Clash Royale CLAN TAG#URR8PPP
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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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7
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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.

1Was 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
add a comment 
7
down vote
favorite
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.
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.

1Was 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
add a comment 

1Was 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
– Dr Xorile
Nov 29 at 4:38
– Dr Xorile
Nov 29 at 4:38
– Bernardo Recamán Santos
Nov 29 at 11:18
– Bernardo Recamán Santos
Nov 29 at 11:18
add a comment 
1 Answer
1
active
oldest
votes
7
down vote
accepted
This appears to work, and I think it’s unique but haven’t checked supercarefully:
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 
1To expand a bit: 2,11,13,14 can each only be adjacent to two numbers, so that part of the sequence is forced, namely 314925 and 10136118. 1 could neighbour 4,6,12 but 6 is now “taken” by 13 and 11, so 1214 is forced. 4 could neighbour 9 but 9 is also taken, 147. 12 could neighbour 7 but that would close off a cycle of length 4. So 12 and 7 are forced: 710 and 512. 3 could neighbour 10 but that’s taken. so 83. 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
add a comment 
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
down vote
accepted
This appears to work, and I think it’s unique but haven’t checked supercarefully:
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 
1To expand a bit: 2,11,13,14 can each only be adjacent to two numbers, so that part of the sequence is forced, namely 314925 and 10136118. 1 could neighbour 4,6,12 but 6 is now “taken” by 13 and 11, so 1214 is forced. 4 could neighbour 9 but 9 is also taken, 147. 12 could neighbour 7 but that would close off a cycle of length 4. So 12 and 7 are forced: 710 and 512. 3 could neighbour 10 but that’s taken. so 83. 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
add a comment 
7
down vote
accepted
This appears to work, and I think it’s unique but haven’t checked supercarefully:
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 
1To expand a bit: 2,11,13,14 can each only be adjacent to two numbers, so that part of the sequence is forced, namely 314925 and 10136118. 1 could neighbour 4,6,12 but 6 is now “taken” by 13 and 11, so 1214 is forced. 4 could neighbour 9 but 9 is also taken, 147. 12 could neighbour 7 but that would close off a cycle of length 4. So 12 and 7 are forced: 710 and 512. 3 could neighbour 10 but that’s taken. so 83. 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
add a comment 
7
down vote
accepted
7
down vote
accepted
This appears to work, and I think it’s unique but haven’t checked supercarefully:
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 supercarefully:
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 
1To expand a bit: 2,11,13,14 can each only be adjacent to two numbers, so that part of the sequence is forced, namely 314925 and 10136118. 1 could neighbour 4,6,12 but 6 is now “taken” by 13 and 11, so 1214 is forced. 4 could neighbour 9 but 9 is also taken, 147. 12 could neighbour 7 but that would close off a cycle of length 4. So 12 and 7 are forced: 710 and 512. 3 could neighbour 10 but that’s taken. so 83. 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
add a comment 

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 
1To expand a bit: 2,11,13,14 can each only be adjacent to two numbers, so that part of the sequence is forced, namely 314925 and 10136118. 1 could neighbour 4,6,12 but 6 is now “taken” by 13 and 11, so 1214 is forced. 4 could neighbour 9 but 9 is also taken, 147. 12 could neighbour 7 but that would close off a cycle of length 4. So 12 and 7 are forced: 710 and 512. 3 could neighbour 10 but that’s taken. so 83. 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
– deep thought
Nov 29 at 3:03
– deep thought
Nov 29 at 3:03
– deep thought
Nov 29 at 3:18
– deep thought
Nov 29 at 3:18
– Gareth McCaughan♦
Nov 29 at 4:04
– Gareth McCaughan♦
Nov 29 at 4:04
add a comment 
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– Dr Xorile
Nov 29 at 4:38
– Bernardo Recamán Santos
Nov 29 at 11:18