TITLE: Fibonacci number that ends with 2014 zeros?
QUESTION [109 upvotes]: 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.

Related question: Fibonacci modular results

REPLY [3 votes]: 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.