Fourteen numbers around a circle











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Place the numbers 1 to 14 around this circle so that both the sum and (absolute) difference of any two neighboring numbers is a prime.



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    Was this an original question? Or do you have a source?
    – Dr Xorile
    Nov 29 at 4:38






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    @DrXorile: It is an original idea inspired on a question of Carlos Zuluaga (Colombia Aprendiendo). It first appeared in Number Play, Gary Antonnick's New York Times puzzle column.
    – Bernardo Recamán Santos
    Nov 29 at 11:18















up vote
7
down vote

favorite












Place the numbers 1 to 14 around this circle so that both the sum and (absolute) difference of any two neighboring numbers is a prime.



enter image description here










share|improve this question


















  • 1




    Was this an original question? Or do you have a source?
    – Dr Xorile
    Nov 29 at 4:38






  • 1




    @DrXorile: It is an original idea inspired on a question of Carlos Zuluaga (Colombia Aprendiendo). It first appeared in Number Play, Gary Antonnick's New York Times puzzle column.
    – Bernardo Recamán Santos
    Nov 29 at 11:18













up vote
7
down vote

favorite









up vote
7
down vote

favorite











Place the numbers 1 to 14 around this circle so that both the sum and (absolute) difference of any two neighboring numbers is a prime.



enter image description here










share|improve this question













Place the numbers 1 to 14 around this circle so that both the sum and (absolute) difference of any two neighboring numbers is a prime.



enter image description here







mathematics number-theory






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share|improve this question










asked Nov 29 at 2:34









Bernardo Recamán Santos

2,3251141




2,3251141








  • 1




    Was this an original question? Or do you have a source?
    – Dr Xorile
    Nov 29 at 4:38






  • 1




    @DrXorile: It is an original idea inspired on a question of Carlos Zuluaga (Colombia Aprendiendo). It first appeared in Number Play, Gary Antonnick's New York Times puzzle column.
    – Bernardo Recamán Santos
    Nov 29 at 11:18














  • 1




    Was this an original question? Or do you have a source?
    – Dr Xorile
    Nov 29 at 4:38






  • 1




    @DrXorile: It is an original idea inspired on a question of Carlos Zuluaga (Colombia Aprendiendo). It first appeared in Number Play, Gary Antonnick's New York Times puzzle column.
    – Bernardo Recamán Santos
    Nov 29 at 11:18








1




1




Was this an original question? Or do you have a source?
– Dr Xorile
Nov 29 at 4:38




Was this an original question? Or do you have a source?
– Dr Xorile
Nov 29 at 4:38




1




1




@DrXorile: It is an original idea inspired on a question of Carlos Zuluaga (Colombia Aprendiendo). It first appeared in Number Play, Gary Antonnick's New York Times puzzle column.
– Bernardo Recamán Santos
Nov 29 at 11:18




@DrXorile: It is an original idea inspired on a question of Carlos Zuluaga (Colombia Aprendiendo). It first appeared in Number Play, Gary Antonnick's New York Times puzzle column.
– Bernardo Recamán Santos
Nov 29 at 11:18










1 Answer
1






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oldest

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up vote
7
down vote



accepted










This appears to work, and I think it's unique but haven't checked super-carefully:




10 -- 13 -- 6 -- 11 -- 8 -- 3 -- 14 -- 9 -- 2 -- 5 -- 12 -- 1 -- 4 -- 7 -- (10)







share|improve this answer





















  • Was about to post this too, and I'm pretty sure it's unique because: there are 2 or 3 possibilities for the neighbours of each number, if 2 then it's forced, and then we can draw a graph for the remaining and there is only one cycle covering all numbers.
    – deep thought
    Nov 29 at 3:03






  • 1




    To expand a bit: 2,11,13,14 can each only be adjacent to two numbers, so that part of the sequence is forced, namely 3-14-9-2-5 and 10-13-6-11-8. 1 could neighbour 4,6,12 but 6 is now "taken" by 13 and 11, so 12-1-4 is forced. 4 could neighbour 9 but 9 is also taken, 1-4-7. 12 could neighbour 7 but that would close off a cycle of length 4. So 12 and 7 are forced: 7-10 and 5-12. 3 could neighbour 10 but that's taken. so 8-3. Done.
    – deep thought
    Nov 29 at 3:18










  • Sorry if I ninjaed you! Your reasons for saying it's unique are very much like mine (unsurprisingly) but since I hadn't bothered to write them out carefully I didn't want to claim too much :-).
    – Gareth McCaughan
    Nov 29 at 4:04











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1 Answer
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1 Answer
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active

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active

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up vote
7
down vote



accepted










This appears to work, and I think it's unique but haven't checked super-carefully:




10 -- 13 -- 6 -- 11 -- 8 -- 3 -- 14 -- 9 -- 2 -- 5 -- 12 -- 1 -- 4 -- 7 -- (10)







share|improve this answer





















  • Was about to post this too, and I'm pretty sure it's unique because: there are 2 or 3 possibilities for the neighbours of each number, if 2 then it's forced, and then we can draw a graph for the remaining and there is only one cycle covering all numbers.
    – deep thought
    Nov 29 at 3:03






  • 1




    To expand a bit: 2,11,13,14 can each only be adjacent to two numbers, so that part of the sequence is forced, namely 3-14-9-2-5 and 10-13-6-11-8. 1 could neighbour 4,6,12 but 6 is now "taken" by 13 and 11, so 12-1-4 is forced. 4 could neighbour 9 but 9 is also taken, 1-4-7. 12 could neighbour 7 but that would close off a cycle of length 4. So 12 and 7 are forced: 7-10 and 5-12. 3 could neighbour 10 but that's taken. so 8-3. Done.
    – deep thought
    Nov 29 at 3:18










  • Sorry if I ninjaed you! Your reasons for saying it's unique are very much like mine (unsurprisingly) but since I hadn't bothered to write them out carefully I didn't want to claim too much :-).
    – Gareth McCaughan
    Nov 29 at 4:04















up vote
7
down vote



accepted










This appears to work, and I think it's unique but haven't checked super-carefully:




10 -- 13 -- 6 -- 11 -- 8 -- 3 -- 14 -- 9 -- 2 -- 5 -- 12 -- 1 -- 4 -- 7 -- (10)







share|improve this answer





















  • Was about to post this too, and I'm pretty sure it's unique because: there are 2 or 3 possibilities for the neighbours of each number, if 2 then it's forced, and then we can draw a graph for the remaining and there is only one cycle covering all numbers.
    – deep thought
    Nov 29 at 3:03






  • 1




    To expand a bit: 2,11,13,14 can each only be adjacent to two numbers, so that part of the sequence is forced, namely 3-14-9-2-5 and 10-13-6-11-8. 1 could neighbour 4,6,12 but 6 is now "taken" by 13 and 11, so 12-1-4 is forced. 4 could neighbour 9 but 9 is also taken, 1-4-7. 12 could neighbour 7 but that would close off a cycle of length 4. So 12 and 7 are forced: 7-10 and 5-12. 3 could neighbour 10 but that's taken. so 8-3. Done.
    – deep thought
    Nov 29 at 3:18










  • Sorry if I ninjaed you! Your reasons for saying it's unique are very much like mine (unsurprisingly) but since I hadn't bothered to write them out carefully I didn't want to claim too much :-).
    – Gareth McCaughan
    Nov 29 at 4:04













up vote
7
down vote



accepted







up vote
7
down vote



accepted






This appears to work, and I think it's unique but haven't checked super-carefully:




10 -- 13 -- 6 -- 11 -- 8 -- 3 -- 14 -- 9 -- 2 -- 5 -- 12 -- 1 -- 4 -- 7 -- (10)







share|improve this answer












This appears to work, and I think it's unique but haven't checked super-carefully:




10 -- 13 -- 6 -- 11 -- 8 -- 3 -- 14 -- 9 -- 2 -- 5 -- 12 -- 1 -- 4 -- 7 -- (10)








share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 29 at 2:57









Gareth McCaughan

59.5k3150230




59.5k3150230












  • Was about to post this too, and I'm pretty sure it's unique because: there are 2 or 3 possibilities for the neighbours of each number, if 2 then it's forced, and then we can draw a graph for the remaining and there is only one cycle covering all numbers.
    – deep thought
    Nov 29 at 3:03






  • 1




    To expand a bit: 2,11,13,14 can each only be adjacent to two numbers, so that part of the sequence is forced, namely 3-14-9-2-5 and 10-13-6-11-8. 1 could neighbour 4,6,12 but 6 is now "taken" by 13 and 11, so 12-1-4 is forced. 4 could neighbour 9 but 9 is also taken, 1-4-7. 12 could neighbour 7 but that would close off a cycle of length 4. So 12 and 7 are forced: 7-10 and 5-12. 3 could neighbour 10 but that's taken. so 8-3. Done.
    – deep thought
    Nov 29 at 3:18










  • Sorry if I ninjaed you! Your reasons for saying it's unique are very much like mine (unsurprisingly) but since I hadn't bothered to write them out carefully I didn't want to claim too much :-).
    – Gareth McCaughan
    Nov 29 at 4:04


















  • Was about to post this too, and I'm pretty sure it's unique because: there are 2 or 3 possibilities for the neighbours of each number, if 2 then it's forced, and then we can draw a graph for the remaining and there is only one cycle covering all numbers.
    – deep thought
    Nov 29 at 3:03






  • 1




    To expand a bit: 2,11,13,14 can each only be adjacent to two numbers, so that part of the sequence is forced, namely 3-14-9-2-5 and 10-13-6-11-8. 1 could neighbour 4,6,12 but 6 is now "taken" by 13 and 11, so 12-1-4 is forced. 4 could neighbour 9 but 9 is also taken, 1-4-7. 12 could neighbour 7 but that would close off a cycle of length 4. So 12 and 7 are forced: 7-10 and 5-12. 3 could neighbour 10 but that's taken. so 8-3. Done.
    – deep thought
    Nov 29 at 3:18










  • Sorry if I ninjaed you! Your reasons for saying it's unique are very much like mine (unsurprisingly) but since I hadn't bothered to write them out carefully I didn't want to claim too much :-).
    – Gareth McCaughan
    Nov 29 at 4:04
















Was about to post this too, and I'm pretty sure it's unique because: there are 2 or 3 possibilities for the neighbours of each number, if 2 then it's forced, and then we can draw a graph for the remaining and there is only one cycle covering all numbers.
– deep thought
Nov 29 at 3:03




Was about to post this too, and I'm pretty sure it's unique because: there are 2 or 3 possibilities for the neighbours of each number, if 2 then it's forced, and then we can draw a graph for the remaining and there is only one cycle covering all numbers.
– deep thought
Nov 29 at 3:03




1




1




To expand a bit: 2,11,13,14 can each only be adjacent to two numbers, so that part of the sequence is forced, namely 3-14-9-2-5 and 10-13-6-11-8. 1 could neighbour 4,6,12 but 6 is now "taken" by 13 and 11, so 12-1-4 is forced. 4 could neighbour 9 but 9 is also taken, 1-4-7. 12 could neighbour 7 but that would close off a cycle of length 4. So 12 and 7 are forced: 7-10 and 5-12. 3 could neighbour 10 but that's taken. so 8-3. Done.
– deep thought
Nov 29 at 3:18




To expand a bit: 2,11,13,14 can each only be adjacent to two numbers, so that part of the sequence is forced, namely 3-14-9-2-5 and 10-13-6-11-8. 1 could neighbour 4,6,12 but 6 is now "taken" by 13 and 11, so 12-1-4 is forced. 4 could neighbour 9 but 9 is also taken, 1-4-7. 12 could neighbour 7 but that would close off a cycle of length 4. So 12 and 7 are forced: 7-10 and 5-12. 3 could neighbour 10 but that's taken. so 8-3. Done.
– deep thought
Nov 29 at 3:18












Sorry if I ninjaed you! Your reasons for saying it's unique are very much like mine (unsurprisingly) but since I hadn't bothered to write them out carefully I didn't want to claim too much :-).
– Gareth McCaughan
Nov 29 at 4:04




Sorry if I ninjaed you! Your reasons for saying it's unique are very much like mine (unsurprisingly) but since I hadn't bothered to write them out carefully I didn't want to claim too much :-).
– Gareth McCaughan
Nov 29 at 4:04


















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