The sum of any ten consecutive numbers from a fibonacci sequence is divisible by 11 [on hold]
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How do I prove that any ten consecutive numbers of a fibonacci sequence is divisible by 11?
divisibility fibonacci-numbers
asked Apr 1 at 14:10
AbigailAbigail
171
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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.
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 Apr 1 at 14:10
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.
$endgroup$
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.
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I mean, just bash? Use the recursive formula to show it.
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– Don Thousand
Apr 1 at 14:12
1
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See also here: math.stackexchange.com/questions/60049/…
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– Jonas Lenz
Apr 1 at 14:12
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How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
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– lab bhattacharjee
Apr 1 at 14:21
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Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
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– Quuxplusone
Apr 1 at 21:25
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 Apr 1 at 14:10
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 Apr 1 at 14:10
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 Apr 1 at 14:10
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 Apr 1 at 14:10
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 Apr 1 at 14:10
AbigailAbigail
171
asked Apr 1 at 14:10
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
Apr 1 at 14:12
1
$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
Apr 1 at 14:12
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How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
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– lab bhattacharjee
Apr 1 at 14:21
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Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
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– Quuxplusone
Apr 1 at 21:25
add a comment |
$begingroup$
I mean, just bash? Use the recursive formula to show it.
$endgroup$
– Don Thousand
Apr 1 at 14:12
1
$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
Apr 1 at 14:12
$begingroup$
How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
$endgroup$
– lab bhattacharjee
Apr 1 at 14:21
$begingroup$
Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
$endgroup$
– Quuxplusone
Apr 1 at 21:25
$begingroup$
I mean, just bash? Use the recursive formula to show it.
$endgroup$
– Don Thousand
Apr 1 at 14:12
$begingroup$
I mean, just bash? Use the recursive formula to show it.
$endgroup$
– Don Thousand
Apr 1 at 14:12
1
1
$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
Apr 1 at 14:12
$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
Apr 1 at 14:12
$begingroup$
How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
$endgroup$
– lab bhattacharjee
Apr 1 at 14:21
$begingroup$
How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
$endgroup$
– lab bhattacharjee
Apr 1 at 14:21
$begingroup$
Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
$endgroup$
– Quuxplusone
Apr 1 at 21:25
$begingroup$
Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
$endgroup$
– Quuxplusone
Apr 1 at 21:25
add a comment |
2 Answers
2
active
oldest
votes
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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 Apr 1 at 14:16
paw88789paw88789
29.6k12351
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9
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For the lazy, the sum is55a + 88b. ;)
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– Mukul Gupta
Apr 1 at 19:14
add a comment |
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If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod{11}$$
And thus $$sum_{k=0}^{9} F_{n+k}equiv8cdot 4^ncdot(-7)cdot(4^{10}-1)-8cdot3^ncdot8cdot((-3)^{10}-1)pmod{11}$$
By Fermat's theorem, $a^{10}equiv1pmod {11}$ for all $anotequiv0pmod{11}$.
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 Apr 1 at 14:46
Saucy O'PathSaucy O'Path
6,3621627
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4
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I can't help but read that as "for all values of 11 which do not divide a".
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– hobbs
Apr 1 at 20:41
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@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
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– J.G.
Apr 1 at 20:59
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@J.G. ... or $anotequiv0pmod{11}$, for that matter.
$endgroup$
– Saucy O'Path
Apr 1 at 21:00
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@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.
Apr 1 at 21:01
add a comment |
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 Apr 1 at 14:16
paw88789paw88789
29.6k12351
$endgroup$
9
$begingroup$
For the lazy, the sum is55a + 88b. ;)
$endgroup$
– Mukul Gupta
Apr 1 at 19:14
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 Apr 1 at 14:16
paw88789paw88789
29.6k12351
$endgroup$
9
$begingroup$
For the lazy, the sum is55a + 88b. ;)
$endgroup$
– Mukul Gupta
Apr 1 at 19:14
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 Apr 1 at 14:16
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 Apr 1 at 14:16
paw88789paw88789
29.6k12351
answered Apr 1 at 14:16
paw88789paw88789
29.6k12351
answered Apr 1 at 14:16
paw88789paw88789
29.6k12351
answered Apr 1 at 14:16
paw88789paw88789
29.6k12351
29.6k12351
9
$begingroup$
For the lazy, the sum is55a + 88b. ;)
$endgroup$
– Mukul Gupta
Apr 1 at 19:14
add a comment |
9
$begingroup$
For the lazy, the sum is55a + 88b. ;)
$endgroup$
– Mukul Gupta
Apr 1 at 19:14
9
9
$begingroup$
For the lazy, the sum is
55a + 88b. ;)$endgroup$
– Mukul Gupta
Apr 1 at 19:14
$begingroup$
For the lazy, the sum is
55a + 88b. ;)$endgroup$
– Mukul Gupta
Apr 1 at 19:14
add a comment |
$begingroup$
If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod{11}$$
And thus $$sum_{k=0}^{9} F_{n+k}equiv8cdot 4^ncdot(-7)cdot(4^{10}-1)-8cdot3^ncdot8cdot((-3)^{10}-1)pmod{11}$$
By Fermat's theorem, $a^{10}equiv1pmod {11}$ for all $anotequiv0pmod{11}$.
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 Apr 1 at 14:46
Saucy O'PathSaucy O'Path
6,3621627
$endgroup$
4
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
Apr 1 at 20:41
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
Apr 1 at 20:59
$begingroup$
@J.G. ... or $anotequiv0pmod{11}$, for that matter.
$endgroup$
– Saucy O'Path
Apr 1 at 21:00
$begingroup$
@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.
Apr 1 at 21:01
add a comment |
$begingroup$
If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod{11}$$
And thus $$sum_{k=0}^{9} F_{n+k}equiv8cdot 4^ncdot(-7)cdot(4^{10}-1)-8cdot3^ncdot8cdot((-3)^{10}-1)pmod{11}$$
By Fermat's theorem, $a^{10}equiv1pmod {11}$ for all $anotequiv0pmod{11}$.
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 Apr 1 at 14:46
Saucy O'PathSaucy O'Path
6,3621627
$endgroup$
4
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
Apr 1 at 20:41
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
Apr 1 at 20:59
$begingroup$
@J.G. ... or $anotequiv0pmod{11}$, for that matter.
$endgroup$
– Saucy O'Path
Apr 1 at 21:00
$begingroup$
@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.
Apr 1 at 21:01
add a comment |
$begingroup$
If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod{11}$$
And thus $$sum_{k=0}^{9} F_{n+k}equiv8cdot 4^ncdot(-7)cdot(4^{10}-1)-8cdot3^ncdot8cdot((-3)^{10}-1)pmod{11}$$
By Fermat's theorem, $a^{10}equiv1pmod {11}$ for all $anotequiv0pmod{11}$.
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 Apr 1 at 14:46
Saucy O'PathSaucy O'Path
6,3621627
$endgroup$
If you use a Binet-like formula for the Fibonacci sequence $mod 11$, you'll notice that $$F_nequiv 8cdot (4^n-(-3)^n)pmod{11}$$
And thus $$sum_{k=0}^{9} F_{n+k}equiv8cdot 4^ncdot(-7)cdot(4^{10}-1)-8cdot3^ncdot8cdot((-3)^{10}-1)pmod{11}$$
By Fermat's theorem, $a^{10}equiv1pmod {11}$ for all $anotequiv0pmod{11}$.
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 Apr 1 at 14:46
Saucy O'PathSaucy O'Path
6,3621627
edited 2 days ago
answered Apr 1 at 14:46
Saucy O'PathSaucy O'Path
6,3621627
answered Apr 1 at 14:46
Saucy O'PathSaucy O'Path
6,3621627
answered Apr 1 at 14:46
Saucy O'PathSaucy O'Path
6,3621627
6,3621627
4
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
Apr 1 at 20:41
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
Apr 1 at 20:59
$begingroup$
@J.G. ... or $anotequiv0pmod{11}$, for that matter.
$endgroup$
– Saucy O'Path
Apr 1 at 21:00
$begingroup$
@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.
Apr 1 at 21:01
add a comment |
4
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
Apr 1 at 20:41
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
Apr 1 at 20:59
$begingroup$
@J.G. ... or $anotequiv0pmod{11}$, for that matter.
$endgroup$
– Saucy O'Path
Apr 1 at 21:00
$begingroup$
@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.
Apr 1 at 21:01
4
4
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
Apr 1 at 20:41
$begingroup$
I can't help but read that as "for all values of 11 which do not divide a".
$endgroup$
– hobbs
Apr 1 at 20:41
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
Apr 1 at 20:59
$begingroup$
@hobbs Agreed; $ainBbb Zsetminus 11Bbb Z$ would be better.
$endgroup$
– J.G.
Apr 1 at 20:59
$begingroup$
@J.G. ... or $anotequiv0pmod{11}$, for that matter.
$endgroup$
– Saucy O'Path
Apr 1 at 21:00
$begingroup$
@J.G. ... or $anotequiv0pmod{11}$, for that matter.
$endgroup$
– Saucy O'Path
Apr 1 at 21:00
$begingroup$
@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.
Apr 1 at 21:01
$begingroup$
@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.
Apr 1 at 21:01
add a comment |
$begingroup$
I mean, just bash? Use the recursive formula to show it.
$endgroup$
– Don Thousand
Apr 1 at 14:12
1
$begingroup$
See also here: math.stackexchange.com/questions/60049/…
$endgroup$
– Jonas Lenz
Apr 1 at 14:12
$begingroup$
How about using artofproblemsolving.com/wiki/index.php?title=Binet%27s_Formula
$endgroup$
– lab bhattacharjee
Apr 1 at 14:21
$begingroup$
Possible duplicate of Why is the sum of any ten consecutive Fibonacci numbers always divisible by $11$?
$endgroup$
– Quuxplusone
Apr 1 at 21:25