The sum of any ten consecutive numbers from a fibonacci sequence is divisible by 11 [on hold]Fibonacci and Lucas identityWhy is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?Fibonacci sequence divisible by 7?How to prove gcd of consecutive Fibonacci numbers is 1?Fibonacci numbers divisible by $9$The sum of $n$ consecutive Fibonacci numbers.Product of consecutive Fibonacci numbers divisibilityWhy is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?Fibonacci sequence divisible by 3?Any non-negative integer can be written as the sum of distinct and non-consecutive Fibonacci numbersThe sum of 8 consecutive Fibonacci numbers is not a Fibonacci numberFibonacci sequence from natural numbers

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The sum of any ten consecutive numbers from a fibonacci sequence is divisible by 11 [on hold]


Fibonacci and Lucas identityWhy is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?Fibonacci sequence divisible by 7?How to prove gcd of consecutive Fibonacci numbers is 1?Fibonacci numbers divisible by $9$The sum of $n$ consecutive Fibonacci numbers.Product of consecutive Fibonacci numbers divisibilityWhy is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?Fibonacci sequence divisible by 3?Any non-negative integer can be written as the sum of distinct and non-consecutive Fibonacci numbersThe sum of 8 consecutive Fibonacci numbers is not a Fibonacci numberFibonacci sequence from natural numbers













2












$begingroup$


How do I prove that any ten consecutive numbers of a fibonacci sequence is divisible by 11?






divisibility fibonacci-numbers







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New contributor




Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







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put on hold as off-topic by Javi, Mike Pierce, John Douma, YiFan, heropup 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Javi, Mike Pierce, John Douma, YiFan, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    I mean, just bash? Use the recursive formula to show it.
    $endgroup$
    – Don Thousand
    2 days ago






  • 1




    $begingroup$
    See also here: math.stackexchange.com/questions/60049/…
    $endgroup$
    – Jonas Lenz
    2 days ago










  • $begingroup$
    How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
    $endgroup$
    – lab bhattacharjee
    2 days ago










  • $begingroup$
    Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
    $endgroup$
    – Quuxplusone
    2 days ago















2












$begingroup$


How do I prove that any ten consecutive numbers of a fibonacci sequence is divisible by 11?






divisibility fibonacci-numbers







share|cite|improve this question







New contributor




Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$



put on hold as off-topic by Javi, Mike Pierce, John Douma, YiFan, heropup 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Javi, Mike Pierce, John Douma, YiFan, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.















  • $begingroup$
    I mean, just bash? Use the recursive formula to show it.
    $endgroup$
    – Don Thousand
    2 days ago






  • 1




    $begingroup$
    See also here: math.stackexchange.com/questions/60049/…
    $endgroup$
    – Jonas Lenz
    2 days ago










  • $begingroup$
    How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
    $endgroup$
    – lab bhattacharjee
    2 days ago










  • $begingroup$
    Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
    $endgroup$
    – Quuxplusone
    2 days ago













2












2








2


2



$begingroup$


How do I prove that any ten consecutive numbers of a fibonacci sequence is divisible by 11?






divisibility fibonacci-numbers







share|cite|improve this question







New contributor




Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




How do I prove that any ten consecutive numbers of a fibonacci sequence is divisible by 11?






divisibility fibonacci-numbers




divisibility fibonacci-numbers






share|cite|improve this question







New contributor




Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question







New contributor




Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question






New contributor




Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 2 days ago









AbigailAbigail

171




171




New contributor




Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




put on hold as off-topic by Javi, Mike Pierce, John Douma, YiFan, heropup 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Javi, Mike Pierce, John Douma, YiFan, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.







put on hold as off-topic by Javi, Mike Pierce, John Douma, YiFan, heropup 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Javi, Mike Pierce, John Douma, YiFan, heropup
If this question can be reworded to fit the rules in the help center, please edit the question.











  • $begingroup$
    I mean, just bash? Use the recursive formula to show it.
    $endgroup$
    – Don Thousand
    2 days ago






  • 1




    $begingroup$
    See also here: math.stackexchange.com/questions/60049/…
    $endgroup$
    – Jonas Lenz
    2 days ago










  • $begingroup$
    How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
    $endgroup$
    – lab bhattacharjee
    2 days ago










  • $begingroup$
    Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
    $endgroup$
    – Quuxplusone
    2 days ago
















  • $begingroup$
    I mean, just bash? Use the recursive formula to show it.
    $endgroup$
    – Don Thousand
    2 days ago






  • 1




    $begingroup$
    See also here: math.stackexchange.com/questions/60049/…
    $endgroup$
    – Jonas Lenz
    2 days ago










  • $begingroup$
    How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
    $endgroup$
    – lab bhattacharjee
    2 days ago










  • $begingroup$
    Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
    $endgroup$
    – Quuxplusone
    2 days ago















$begingroup$
I mean, just bash? Use the recursive formula to show it.
$endgroup$
– Don Thousand
2 days ago




$begingroup$
I mean, just bash? Use the recursive formula to show it.
$endgroup$
– Don Thousand
2 days ago




1




1




$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
2 days ago




$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
2 days ago












$begingroup$
How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
$endgroup$
– lab bhattacharjee
2 days ago




$begingroup$
How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
$endgroup$
– lab bhattacharjee
2 days ago












$begingroup$
Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
$endgroup$
– Quuxplusone
2 days ago




$begingroup$
Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
$endgroup$
– Quuxplusone
2 days ago










2 Answers
2






active

oldest

votes


















16












$begingroup$

Suggestion:
If the first two numbers in the sequences are $a, b$, then you can use the Fibonacci recursion to generate the next $8$ numbers. Then add them all up and see what you get. (Your answer will be in terms of $a$ and $b$.)






share|cite|improve this answer









$endgroup$








  • 9




    $begingroup$
    For the lazy, the sum is 55a + 88b. ;)
    $endgroup$
    – Mukul Gupta
    2 days ago


















3












$begingroup$

If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod11$$



And thus $$sum_k=0^9 F_n+kequiv8cdot 4^ncdot(-7)cdot(4^10-1)-8cdot3^ncdot8cdot((-3)^10-1)pmod11$$



By Fermat's theorem, $a^10equiv1pmod 11$ for all $anotequiv0pmod11$.



More generally, the sum of $p-1$ consecutive Fibonacci numbers is divisible by the prime $p$ as soon as the polynomial $x^2-x-1$ is reducible in $Bbb F_p[x]$ (and $1$ is not a root, which can never occur).






share|cite|improve this answer











$endgroup$








  • 4




    $begingroup$
    I can't help but read that as "for all values of 11 which do not divide a".
    $endgroup$
    – hobbs
    2 days ago










  • $begingroup$
    @hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
    $endgroup$
    – J.G.
    2 days ago











  • $begingroup$
    @J.G. ... or $anotequiv0pmod11$, for that matter.
    $endgroup$
    – Saucy O'Path
    2 days ago










  • $begingroup$
    @SaucyO'Path It'd certainly be more consistent with the rest of your formalism. You may as well edit it in, then.
    $endgroup$
    – J.G.
    2 days ago

















2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









16












$begingroup$

Suggestion:
If the first two numbers in the sequences are $a, b$, then you can use the Fibonacci recursion to generate the next $8$ numbers. Then add them all up and see what you get. (Your answer will be in terms of $a$ and $b$.)






share|cite|improve this answer









$endgroup$








  • 9




    $begingroup$
    For the lazy, the sum is 55a + 88b. ;)
    $endgroup$
    – Mukul Gupta
    2 days ago















16












$begingroup$

Suggestion:
If the first two numbers in the sequences are $a, b$, then you can use the Fibonacci recursion to generate the next $8$ numbers. Then add them all up and see what you get. (Your answer will be in terms of $a$ and $b$.)






share|cite|improve this answer









$endgroup$








  • 9




    $begingroup$
    For the lazy, the sum is 55a + 88b. ;)
    $endgroup$
    – Mukul Gupta
    2 days ago













16












16








16





$begingroup$

Suggestion:
If the first two numbers in the sequences are $a, b$, then you can use the Fibonacci recursion to generate the next $8$ numbers. Then add them all up and see what you get. (Your answer will be in terms of $a$ and $b$.)






share|cite|improve this answer









$endgroup$



Suggestion:
If the first two numbers in the sequences are $a, b$, then you can use the Fibonacci recursion to generate the next $8$ numbers. Then add them all up and see what you get. (Your answer will be in terms of $a$ and $b$.)







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 2 days ago









paw88789paw88789

29.6k12351




29.6k12351







  • 9




    $begingroup$
    For the lazy, the sum is 55a + 88b. ;)
    $endgroup$
    – Mukul Gupta
    2 days ago












  • 9




    $begingroup$
    For the lazy, the sum is 55a + 88b. ;)
    $endgroup$
    – Mukul Gupta
    2 days ago







9




9




$begingroup$
For the lazy, the sum is 55a + 88b. ;)
$endgroup$
– Mukul Gupta
2 days ago




$begingroup$
For the lazy, the sum is 55a + 88b. ;)
$endgroup$
– Mukul Gupta
2 days ago











3












$begingroup$

If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod11$$



And thus $$sum_k=0^9 F_n+kequiv8cdot 4^ncdot(-7)cdot(4^10-1)-8cdot3^ncdot8cdot((-3)^10-1)pmod11$$



By Fermat's theorem, $a^10equiv1pmod 11$ for all $anotequiv0pmod11$.



More generally, the sum of $p-1$ consecutive Fibonacci numbers is divisible by the prime $p$ as soon as the polynomial $x^2-x-1$ is reducible in $Bbb F_p[x]$ (and $1$ is not a root, which can never occur).






share|cite|improve this answer











$endgroup$








  • 4




    $begingroup$
    I can't help but read that as "for all values of 11 which do not divide a".
    $endgroup$
    – hobbs
    2 days ago










  • $begingroup$
    @hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
    $endgroup$
    – J.G.
    2 days ago











  • $begingroup$
    @J.G. ... or $anotequiv0pmod11$, for that matter.
    $endgroup$
    – Saucy O'Path
    2 days ago










  • $begingroup$
    @SaucyO'Path It'd certainly be more consistent with the rest of your formalism. You may as well edit it in, then.
    $endgroup$
    – J.G.
    2 days ago















3












$begingroup$

If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod11$$



And thus $$sum_k=0^9 F_n+kequiv8cdot 4^ncdot(-7)cdot(4^10-1)-8cdot3^ncdot8cdot((-3)^10-1)pmod11$$



By Fermat's theorem, $a^10equiv1pmod 11$ for all $anotequiv0pmod11$.



More generally, the sum of $p-1$ consecutive Fibonacci numbers is divisible by the prime $p$ as soon as the polynomial $x^2-x-1$ is reducible in $Bbb F_p[x]$ (and $1$ is not a root, which can never occur).






share|cite|improve this answer











$endgroup$








  • 4




    $begingroup$
    I can't help but read that as "for all values of 11 which do not divide a".
    $endgroup$
    – hobbs
    2 days ago










  • $begingroup$
    @hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
    $endgroup$
    – J.G.
    2 days ago











  • $begingroup$
    @J.G. ... or $anotequiv0pmod11$, for that matter.
    $endgroup$
    – Saucy O'Path
    2 days ago










  • $begingroup$
    @SaucyO'Path It'd certainly be more consistent with the rest of your formalism. You may as well edit it in, then.
    $endgroup$
    – J.G.
    2 days ago













3












3








3





$begingroup$

If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod11$$



And thus $$sum_k=0^9 F_n+kequiv8cdot 4^ncdot(-7)cdot(4^10-1)-8cdot3^ncdot8cdot((-3)^10-1)pmod11$$



By Fermat's theorem, $a^10equiv1pmod 11$ for all $anotequiv0pmod11$.



More generally, the sum of $p-1$ consecutive Fibonacci numbers is divisible by the prime $p$ as soon as the polynomial $x^2-x-1$ is reducible in $Bbb F_p[x]$ (and $1$ is not a root, which can never occur).






share|cite|improve this answer











$endgroup$



If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod11$$



And thus $$sum_k=0^9 F_n+kequiv8cdot 4^ncdot(-7)cdot(4^10-1)-8cdot3^ncdot8cdot((-3)^10-1)pmod11$$



By Fermat's theorem, $a^10equiv1pmod 11$ for all $anotequiv0pmod11$.



More generally, the sum of $p-1$ consecutive Fibonacci numbers is divisible by the prime $p$ as soon as the polynomial $x^2-x-1$ is reducible in $Bbb F_p[x]$ (and $1$ is not a root, which can never occur).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









Saucy O'PathSaucy O'Path

6,3521627




6,3521627







  • 4




    $begingroup$
    I can't help but read that as "for all values of 11 which do not divide a".
    $endgroup$
    – hobbs
    2 days ago










  • $begingroup$
    @hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
    $endgroup$
    – J.G.
    2 days ago











  • $begingroup$
    @J.G. ... or $anotequiv0pmod11$, for that matter.
    $endgroup$
    – Saucy O'Path
    2 days ago










  • $begingroup$
    @SaucyO'Path It'd certainly be more consistent with the rest of your formalism. You may as well edit it in, then.
    $endgroup$
    – J.G.
    2 days ago












  • 4




    $begingroup$
    I can't help but read that as "for all values of 11 which do not divide a".
    $endgroup$
    – hobbs
    2 days ago










  • $begingroup$
    @hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
    $endgroup$
    – J.G.
    2 days ago











  • $begingroup$
    @J.G. ... or $anotequiv0pmod11$, for that matter.
    $endgroup$
    – Saucy O'Path
    2 days ago










  • $begingroup$
    @SaucyO'Path It'd certainly be more consistent with the rest of your formalism. You may as well edit it in, then.
    $endgroup$
    – J.G.
    2 days ago







4




4




$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
2 days ago




$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
2 days ago












$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
2 days ago





$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
2 days ago













$begingroup$
@J.G. ... or $anotequiv0pmod11$, for that matter.
$endgroup$
– Saucy O'Path
2 days ago




$begingroup$
@J.G. ... or $anotequiv0pmod11$, for that matter.
$endgroup$
– Saucy O'Path
2 days ago












$begingroup$
@SaucyO'Path It'd certainly be more consistent with the rest of your formalism. You may as well edit it in, then.
$endgroup$
– J.G.
2 days ago




$begingroup$
@SaucyO'Path It'd certainly be more consistent with the rest of your formalism. You may as well edit it in, then.
$endgroup$
– J.G.
2 days ago



-

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Marilyn Monroe Ny fiainany manokana | Jereo koa | Meny fitetezanafanitarana azy.

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OlonaBiôgrafiaVangovangoMpilalaoMpilalao amin'ny horonantsaryMpamoaka horonantsaryTeraka tamin'ny taona 1926Maty tamin'ny taona 1962 EtazoniaLos Angeles fanitarana azy . Marilyn Monroe Avy amin'i Wikipedia Sauter à la navigation Sauter à la recherche Marilyn Monroe Ankapobeny Anarana Marilyn Monroe Teraka 1 Jiona 1926 Maty 5 Aogositra 1962 Fiaviana sy ny andraikitra Firenena Etazonia Asa : mpilalao mpilalao amin'ny horonantsary mpamoaka horonantsary Fiainana manokana Marilyn Monroe dia mpilalao, mpilalao amin'ny horonantsary, mpamoaka horonantsary mizaka ny zom-pirenen'i Etazonia teraka ny 1 Jiona 1926 tao Los Angeles ary maty ny 5 Aogositra 1962 Ny fiainany manokana | Ny vadiny dia James Dougherty, Joe DiMaggio, Arthur Miller. Jereo koa | Biôgrafia Rohy ivelany | Ao amin'i Freebase: [1] Mbola ambangovangony ity lahatsoratra ity ary tokony hofenoina . Azonao atao ny mandray anjara eto amin'ny Wikipedia amin...
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