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Math Help - Fibbonaci Puzzler

  1. #1
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    Fibbonaci Puzzler

    Prove that there exists a fibbonaci number with 2007 zeros at the end.
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  2. #2
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    Re: Fibbonaci Puzzler

    Surprised no one answered this. Basically the sequence is periodic for mod 10^k. I'm not great at "mathematical" proofs but I'll try to explain why the

    81654467027036406333526832763551927773679097587863 01828103274902654645800068375700423192578874801138 49896633172886417559644153347373906530476511639799 03470572826662271659037360447273031780400848575549 03508099075491216597278597834450044376476326281932 72152447517016200655258545132287805391608994236702 32370210418870367835844101551100928971591752027793 74607868568852438352189365382779980690816983216280 47397573781719260414198780609712500413245349971425 87004390684924423594258925625019876096653599085609 00833537023852992689698182781120930012927796187150 19307793377582295367811015065238563352113043140314 90944085671313823277974734958981557281292085601890 06744176078944577779348337171652625983474768859811 57299089487587648844548714667589147775120730050057 94136877675483582484025033771770603207595920161761 00860244176010390835538357799347775384097791851502 58580435803764905167388254314601808367655065706903 17145676009101910237591110135982587999542354257337 69865643686884677779609681580181367794399263876437 47119348248768178379895525253396714698402834275124 89595937485206340807420157549668224663572188043588 91996801563311872885991207304462688534842842899765 28900852819220986366298798374162295530625566277432 64488853168463514650779309887611114310442114751605 26624148697040922363425273844106276196773808206444 55020825499976294193040707383269299795703071190607 99176489670245238094992146216100081801414489746093 7501st term

    is the one you're looking for.

    (starting n = 1 is 0, n = 2 is 1, n = 3 is 1, n = 4 is 2, n = 5 is 3, n = 6 is 5.... )


    The entire Fibonacci series (F) is determined by two numbers (pairs), and F_n mod 10^k for those numbers also determines the sequence. Because there are a finite number of pairings for x mod 10^k, the sequence created by F_n mod 10^k is periodic.

    The periodicity of the sequence F_n mod 10^k can be determined:
    at k=1 it's 60, k=2 it's 300, and k=3 it's 1500.

    The F_n term immediately following the end of the period in effect "restarts" the mod sequence at 0.

    or in other words, the F_n term where n = m*12*5^k + 1 (where m is any positive integer) will always have k zeros. ( F_[m*12*5^k + 1] mod 10^k = 0 )

    If someone wants to form an actual proof out of that I'd be interested. Note: There are other terms with k zeros at the end besides the (m*12*5^k + 1)th terms.

    Also if any of this is unclear please let me know and I'll do my best.
    Last edited by Stro; June 24th 2011 at 12:51 AM.
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  3. #3
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    Re: Fibbonaci Puzzler

    Actually, wolfram alpha disagrees with me.

    Their 37,500th term (n=1, F_n=1) only has 4 zeroes, and my thinking meant it ought to have had 5. Couldn't edit out the above but I'll add more info if I figure this out. I don't think I'm completely wrong.

    Spoiler:
    48643450298934991745127373431498126302650284091297 91729250358638528232068566519689555390474974235878 55014446731404580526452998299867450933484548818010 03862868708185634546033125178185889015208954813005 99252276278396311090943367609139320905575849677030 69685599206853418989813687719877545331484127152890 55494240017609436553449417214039291977823910697452 91351062035075654058038378141261524647111091210562 30346402557362277562398237342714448598750725903701 91123051892344255548507392123929062693276618677895 77747885080863742556721758941996301755724460709854 57818478754063687825040458950890840607608257639996 04316769791880533421346204730863411877227017205815 33585295803343177142855465828387063146632276646229 76545443966204704581957236386624052272387752470368 71951252303960567706071701405401447635564761904148 41412399265591583848920155868229448707787540020438 62585812279845687068397994578841608956160614872199 35966279130129948385593505374582603400942714638344 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  4. #4
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    Re: Fibbonaci Puzzler

    Quote Originally Posted by Chris11 View Post
    Prove that there exists a Fibonacci (sp!) number with 2007 zeros at the end.
    You want to find N such that 10^n divides the Nth Fibonacci number F_N, where n=2007.

    I haven't tried to prove any of this, but a quick trawl through a table of Fibonacci numbers shows that (for values of N up to 300) 2^n divides F_N whenever N is a multiple of 3\times2^{n-1}, and 5^n divides F_N whenever N is a multiple of 5^n.

    If you can prove that both of those observations hold in general, then it will follow that 10^n divides F_N when N = \tfrac32\times10^n. When n=2007, that is a large number even in comparison with those calculated by Stro. But in fact this approach is just a slight modification of the one proposed by Stro in his first comment.
    Last edited by Opalg; June 26th 2011 at 11:41 PM. Reason: typos
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