# Fourteen numbers around a circle

Clash Royale CLAN TAG#URR8PPP

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.

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

• @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

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.

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

• @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

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.

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.

mathematics number-theory

Bernardo Recamán Santos

2,3251141

2,3251141

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

• @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

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

• @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

– Dr Xorile
Nov 29 at 4:38

– 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

@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

active

oldest

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)

• 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

• 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

active

oldest

active

oldest

active

oldest

active

oldest

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)

• 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

• 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

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)

• 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

• 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

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)

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)

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

• 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

• 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

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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