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Define the sequence of Fibonacci numbers as: $F_1$ = $F_2$ = 1 and $F_n$ = $F_{n−1}$ + $F_{n−2}$ for every n > 2. Prove that, for any positive integer $k$, there is a Fibonacci number ending with $k$ zeroes

I am able to figure out that this problem is a application of PigeonHole Principle. I can't figure out how to approach it or what will be the reasoning. Can someone help me out? A little bit wordy and rigorous proof/hint is appreciable.

Maths lover
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    I don't see how php would help. Nonetheless, this is an interesting problem! – Prasun Biswas Dec 06 '20 at 05:29
  • I may be wrong.... If there is any other way out... Kindly specify – Maths lover Dec 06 '20 at 05:33
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    Checking with a python script gives the first instance with 1 trailing zero as $F_{15}$, with 2 trailing zeros as $F_{150}$, with 3 trailing zeros as $F_{750}$, with 4 trailing zeros as $F_{7500}$, with 5 trailing zeros as $F_{75000}$,... – Prasun Biswas Dec 06 '20 at 05:40
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    Perhaps induction might help. – Prasun Biswas Dec 06 '20 at 05:46
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    @Prasun Biswas can you mention a rigorous proof in the answèr secton....I am eager to know how this probkem will go –  Dec 06 '20 at 05:47
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    @PrasunBiswas: I don’t know whether it’s the first one with $6$ trailing zeroes, but this tool says that $F_{750000}$ does have $6$. – Brian M. Scott Dec 06 '20 at 06:00
  • How do we generalise – Maths lover Dec 06 '20 at 06:01
  • @BrianM.Scott: I think we can show using Binet's fibonacci formula that $F_{75\cdot 10^k}$ has $k+2$ trailing zeros for $k\geq 1$ and use $F_{15}$ and $F_{150}$ for the example with 1 and 2 trailing zero respectively. But this would surely be tedious... – Prasun Biswas Dec 06 '20 at 06:05
  • Can you kindly do the same...i need to learn...I mean rhe Binets formula part –  Dec 06 '20 at 06:10
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    I think u can read https://math.stackexchange.com/q/872955/770472 to explore more. – kuspia Dec 06 '20 at 06:16
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    https://math.stackexchange.com/q/872071/851269....this can be helpful @Prasun Biswas – Maths lover Dec 06 '20 at 06:16
  • @solver: that thread basically solves this question and more: it generalizes to a stronger result that $k\mid F_n\implies k^d\mid F_{n\cdot k^{d-1}}~\forall~d\geq 1$. The problem here is the special case $k=10,n=15$; pretty neat result! Thanks for finding this! – Prasun Biswas Dec 06 '20 at 06:26
  • @BrianM.Scott: would it be appropriate to close this as a dupe? The linked thread by solver essentially solves this problem but since it is not an exact duplicate but a stronger result, I wonder if closing as dupe is the right call... – Prasun Biswas Dec 06 '20 at 06:30
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    @PrasunBiswas: That’s a hard call. I’m generally reluctant to close when there’s that much difference between the actual questions, but between that one and the one that you found, this question is certainly answered. For now I think that I’ll wait and see how others feel about, since my vote would close it automatically. – Brian M. Scott Dec 06 '20 at 07:27

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