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.

enter image description here

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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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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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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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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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