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.
mathematics number-theory
asked Nov 29 at 2:34
Bernardo Recamán Santos
2,2761141
add a comment |
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.
mathematics number-theory
asked Nov 29 at 2:34
Bernardo Recamán Santos
2,2761141
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
add a comment |
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.
mathematics number-theory
asked Nov 29 at 2:34
Bernardo Recamán Santos
2,2761141
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
mathematics number-theory
asked Nov 29 at 2:34
Bernardo Recamán Santos
2,2761141
asked Nov 29 at 2:34
Bernardo Recamán Santos
2,2761141
asked Nov 29 at 2:34
Bernardo Recamán Santos
2,2761141
asked Nov 29 at 2:34
Bernardo Recamán Santos
2,2761141
asked Nov 29 at 2:34
Bernardo Recamán Santos
2,2761141
2,2761141
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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)
answered Nov 29 at 2:57
Gareth McCaughan♦
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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)
answered Nov 29 at 2:57
Gareth McCaughan♦
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
add a comment |
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)
answered Nov 29 at 2:57
Gareth McCaughan♦
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
add a comment |
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)
answered Nov 29 at 2:57
Gareth McCaughan♦
59.5k3150230
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)
answered Nov 29 at 2:57
Gareth McCaughan♦
59.5k3150230
answered Nov 29 at 2:57
Gareth McCaughan♦
59.5k3150230
answered Nov 29 at 2:57
Gareth McCaughan♦
59.5k3150230
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
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
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
add a comment |
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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