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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. 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 $$ u\leq \frac{3}{4}10^{2014}.$$

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. 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 $$ u\leq \frac{3}{4}10^{2014}.$$

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. 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 lemma we get $$ u\leq \frac{3}{4}10^{2014}.$$

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. 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 $$ u\leq \frac{3}{4}10^{2014}.$$

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 $$u\leq 2^{2048}\cdot 5^{2014}=20^{2014}$$ 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. 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 lemma we get $$ u\leq \frac{3}{4}10^{2014}.$$

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. 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 lemma we get $$ u\leq \frac{3}{4}10^{2014}.$$