The sum of any ten consecutive numbers from a fibonacci sequence is divisible by 11 The Next CEO of Stack OverflowFibonacci 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
The Next CEO of Stack OverflowFibonacci 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
$begingroup$
How do I prove that any ten consecutive numbers of a fibonacci sequence is divisible by 11?
divisibility fibonacci-numbers
asked 9 hours ago
AbigailAbigail
171
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add a comment |
$begingroup$
How do I prove that any ten consecutive numbers of a fibonacci sequence is divisible by 11?
divisibility fibonacci-numbers
asked 9 hours ago
AbigailAbigail
171
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Abigail is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
$begingroup$
I mean, just bash? Use the recursive formula to show it.
$endgroup$
– Don Thousand
9 hours ago
1
$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
9 hours ago
$begingroup$
How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
$endgroup$
– lab bhattacharjee
9 hours ago
$begingroup$
Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
$endgroup$
– Quuxplusone
2 hours ago
add a comment |
$begingroup$
How do I prove that any ten consecutive numbers of a fibonacci sequence is divisible by 11?
divisibility fibonacci-numbers
asked 9 hours ago
AbigailAbigail
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.
$endgroup$
How do I prove that any ten consecutive numbers of a fibonacci sequence is divisible by 11?
divisibility fibonacci-numbers
divisibility fibonacci-numbers
asked 9 hours ago
AbigailAbigail
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.
asked 9 hours ago
AbigailAbigail
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.
asked 9 hours ago
AbigailAbigail
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.
asked 9 hours ago
AbigailAbigail
171
asked 9 hours 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.
$begingroup$
I mean, just bash? Use the recursive formula to show it.
$endgroup$
– Don Thousand
9 hours ago
1
$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
9 hours ago
$begingroup$
How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
$endgroup$
– lab bhattacharjee
9 hours ago
$begingroup$
Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
$endgroup$
– Quuxplusone
2 hours ago
add a comment |
$begingroup$
I mean, just bash? Use the recursive formula to show it.
$endgroup$
– Don Thousand
9 hours ago
1
$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
9 hours ago
$begingroup$
How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
$endgroup$
– lab bhattacharjee
9 hours ago
$begingroup$
Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
$endgroup$
– Quuxplusone
2 hours ago
$begingroup$
I mean, just bash? Use the recursive formula to show it.
$endgroup$
– Don Thousand
9 hours ago
$begingroup$
I mean, just bash? Use the recursive formula to show it.
$endgroup$
– Don Thousand
9 hours ago
1
1
$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
9 hours ago
$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
9 hours ago
$begingroup$
How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
$endgroup$
– lab bhattacharjee
9 hours ago
$begingroup$
How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
$endgroup$
– lab bhattacharjee
9 hours ago
$begingroup$
Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
$endgroup$
– Quuxplusone
2 hours ago
$begingroup$
Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
$endgroup$
– Quuxplusone
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $b$.)
answered 9 hours ago
paw88789paw88789
29.6k12351
$endgroup$
7
$begingroup$
For the lazy, the sum is55a + 88b. ;)
$endgroup$
– Mukul Gupta
4 hours ago
add a comment |
$begingroup$
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 $11nmid a$.
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).
answered 9 hours ago
Saucy O'PathSaucy O'Path
6,2641627
$endgroup$
3
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
3 hours ago
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
2 hours ago
$begingroup$
@J.G. ... or $anotequiv0pmod11$, for that matter.
$endgroup$
– Saucy O'Path
2 hours 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 hours ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $b$.)
answered 9 hours ago
paw88789paw88789
29.6k12351
$endgroup$
7
$begingroup$
For the lazy, the sum is55a + 88b. ;)
$endgroup$
– Mukul Gupta
4 hours ago
add a comment |
$begingroup$
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 $b$.)
answered 9 hours ago
paw88789paw88789
29.6k12351
$endgroup$
7
$begingroup$
For the lazy, the sum is55a + 88b. ;)
$endgroup$
– Mukul Gupta
4 hours ago
add a comment |
$begingroup$
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 $b$.)
answered 9 hours ago
paw88789paw88789
29.6k12351
$endgroup$
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 $b$.)
answered 9 hours ago
paw88789paw88789
29.6k12351
answered 9 hours ago
paw88789paw88789
29.6k12351
answered 9 hours ago
paw88789paw88789
29.6k12351
answered 9 hours ago
paw88789paw88789
29.6k12351
29.6k12351
7
$begingroup$
For the lazy, the sum is55a + 88b. ;)
$endgroup$
– Mukul Gupta
4 hours ago
add a comment |
7
$begingroup$
For the lazy, the sum is55a + 88b. ;)
$endgroup$
– Mukul Gupta
4 hours ago
7
7
$begingroup$
For the lazy, the sum is
55a + 88b. ;)$endgroup$
– Mukul Gupta
4 hours ago
$begingroup$
For the lazy, the sum is
55a + 88b. ;)$endgroup$
– Mukul Gupta
4 hours ago
add a comment |
$begingroup$
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 $11nmid a$.
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).
answered 9 hours ago
Saucy O'PathSaucy O'Path
6,2641627
$endgroup$
3
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
3 hours ago
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
2 hours ago
$begingroup$
@J.G. ... or $anotequiv0pmod11$, for that matter.
$endgroup$
– Saucy O'Path
2 hours 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 hours ago
add a comment |
$begingroup$
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 $11nmid a$.
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).
answered 9 hours ago
Saucy O'PathSaucy O'Path
6,2641627
$endgroup$
3
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
3 hours ago
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
2 hours ago
$begingroup$
@J.G. ... or $anotequiv0pmod11$, for that matter.
$endgroup$
– Saucy O'Path
2 hours 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 hours ago
add a comment |
$begingroup$
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 $11nmid a$.
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).
answered 9 hours ago
Saucy O'PathSaucy O'Path
6,2641627
$endgroup$
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 $11nmid a$.
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).
answered 9 hours ago
Saucy O'PathSaucy O'Path
6,2641627
edited 9 hours ago
answered 9 hours ago
Saucy O'PathSaucy O'Path
6,2641627
answered 9 hours ago
Saucy O'PathSaucy O'Path
6,2641627
answered 9 hours ago
Saucy O'PathSaucy O'Path
6,2641627
6,2641627
3
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
3 hours ago
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
2 hours ago
$begingroup$
@J.G. ... or $anotequiv0pmod11$, for that matter.
$endgroup$
– Saucy O'Path
2 hours 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 hours ago
add a comment |
3
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
3 hours ago
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
2 hours ago
$begingroup$
@J.G. ... or $anotequiv0pmod11$, for that matter.
$endgroup$
– Saucy O'Path
2 hours 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 hours ago
3
3
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
3 hours ago
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
3 hours ago
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
2 hours ago
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
2 hours ago
$begingroup$
@J.G. ... or $anotequiv0pmod11$, for that matter.
$endgroup$
– Saucy O'Path
2 hours ago
$begingroup$
@J.G. ... or $anotequiv0pmod11$, for that matter.
$endgroup$
– Saucy O'Path
2 hours 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 hours 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 hours ago
add a comment |
Abigail is a new contributor. Be nice, and check out our Code of Conduct.
Abigail is a new contributor. Be nice, and check out our Code of Conduct.
Abigail is a new contributor. Be nice, and check out our Code of Conduct.
Abigail is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
I mean, just bash? Use the recursive formula to show it.
$endgroup$
– Don Thousand
9 hours ago
1
$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
9 hours ago
$begingroup$
How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
$endgroup$
– lab bhattacharjee
9 hours ago
$begingroup$
Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
$endgroup$
– Quuxplusone
2 hours ago