Is there among first $100000001$ Fibonacci numbers one that ends with $0000$?
$begingroup$
This looks like a difficult problem:
Is there among first $100000001$ Fibonacci numbers one that ends with
$0000$?
(it is from a competition training; trainer suggests using pigeonhole principle)
fibonacci-numbers pigeonhole-principle
asked Sep 21 '14 at 22:32
VividDVividD
8,37254793
$endgroup$
add a comment |
$begingroup$
This looks like a difficult problem:
Is there among first $100000001$ Fibonacci numbers one that ends with
$0000$?
(it is from a competition training; trainer suggests using pigeonhole principle)
fibonacci-numbers pigeonhole-principle
asked Sep 21 '14 at 22:32
VividDVividD
8,37254793
$endgroup$
2
$begingroup$
Then ignore the trainer. Its advice can be useful if you get stuck after you've gotten somewhere into the problem, but it's worthless for actually starting the problem.
$endgroup$
– Hurkyl
Sep 21 '14 at 22:42
1
$begingroup$
Hint: $100000001=1, 0000^2+1$. This is no coincidence.
$endgroup$
– Yves Daoust
May 7 '15 at 7:52
add a comment |
$begingroup$
This looks like a difficult problem:
Is there among first $100000001$ Fibonacci numbers one that ends with
$0000$?
(it is from a competition training; trainer suggests using pigeonhole principle)
fibonacci-numbers pigeonhole-principle
asked Sep 21 '14 at 22:32
VividDVividD
8,37254793
$endgroup$
This looks like a difficult problem:
Is there among first $100000001$ Fibonacci numbers one that ends with
$0000$?
(it is from a competition training; trainer suggests using pigeonhole principle)
fibonacci-numbers pigeonhole-principle
fibonacci-numbers pigeonhole-principle
asked Sep 21 '14 at 22:32
VividDVividD
8,37254793
asked Sep 21 '14 at 22:32
VividDVividD
8,37254793
edited Dec 16 '15 at 16:01
VividD
asked Sep 21 '14 at 22:32
VividDVividD
8,37254793
asked Sep 21 '14 at 22:32
VividDVividD
8,37254793
asked Sep 21 '14 at 22:32
VividDVividD
8,37254793
8,37254793
2
$begingroup$
Then ignore the trainer. Its advice can be useful if you get stuck after you've gotten somewhere into the problem, but it's worthless for actually starting the problem.
$endgroup$
– Hurkyl
Sep 21 '14 at 22:42
1
$begingroup$
Hint: $100000001=1, 0000^2+1$. This is no coincidence.
$endgroup$
– Yves Daoust
May 7 '15 at 7:52
add a comment |
2
$begingroup$
Then ignore the trainer. Its advice can be useful if you get stuck after you've gotten somewhere into the problem, but it's worthless for actually starting the problem.
$endgroup$
– Hurkyl
Sep 21 '14 at 22:42
1
$begingroup$
Hint: $100000001=1, 0000^2+1$. This is no coincidence.
$endgroup$
– Yves Daoust
May 7 '15 at 7:52
2
2
$begingroup$
Then ignore the trainer. Its advice can be useful if you get stuck after you've gotten somewhere into the problem, but it's worthless for actually starting the problem.
$endgroup$
– Hurkyl
Sep 21 '14 at 22:42
$begingroup$
Then ignore the trainer. Its advice can be useful if you get stuck after you've gotten somewhere into the problem, but it's worthless for actually starting the problem.
$endgroup$
– Hurkyl
Sep 21 '14 at 22:42
1
1
$begingroup$
Hint: $100000001=1, 0000^2+1$. This is no coincidence.
$endgroup$
– Yves Daoust
May 7 '15 at 7:52
$begingroup$
Hint: $100000001=1, 0000^2+1$. This is no coincidence.
$endgroup$
– Yves Daoust
May 7 '15 at 7:52
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Consider Fibonacci numbers $mod 10000$. The sequence begins: $F_0=0, 1, 1, ldots$ and continues until $F_{100000001}$. Consider the set of $100000001$ ordered pairs $(F_n, F_{n+1})$. By the pigeonhole principle, at least one of these ordered pairs occurs twice in the sequence.
Now note that the entire sequence of Fibonacci numbers $mod 10000$ are uniquely determined by any two consecutive values, as the sequence can be constructed both forwards and backwards. ($F_nequiv F_{n+2}-F_{n+1} mod 10000$).
So if the ordered pair $(F_n, F_{n+1})$ occurs at both $n=m$ and $n=m+t$ for $m,t in mathbb{N}$, then the sequence is recurrent ($F_n equiv F_{n+t} mod 10000$ for all $n$).
Hence the ordered pair $(F_n=0, F_{n+1}=1)$ must also occur at both $n=0$ and $n=t$. And since $m+t$ is among the first 100000001 Fibonacci numbers, then $t$ must also be among them.
answered Sep 21 '14 at 22:46
Fengyang WangFengyang Wang
1,2891920
$endgroup$
$begingroup$
Excellent, thanks! It looks now not that difficult at all.
$endgroup$
– VividD
Sep 21 '14 at 23:03
$begingroup$
There's an important detail this leaves out; you show that the sequence is eventually cyclic - that is $F_nequiv F_{n+t}mod 10000$ for all large enough $n$ - but you then use that it holds for $n=1$. You need to notice that the map $(a,b)mapsto (b,a+b)$ (which transitions $(F_{n-1},F_n)$ to $(F_n,F_{n+1})$) is a bijection (working mod 10000) in order to conclude that it is periodic. Otherwise, repeated applications of that map might enter a cycle, but one which does not include $(0,1)$. (e.g. like how the map $f(x)=max(x-1,0)$ acts - eventually cyclic, but not a bijection)
$endgroup$
– Milo Brandt
Apr 14 '15 at 23:30
add a comment |
$begingroup$
This is slightly irrelevant to what you are looking for, but I'm sure you'll be happy to know that for $n = 7500$,
f(n) = 11423965231520587047220488928656904198487186633317560797959030595738263643588305263964321080516991429937628886229555340146644442744473185460778302934743807002248109695741208782411159189994651520930091202035101269350523609417276542209682261168150544790025062794209091503702088574338650460569295592498666443239807989522593072562158640947468656887645879356201301594841872491497556389555817277508349058330498007583814270123329724353233156029127910968370052734811192660492733375394472692191584489489590970254440914222778382439339334175624660291588778456250479185237898309112318829984358216337347549014336517486496643224502773380042071174360597192343056318489287038447004730922073980870072990706067508624038407888471294048912294153491398930715643640170172837379127969101176561450586945715460276780809807889664272818316865711724985646554559305334340318994612185260719042008960311269000122672589731283419608098303367260382379660402261886574952211783683104453334281684425994447306306414660032519055079504313562694958935754118796157632978970220780288168992181699708922971417067735144929461193639081445200786881549331150381216073705417531166786634690469206418611524663013854198045284806720735273715046888704916821855277543026346215355286395854263168251068150374988851620501196943905031285049077628443804052134507022504682483293396215268186620124762379744668092166035314553541731537245946256422861852573006230492322259630342294350827184840607509969289328320360093204783447860955806396350723341261564285649453007949089154165288839814442677339344794691881510389855765582716774490000,
so there is at least one!
edited May 7 '15 at 7:35
VividD
8,37254793
answered Sep 22 '14 at 10:24
user177886user177886
412
$endgroup$
add a comment |
$begingroup$
Consider $a_n$ as the the terminal four digits of nth term now pair up $(a_n,a_{n+1})$ ..... now we end up having 100000001 pairs now each $a_n$ is four digit only we are looking into numbers less than 10000 the pairs of them will be 10 *4*10*4=10*8 by PPP we have two set identical or pairs identical now apply it periodically we end up $a_0 =a_{n+1}$ thus concluded
edited Jan 16 at 14:37
Dmitry
716618
answered Jan 16 at 13:48
JishnuJishnu
1
$endgroup$
add a comment |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider Fibonacci numbers $mod 10000$. The sequence begins: $F_0=0, 1, 1, ldots$ and continues until $F_{100000001}$. Consider the set of $100000001$ ordered pairs $(F_n, F_{n+1})$. By the pigeonhole principle, at least one of these ordered pairs occurs twice in the sequence.
Now note that the entire sequence of Fibonacci numbers $mod 10000$ are uniquely determined by any two consecutive values, as the sequence can be constructed both forwards and backwards. ($F_nequiv F_{n+2}-F_{n+1} mod 10000$).
So if the ordered pair $(F_n, F_{n+1})$ occurs at both $n=m$ and $n=m+t$ for $m,t in mathbb{N}$, then the sequence is recurrent ($F_n equiv F_{n+t} mod 10000$ for all $n$).
Hence the ordered pair $(F_n=0, F_{n+1}=1)$ must also occur at both $n=0$ and $n=t$. And since $m+t$ is among the first 100000001 Fibonacci numbers, then $t$ must also be among them.
answered Sep 21 '14 at 22:46
Fengyang WangFengyang Wang
1,2891920
$endgroup$
$begingroup$
Excellent, thanks! It looks now not that difficult at all.
$endgroup$
– VividD
Sep 21 '14 at 23:03
$begingroup$
There's an important detail this leaves out; you show that the sequence is eventually cyclic - that is $F_nequiv F_{n+t}mod 10000$ for all large enough $n$ - but you then use that it holds for $n=1$. You need to notice that the map $(a,b)mapsto (b,a+b)$ (which transitions $(F_{n-1},F_n)$ to $(F_n,F_{n+1})$) is a bijection (working mod 10000) in order to conclude that it is periodic. Otherwise, repeated applications of that map might enter a cycle, but one which does not include $(0,1)$. (e.g. like how the map $f(x)=max(x-1,0)$ acts - eventually cyclic, but not a bijection)
$endgroup$
– Milo Brandt
Apr 14 '15 at 23:30
add a comment |
$begingroup$
Consider Fibonacci numbers $mod 10000$. The sequence begins: $F_0=0, 1, 1, ldots$ and continues until $F_{100000001}$. Consider the set of $100000001$ ordered pairs $(F_n, F_{n+1})$. By the pigeonhole principle, at least one of these ordered pairs occurs twice in the sequence.
Now note that the entire sequence of Fibonacci numbers $mod 10000$ are uniquely determined by any two consecutive values, as the sequence can be constructed both forwards and backwards. ($F_nequiv F_{n+2}-F_{n+1} mod 10000$).
So if the ordered pair $(F_n, F_{n+1})$ occurs at both $n=m$ and $n=m+t$ for $m,t in mathbb{N}$, then the sequence is recurrent ($F_n equiv F_{n+t} mod 10000$ for all $n$).
Hence the ordered pair $(F_n=0, F_{n+1}=1)$ must also occur at both $n=0$ and $n=t$. And since $m+t$ is among the first 100000001 Fibonacci numbers, then $t$ must also be among them.
answered Sep 21 '14 at 22:46
Fengyang WangFengyang Wang
1,2891920
$endgroup$
$begingroup$
Excellent, thanks! It looks now not that difficult at all.
$endgroup$
– VividD
Sep 21 '14 at 23:03
$begingroup$
There's an important detail this leaves out; you show that the sequence is eventually cyclic - that is $F_nequiv F_{n+t}mod 10000$ for all large enough $n$ - but you then use that it holds for $n=1$. You need to notice that the map $(a,b)mapsto (b,a+b)$ (which transitions $(F_{n-1},F_n)$ to $(F_n,F_{n+1})$) is a bijection (working mod 10000) in order to conclude that it is periodic. Otherwise, repeated applications of that map might enter a cycle, but one which does not include $(0,1)$. (e.g. like how the map $f(x)=max(x-1,0)$ acts - eventually cyclic, but not a bijection)
$endgroup$
– Milo Brandt
Apr 14 '15 at 23:30
add a comment |
$begingroup$
Consider Fibonacci numbers $mod 10000$. The sequence begins: $F_0=0, 1, 1, ldots$ and continues until $F_{100000001}$. Consider the set of $100000001$ ordered pairs $(F_n, F_{n+1})$. By the pigeonhole principle, at least one of these ordered pairs occurs twice in the sequence.
Now note that the entire sequence of Fibonacci numbers $mod 10000$ are uniquely determined by any two consecutive values, as the sequence can be constructed both forwards and backwards. ($F_nequiv F_{n+2}-F_{n+1} mod 10000$).
So if the ordered pair $(F_n, F_{n+1})$ occurs at both $n=m$ and $n=m+t$ for $m,t in mathbb{N}$, then the sequence is recurrent ($F_n equiv F_{n+t} mod 10000$ for all $n$).
Hence the ordered pair $(F_n=0, F_{n+1}=1)$ must also occur at both $n=0$ and $n=t$. And since $m+t$ is among the first 100000001 Fibonacci numbers, then $t$ must also be among them.
answered Sep 21 '14 at 22:46
Fengyang WangFengyang Wang
1,2891920
$endgroup$
Consider Fibonacci numbers $mod 10000$. The sequence begins: $F_0=0, 1, 1, ldots$ and continues until $F_{100000001}$. Consider the set of $100000001$ ordered pairs $(F_n, F_{n+1})$. By the pigeonhole principle, at least one of these ordered pairs occurs twice in the sequence.
Now note that the entire sequence of Fibonacci numbers $mod 10000$ are uniquely determined by any two consecutive values, as the sequence can be constructed both forwards and backwards. ($F_nequiv F_{n+2}-F_{n+1} mod 10000$).
So if the ordered pair $(F_n, F_{n+1})$ occurs at both $n=m$ and $n=m+t$ for $m,t in mathbb{N}$, then the sequence is recurrent ($F_n equiv F_{n+t} mod 10000$ for all $n$).
Hence the ordered pair $(F_n=0, F_{n+1}=1)$ must also occur at both $n=0$ and $n=t$. And since $m+t$ is among the first 100000001 Fibonacci numbers, then $t$ must also be among them.
answered Sep 21 '14 at 22:46
Fengyang WangFengyang Wang
1,2891920
answered Sep 21 '14 at 22:46
Fengyang WangFengyang Wang
1,2891920
answered Sep 21 '14 at 22:46
Fengyang WangFengyang Wang
1,2891920
answered Sep 21 '14 at 22:46
Fengyang WangFengyang Wang
1,2891920
1,2891920
$begingroup$
Excellent, thanks! It looks now not that difficult at all.
$endgroup$
– VividD
Sep 21 '14 at 23:03
$begingroup$
There's an important detail this leaves out; you show that the sequence is eventually cyclic - that is $F_nequiv F_{n+t}mod 10000$ for all large enough $n$ - but you then use that it holds for $n=1$. You need to notice that the map $(a,b)mapsto (b,a+b)$ (which transitions $(F_{n-1},F_n)$ to $(F_n,F_{n+1})$) is a bijection (working mod 10000) in order to conclude that it is periodic. Otherwise, repeated applications of that map might enter a cycle, but one which does not include $(0,1)$. (e.g. like how the map $f(x)=max(x-1,0)$ acts - eventually cyclic, but not a bijection)
$endgroup$
– Milo Brandt
Apr 14 '15 at 23:30
add a comment |
$begingroup$
Excellent, thanks! It looks now not that difficult at all.
$endgroup$
– VividD
Sep 21 '14 at 23:03
$begingroup$
There's an important detail this leaves out; you show that the sequence is eventually cyclic - that is $F_nequiv F_{n+t}mod 10000$ for all large enough $n$ - but you then use that it holds for $n=1$. You need to notice that the map $(a,b)mapsto (b,a+b)$ (which transitions $(F_{n-1},F_n)$ to $(F_n,F_{n+1})$) is a bijection (working mod 10000) in order to conclude that it is periodic. Otherwise, repeated applications of that map might enter a cycle, but one which does not include $(0,1)$. (e.g. like how the map $f(x)=max(x-1,0)$ acts - eventually cyclic, but not a bijection)
$endgroup$
– Milo Brandt
Apr 14 '15 at 23:30
$begingroup$
Excellent, thanks! It looks now not that difficult at all.
$endgroup$
– VividD
Sep 21 '14 at 23:03
$begingroup$
Excellent, thanks! It looks now not that difficult at all.
$endgroup$
– VividD
Sep 21 '14 at 23:03
$begingroup$
There's an important detail this leaves out; you show that the sequence is eventually cyclic - that is $F_nequiv F_{n+t}mod 10000$ for all large enough $n$ - but you then use that it holds for $n=1$. You need to notice that the map $(a,b)mapsto (b,a+b)$ (which transitions $(F_{n-1},F_n)$ to $(F_n,F_{n+1})$) is a bijection (working mod 10000) in order to conclude that it is periodic. Otherwise, repeated applications of that map might enter a cycle, but one which does not include $(0,1)$. (e.g. like how the map $f(x)=max(x-1,0)$ acts - eventually cyclic, but not a bijection)
$endgroup$
– Milo Brandt
Apr 14 '15 at 23:30
$begingroup$
There's an important detail this leaves out; you show that the sequence is eventually cyclic - that is $F_nequiv F_{n+t}mod 10000$ for all large enough $n$ - but you then use that it holds for $n=1$. You need to notice that the map $(a,b)mapsto (b,a+b)$ (which transitions $(F_{n-1},F_n)$ to $(F_n,F_{n+1})$) is a bijection (working mod 10000) in order to conclude that it is periodic. Otherwise, repeated applications of that map might enter a cycle, but one which does not include $(0,1)$. (e.g. like how the map $f(x)=max(x-1,0)$ acts - eventually cyclic, but not a bijection)
$endgroup$
– Milo Brandt
Apr 14 '15 at 23:30
add a comment |
$begingroup$
This is slightly irrelevant to what you are looking for, but I'm sure you'll be happy to know that for $n = 7500$,
f(n) = 11423965231520587047220488928656904198487186633317560797959030595738263643588305263964321080516991429937628886229555340146644442744473185460778302934743807002248109695741208782411159189994651520930091202035101269350523609417276542209682261168150544790025062794209091503702088574338650460569295592498666443239807989522593072562158640947468656887645879356201301594841872491497556389555817277508349058330498007583814270123329724353233156029127910968370052734811192660492733375394472692191584489489590970254440914222778382439339334175624660291588778456250479185237898309112318829984358216337347549014336517486496643224502773380042071174360597192343056318489287038447004730922073980870072990706067508624038407888471294048912294153491398930715643640170172837379127969101176561450586945715460276780809807889664272818316865711724985646554559305334340318994612185260719042008960311269000122672589731283419608098303367260382379660402261886574952211783683104453334281684425994447306306414660032519055079504313562694958935754118796157632978970220780288168992181699708922971417067735144929461193639081445200786881549331150381216073705417531166786634690469206418611524663013854198045284806720735273715046888704916821855277543026346215355286395854263168251068150374988851620501196943905031285049077628443804052134507022504682483293396215268186620124762379744668092166035314553541731537245946256422861852573006230492322259630342294350827184840607509969289328320360093204783447860955806396350723341261564285649453007949089154165288839814442677339344794691881510389855765582716774490000,
so there is at least one!
edited May 7 '15 at 7:35
VividD
8,37254793
answered Sep 22 '14 at 10:24
user177886user177886
412
$endgroup$
add a comment |
$begingroup$
This is slightly irrelevant to what you are looking for, but I'm sure you'll be happy to know that for $n = 7500$,
f(n) = 11423965231520587047220488928656904198487186633317560797959030595738263643588305263964321080516991429937628886229555340146644442744473185460778302934743807002248109695741208782411159189994651520930091202035101269350523609417276542209682261168150544790025062794209091503702088574338650460569295592498666443239807989522593072562158640947468656887645879356201301594841872491497556389555817277508349058330498007583814270123329724353233156029127910968370052734811192660492733375394472692191584489489590970254440914222778382439339334175624660291588778456250479185237898309112318829984358216337347549014336517486496643224502773380042071174360597192343056318489287038447004730922073980870072990706067508624038407888471294048912294153491398930715643640170172837379127969101176561450586945715460276780809807889664272818316865711724985646554559305334340318994612185260719042008960311269000122672589731283419608098303367260382379660402261886574952211783683104453334281684425994447306306414660032519055079504313562694958935754118796157632978970220780288168992181699708922971417067735144929461193639081445200786881549331150381216073705417531166786634690469206418611524663013854198045284806720735273715046888704916821855277543026346215355286395854263168251068150374988851620501196943905031285049077628443804052134507022504682483293396215268186620124762379744668092166035314553541731537245946256422861852573006230492322259630342294350827184840607509969289328320360093204783447860955806396350723341261564285649453007949089154165288839814442677339344794691881510389855765582716774490000,
so there is at least one!
edited May 7 '15 at 7:35
VividD
8,37254793
answered Sep 22 '14 at 10:24
user177886user177886
412
$endgroup$
add a comment |
$begingroup$
This is slightly irrelevant to what you are looking for, but I'm sure you'll be happy to know that for $n = 7500$,
f(n) = 11423965231520587047220488928656904198487186633317560797959030595738263643588305263964321080516991429937628886229555340146644442744473185460778302934743807002248109695741208782411159189994651520930091202035101269350523609417276542209682261168150544790025062794209091503702088574338650460569295592498666443239807989522593072562158640947468656887645879356201301594841872491497556389555817277508349058330498007583814270123329724353233156029127910968370052734811192660492733375394472692191584489489590970254440914222778382439339334175624660291588778456250479185237898309112318829984358216337347549014336517486496643224502773380042071174360597192343056318489287038447004730922073980870072990706067508624038407888471294048912294153491398930715643640170172837379127969101176561450586945715460276780809807889664272818316865711724985646554559305334340318994612185260719042008960311269000122672589731283419608098303367260382379660402261886574952211783683104453334281684425994447306306414660032519055079504313562694958935754118796157632978970220780288168992181699708922971417067735144929461193639081445200786881549331150381216073705417531166786634690469206418611524663013854198045284806720735273715046888704916821855277543026346215355286395854263168251068150374988851620501196943905031285049077628443804052134507022504682483293396215268186620124762379744668092166035314553541731537245946256422861852573006230492322259630342294350827184840607509969289328320360093204783447860955806396350723341261564285649453007949089154165288839814442677339344794691881510389855765582716774490000,
so there is at least one!
edited May 7 '15 at 7:35
VividD
8,37254793
answered Sep 22 '14 at 10:24
user177886user177886
412
$endgroup$
This is slightly irrelevant to what you are looking for, but I'm sure you'll be happy to know that for $n = 7500$,
f(n) = 11423965231520587047220488928656904198487186633317560797959030595738263643588305263964321080516991429937628886229555340146644442744473185460778302934743807002248109695741208782411159189994651520930091202035101269350523609417276542209682261168150544790025062794209091503702088574338650460569295592498666443239807989522593072562158640947468656887645879356201301594841872491497556389555817277508349058330498007583814270123329724353233156029127910968370052734811192660492733375394472692191584489489590970254440914222778382439339334175624660291588778456250479185237898309112318829984358216337347549014336517486496643224502773380042071174360597192343056318489287038447004730922073980870072990706067508624038407888471294048912294153491398930715643640170172837379127969101176561450586945715460276780809807889664272818316865711724985646554559305334340318994612185260719042008960311269000122672589731283419608098303367260382379660402261886574952211783683104453334281684425994447306306414660032519055079504313562694958935754118796157632978970220780288168992181699708922971417067735144929461193639081445200786881549331150381216073705417531166786634690469206418611524663013854198045284806720735273715046888704916821855277543026346215355286395854263168251068150374988851620501196943905031285049077628443804052134507022504682483293396215268186620124762379744668092166035314553541731537245946256422861852573006230492322259630342294350827184840607509969289328320360093204783447860955806396350723341261564285649453007949089154165288839814442677339344794691881510389855765582716774490000,
so there is at least one!
edited May 7 '15 at 7:35
VividD
8,37254793
answered Sep 22 '14 at 10:24
user177886user177886
412
edited May 7 '15 at 7:35
VividD
8,37254793
edited May 7 '15 at 7:35
VividD
8,37254793
edited May 7 '15 at 7:35
VividD
8,37254793
8,37254793
answered Sep 22 '14 at 10:24
user177886user177886
412
answered Sep 22 '14 at 10:24
user177886user177886
412
answered Sep 22 '14 at 10:24
user177886user177886
412
412
add a comment |
add a comment |
$begingroup$
Consider $a_n$ as the the terminal four digits of nth term now pair up $(a_n,a_{n+1})$ ..... now we end up having 100000001 pairs now each $a_n$ is four digit only we are looking into numbers less than 10000 the pairs of them will be 10 *4*10*4=10*8 by PPP we have two set identical or pairs identical now apply it periodically we end up $a_0 =a_{n+1}$ thus concluded
edited Jan 16 at 14:37
Dmitry
716618
answered Jan 16 at 13:48
JishnuJishnu
1
$endgroup$
add a comment |
$begingroup$
Consider $a_n$ as the the terminal four digits of nth term now pair up $(a_n,a_{n+1})$ ..... now we end up having 100000001 pairs now each $a_n$ is four digit only we are looking into numbers less than 10000 the pairs of them will be 10 *4*10*4=10*8 by PPP we have two set identical or pairs identical now apply it periodically we end up $a_0 =a_{n+1}$ thus concluded
edited Jan 16 at 14:37
Dmitry
716618
answered Jan 16 at 13:48
JishnuJishnu
1
$endgroup$
add a comment |
$begingroup$
Consider $a_n$ as the the terminal four digits of nth term now pair up $(a_n,a_{n+1})$ ..... now we end up having 100000001 pairs now each $a_n$ is four digit only we are looking into numbers less than 10000 the pairs of them will be 10 *4*10*4=10*8 by PPP we have two set identical or pairs identical now apply it periodically we end up $a_0 =a_{n+1}$ thus concluded
edited Jan 16 at 14:37
Dmitry
716618
answered Jan 16 at 13:48
JishnuJishnu
1
$endgroup$
Consider $a_n$ as the the terminal four digits of nth term now pair up $(a_n,a_{n+1})$ ..... now we end up having 100000001 pairs now each $a_n$ is four digit only we are looking into numbers less than 10000 the pairs of them will be 10 *4*10*4=10*8 by PPP we have two set identical or pairs identical now apply it periodically we end up $a_0 =a_{n+1}$ thus concluded
edited Jan 16 at 14:37
Dmitry
716618
answered Jan 16 at 13:48
JishnuJishnu
1
edited Jan 16 at 14:37
Dmitry
716618
edited Jan 16 at 14:37
Dmitry
716618
edited Jan 16 at 14:37
Dmitry
716618
716618
answered Jan 16 at 13:48
JishnuJishnu
1
answered Jan 16 at 13:48
JishnuJishnu
1
answered Jan 16 at 13:48
JishnuJishnu
1
1
add a comment |
add a comment |
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$begingroup$
Then ignore the trainer. Its advice can be useful if you get stuck after you've gotten somewhere into the problem, but it's worthless for actually starting the problem.
$endgroup$
– Hurkyl
Sep 21 '14 at 22:42
1
$begingroup$
Hint: $100000001=1, 0000^2+1$. This is no coincidence.
$endgroup$
– Yves Daoust
May 7 '15 at 7:52