elements in $\mathbb{Z}\_m\times\mathbb{Z}\_m$, so for two distinct integers $a,b$ we must have $\langle F\_a,F\_{a+1}\rangle\equiv\langle F\_b,F\_{b+1}\rangle \pmod{m}$ as a consequence of the Dirichlet box principle. However, the last condition implies $F\_{a+2}\equiv F\_{b+2}\pmod{m}$ and, by induction, $F\_{a+k}\equiv F\_{b+k}\pmod{m}$. Hence the period of the Fibonacci sequence $\pmod{m}$ is bounded by $m^2$ ($m^2-1$ if we are careful enough to notice that $\langle F\_c,F\_{c+1}\rangle\equiv\langle0,0\rangle\pmod{m}$ is not possible since two consecutive Fibonacci numbers are always coprime). Now it suffices to take $m=10^{2014}$ and notice that $F\_0=0$ to prove that there exists an integer $u\leq 10^{4028}$ such that $F\_u\equiv 0\pmod{m}$, i.e. $F\_u$ ends with at least $2014$ zeroes.

It is also possible to give better estimates for $u$. 

Since $F\_k$ is divisible by $5$ only when $k$ is divisible by $5$ and:

$$ F\_{5k} = F\_k(25 F\_k^4 + 25(-1)^k F\_k^2 + 5),$$
it follows that:
$$ \nu\_5(F\_k) = \nu\_5(k), $$
so $$u\leq 2^{4028}\cdot 5^{2014}=20^{2014}$$ by the Chinese theorem. I put a proof of the Oleg567 conjecture:
$$ k\mid F\_n \quad\Longrightarrow\quad k^d\mid F\_{k^{d-1}n} $$
in a [separate question](https://math.stackexchange.com/questions/872955/an-analogue-of-hensels-lifting-for-fibonacci-numbers). Since $8=F\_6\mid F\_{750}$ (because $6\mid 750$) and $\nu\_5(750)=3$, we have that $1000|F\_{750}$ and through the Oleg567's lemma we get
$$ u\leq \frac{3}{4}10^{2014}.$$
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.