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: 
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)
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.