A partition of 80 into seven parts












6














The sum of seven positive integers, not necessarily distinct, is 80. If placed appropriately on the vertices of this graph, two of them will be joined by an edge if and only if they have a common divisor greater than 1 (that is, they are not relatively prime). In non-decreasing order, what are those seven integers?



enter image description here










share|improve this question






















  • For those interested, this is were it all began:youtube.com/watch?v=6Giujua_s80&t=18s
    – Bernardo Recamán Santos
    Dec 1 at 18:29










  • Is there any reason to the strange phrasing non-decreasing order? You can just say "increasing order".
    – Hugh
    Dec 1 at 21:15










  • @Hugh It's maths-y language... Strictly speaking, the number sequence does not always increase, since the second 14 isn't larger than the first. But it never decreases.
    – Angkor
    Dec 1 at 22:06










  • @Angkor ah, yes. I had not yet looked at the solution provided by Dr. Xorile and so I did not see that there is a duplicate number. The choice of words makes a lot more sense now.
    – Hugh
    Dec 1 at 23:12
















6














The sum of seven positive integers, not necessarily distinct, is 80. If placed appropriately on the vertices of this graph, two of them will be joined by an edge if and only if they have a common divisor greater than 1 (that is, they are not relatively prime). In non-decreasing order, what are those seven integers?



enter image description here










share|improve this question






















  • For those interested, this is were it all began:youtube.com/watch?v=6Giujua_s80&t=18s
    – Bernardo Recamán Santos
    Dec 1 at 18:29










  • Is there any reason to the strange phrasing non-decreasing order? You can just say "increasing order".
    – Hugh
    Dec 1 at 21:15










  • @Hugh It's maths-y language... Strictly speaking, the number sequence does not always increase, since the second 14 isn't larger than the first. But it never decreases.
    – Angkor
    Dec 1 at 22:06










  • @Angkor ah, yes. I had not yet looked at the solution provided by Dr. Xorile and so I did not see that there is a duplicate number. The choice of words makes a lot more sense now.
    – Hugh
    Dec 1 at 23:12














6












6








6


1





The sum of seven positive integers, not necessarily distinct, is 80. If placed appropriately on the vertices of this graph, two of them will be joined by an edge if and only if they have a common divisor greater than 1 (that is, they are not relatively prime). In non-decreasing order, what are those seven integers?



enter image description here










share|improve this question













The sum of seven positive integers, not necessarily distinct, is 80. If placed appropriately on the vertices of this graph, two of them will be joined by an edge if and only if they have a common divisor greater than 1 (that is, they are not relatively prime). In non-decreasing order, what are those seven integers?



enter image description here







arithmetic






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Dec 1 at 16:20









Bernardo Recamán Santos

2,3281141




2,3281141












  • For those interested, this is were it all began:youtube.com/watch?v=6Giujua_s80&t=18s
    – Bernardo Recamán Santos
    Dec 1 at 18:29










  • Is there any reason to the strange phrasing non-decreasing order? You can just say "increasing order".
    – Hugh
    Dec 1 at 21:15










  • @Hugh It's maths-y language... Strictly speaking, the number sequence does not always increase, since the second 14 isn't larger than the first. But it never decreases.
    – Angkor
    Dec 1 at 22:06










  • @Angkor ah, yes. I had not yet looked at the solution provided by Dr. Xorile and so I did not see that there is a duplicate number. The choice of words makes a lot more sense now.
    – Hugh
    Dec 1 at 23:12


















  • For those interested, this is were it all began:youtube.com/watch?v=6Giujua_s80&t=18s
    – Bernardo Recamán Santos
    Dec 1 at 18:29










  • Is there any reason to the strange phrasing non-decreasing order? You can just say "increasing order".
    – Hugh
    Dec 1 at 21:15










  • @Hugh It's maths-y language... Strictly speaking, the number sequence does not always increase, since the second 14 isn't larger than the first. But it never decreases.
    – Angkor
    Dec 1 at 22:06










  • @Angkor ah, yes. I had not yet looked at the solution provided by Dr. Xorile and so I did not see that there is a duplicate number. The choice of words makes a lot more sense now.
    – Hugh
    Dec 1 at 23:12
















For those interested, this is were it all began:youtube.com/watch?v=6Giujua_s80&t=18s
– Bernardo Recamán Santos
Dec 1 at 18:29




For those interested, this is were it all began:youtube.com/watch?v=6Giujua_s80&t=18s
– Bernardo Recamán Santos
Dec 1 at 18:29












Is there any reason to the strange phrasing non-decreasing order? You can just say "increasing order".
– Hugh
Dec 1 at 21:15




Is there any reason to the strange phrasing non-decreasing order? You can just say "increasing order".
– Hugh
Dec 1 at 21:15












@Hugh It's maths-y language... Strictly speaking, the number sequence does not always increase, since the second 14 isn't larger than the first. But it never decreases.
– Angkor
Dec 1 at 22:06




@Hugh It's maths-y language... Strictly speaking, the number sequence does not always increase, since the second 14 isn't larger than the first. But it never decreases.
– Angkor
Dec 1 at 22:06












@Angkor ah, yes. I had not yet looked at the solution provided by Dr. Xorile and so I did not see that there is a duplicate number. The choice of words makes a lot more sense now.
– Hugh
Dec 1 at 23:12




@Angkor ah, yes. I had not yet looked at the solution provided by Dr. Xorile and so I did not see that there is a duplicate number. The choice of words makes a lot more sense now.
– Hugh
Dec 1 at 23:12










2 Answers
2






active

oldest

votes


















6














I don't know if this answer is unique, but it seems to work




1,5,7,9,14,14,30




Here it is in place:




Solution




There seems to be some confusion in the other answers, so let me spell it out my thinking a little. OP specified that the numbers would be "joined by an edge if and only if they have a common divisor greater than 1".



In other words, if they are joined, then they have a common divisor greater than 1. And not otherwise.



So:




  • 7 and 14 are joined (gcd (greatest common divisor) is 7)

  • 7 and 30 are not joined (gcd is 1)

  • 7 and 9 are not joined (gcd is 1)

  • 7 and 5 are not joined (gcd is 1)

  • 14 and 14 are joined (gcd is 14)

  • 30 and 14 are joined (gcd is 2)

  • 30 and 9 are joined (gcd is 3)

  • 14 and 9 are not joined (gcd is 1)

  • 1 is not joined to anything (gcd is always 1)

  • etc.


In order to solve this, I realized that the number in the middle must have at least 3 distinct prime factors, so that it could share those factors with the different arms and have those arms not share a factor.



2 x 3 x 5 is the smallest way to do this. The challenge was going to be keeping it below 80. So I made the 2 go to the left where more factors would come into play and the 3 and 5 go right. That gets us to here:




partial solution




Now $x$ needs to share a divisor with the leftmost position. It cannot be 2, 3, or 5, so 7 is a reasonable guess:




partial solution 2




Now we add that up so far and it comes to 73. Mmm. That would make the final number 7, which would be sharing a factor with some of the numbers on the board already. In fact, any number less than 7 would share a factor with something... except 1. So I changed 3 to 9 ($3^2$). This doesn't change any of the divisor logic - numbers will still either be relatively prime or not. But it leaves the final number equal to 1. Hence the solution. Probably unique, looking through this logic.






share|improve this answer























  • Quick! I think it'd be nice to show how they fit on the graph. Could you also explain how you got to this answer?
    – Angkor
    Dec 1 at 17:09










  • The answer is unique.
    – Bernardo Recamán Santos
    Dec 1 at 18:30



















-2














Seems like there are a lot of possible solutions - but assuming the two connected numbers common divisor cannot be one of the two numbers, and that you cannot repeat a number, I got this:




      6                 10

4 12 36

8 14



Every number shares a divisor of 2, which is more than 1.



So my final answer is:




4, 6, 8, 10, 12, 14, 36.







share|improve this answer

















  • 1




    12 and 36 are not relatively prime, so should be connected. Also 4 and 12. 6 and 10. 6 and 10, etc etc
    – Dr Xorile
    Dec 1 at 17:52











Your Answer





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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









6














I don't know if this answer is unique, but it seems to work




1,5,7,9,14,14,30




Here it is in place:




Solution




There seems to be some confusion in the other answers, so let me spell it out my thinking a little. OP specified that the numbers would be "joined by an edge if and only if they have a common divisor greater than 1".



In other words, if they are joined, then they have a common divisor greater than 1. And not otherwise.



So:




  • 7 and 14 are joined (gcd (greatest common divisor) is 7)

  • 7 and 30 are not joined (gcd is 1)

  • 7 and 9 are not joined (gcd is 1)

  • 7 and 5 are not joined (gcd is 1)

  • 14 and 14 are joined (gcd is 14)

  • 30 and 14 are joined (gcd is 2)

  • 30 and 9 are joined (gcd is 3)

  • 14 and 9 are not joined (gcd is 1)

  • 1 is not joined to anything (gcd is always 1)

  • etc.


In order to solve this, I realized that the number in the middle must have at least 3 distinct prime factors, so that it could share those factors with the different arms and have those arms not share a factor.



2 x 3 x 5 is the smallest way to do this. The challenge was going to be keeping it below 80. So I made the 2 go to the left where more factors would come into play and the 3 and 5 go right. That gets us to here:




partial solution




Now $x$ needs to share a divisor with the leftmost position. It cannot be 2, 3, or 5, so 7 is a reasonable guess:




partial solution 2




Now we add that up so far and it comes to 73. Mmm. That would make the final number 7, which would be sharing a factor with some of the numbers on the board already. In fact, any number less than 7 would share a factor with something... except 1. So I changed 3 to 9 ($3^2$). This doesn't change any of the divisor logic - numbers will still either be relatively prime or not. But it leaves the final number equal to 1. Hence the solution. Probably unique, looking through this logic.






share|improve this answer























  • Quick! I think it'd be nice to show how they fit on the graph. Could you also explain how you got to this answer?
    – Angkor
    Dec 1 at 17:09










  • The answer is unique.
    – Bernardo Recamán Santos
    Dec 1 at 18:30
















6














I don't know if this answer is unique, but it seems to work




1,5,7,9,14,14,30




Here it is in place:




Solution




There seems to be some confusion in the other answers, so let me spell it out my thinking a little. OP specified that the numbers would be "joined by an edge if and only if they have a common divisor greater than 1".



In other words, if they are joined, then they have a common divisor greater than 1. And not otherwise.



So:




  • 7 and 14 are joined (gcd (greatest common divisor) is 7)

  • 7 and 30 are not joined (gcd is 1)

  • 7 and 9 are not joined (gcd is 1)

  • 7 and 5 are not joined (gcd is 1)

  • 14 and 14 are joined (gcd is 14)

  • 30 and 14 are joined (gcd is 2)

  • 30 and 9 are joined (gcd is 3)

  • 14 and 9 are not joined (gcd is 1)

  • 1 is not joined to anything (gcd is always 1)

  • etc.


In order to solve this, I realized that the number in the middle must have at least 3 distinct prime factors, so that it could share those factors with the different arms and have those arms not share a factor.



2 x 3 x 5 is the smallest way to do this. The challenge was going to be keeping it below 80. So I made the 2 go to the left where more factors would come into play and the 3 and 5 go right. That gets us to here:




partial solution




Now $x$ needs to share a divisor with the leftmost position. It cannot be 2, 3, or 5, so 7 is a reasonable guess:




partial solution 2




Now we add that up so far and it comes to 73. Mmm. That would make the final number 7, which would be sharing a factor with some of the numbers on the board already. In fact, any number less than 7 would share a factor with something... except 1. So I changed 3 to 9 ($3^2$). This doesn't change any of the divisor logic - numbers will still either be relatively prime or not. But it leaves the final number equal to 1. Hence the solution. Probably unique, looking through this logic.






share|improve this answer























  • Quick! I think it'd be nice to show how they fit on the graph. Could you also explain how you got to this answer?
    – Angkor
    Dec 1 at 17:09










  • The answer is unique.
    – Bernardo Recamán Santos
    Dec 1 at 18:30














6












6








6






I don't know if this answer is unique, but it seems to work




1,5,7,9,14,14,30




Here it is in place:




Solution




There seems to be some confusion in the other answers, so let me spell it out my thinking a little. OP specified that the numbers would be "joined by an edge if and only if they have a common divisor greater than 1".



In other words, if they are joined, then they have a common divisor greater than 1. And not otherwise.



So:




  • 7 and 14 are joined (gcd (greatest common divisor) is 7)

  • 7 and 30 are not joined (gcd is 1)

  • 7 and 9 are not joined (gcd is 1)

  • 7 and 5 are not joined (gcd is 1)

  • 14 and 14 are joined (gcd is 14)

  • 30 and 14 are joined (gcd is 2)

  • 30 and 9 are joined (gcd is 3)

  • 14 and 9 are not joined (gcd is 1)

  • 1 is not joined to anything (gcd is always 1)

  • etc.


In order to solve this, I realized that the number in the middle must have at least 3 distinct prime factors, so that it could share those factors with the different arms and have those arms not share a factor.



2 x 3 x 5 is the smallest way to do this. The challenge was going to be keeping it below 80. So I made the 2 go to the left where more factors would come into play and the 3 and 5 go right. That gets us to here:




partial solution




Now $x$ needs to share a divisor with the leftmost position. It cannot be 2, 3, or 5, so 7 is a reasonable guess:




partial solution 2




Now we add that up so far and it comes to 73. Mmm. That would make the final number 7, which would be sharing a factor with some of the numbers on the board already. In fact, any number less than 7 would share a factor with something... except 1. So I changed 3 to 9 ($3^2$). This doesn't change any of the divisor logic - numbers will still either be relatively prime or not. But it leaves the final number equal to 1. Hence the solution. Probably unique, looking through this logic.






share|improve this answer














I don't know if this answer is unique, but it seems to work




1,5,7,9,14,14,30




Here it is in place:




Solution




There seems to be some confusion in the other answers, so let me spell it out my thinking a little. OP specified that the numbers would be "joined by an edge if and only if they have a common divisor greater than 1".



In other words, if they are joined, then they have a common divisor greater than 1. And not otherwise.



So:




  • 7 and 14 are joined (gcd (greatest common divisor) is 7)

  • 7 and 30 are not joined (gcd is 1)

  • 7 and 9 are not joined (gcd is 1)

  • 7 and 5 are not joined (gcd is 1)

  • 14 and 14 are joined (gcd is 14)

  • 30 and 14 are joined (gcd is 2)

  • 30 and 9 are joined (gcd is 3)

  • 14 and 9 are not joined (gcd is 1)

  • 1 is not joined to anything (gcd is always 1)

  • etc.


In order to solve this, I realized that the number in the middle must have at least 3 distinct prime factors, so that it could share those factors with the different arms and have those arms not share a factor.



2 x 3 x 5 is the smallest way to do this. The challenge was going to be keeping it below 80. So I made the 2 go to the left where more factors would come into play and the 3 and 5 go right. That gets us to here:




partial solution




Now $x$ needs to share a divisor with the leftmost position. It cannot be 2, 3, or 5, so 7 is a reasonable guess:




partial solution 2




Now we add that up so far and it comes to 73. Mmm. That would make the final number 7, which would be sharing a factor with some of the numbers on the board already. In fact, any number less than 7 would share a factor with something... except 1. So I changed 3 to 9 ($3^2$). This doesn't change any of the divisor logic - numbers will still either be relatively prime or not. But it leaves the final number equal to 1. Hence the solution. Probably unique, looking through this logic.







share|improve this answer














share|improve this answer



share|improve this answer








edited Dec 1 at 18:13

























answered Dec 1 at 16:35









Dr Xorile

11.5k12363




11.5k12363












  • Quick! I think it'd be nice to show how they fit on the graph. Could you also explain how you got to this answer?
    – Angkor
    Dec 1 at 17:09










  • The answer is unique.
    – Bernardo Recamán Santos
    Dec 1 at 18:30


















  • Quick! I think it'd be nice to show how they fit on the graph. Could you also explain how you got to this answer?
    – Angkor
    Dec 1 at 17:09










  • The answer is unique.
    – Bernardo Recamán Santos
    Dec 1 at 18:30
















Quick! I think it'd be nice to show how they fit on the graph. Could you also explain how you got to this answer?
– Angkor
Dec 1 at 17:09




Quick! I think it'd be nice to show how they fit on the graph. Could you also explain how you got to this answer?
– Angkor
Dec 1 at 17:09












The answer is unique.
– Bernardo Recamán Santos
Dec 1 at 18:30




The answer is unique.
– Bernardo Recamán Santos
Dec 1 at 18:30











-2














Seems like there are a lot of possible solutions - but assuming the two connected numbers common divisor cannot be one of the two numbers, and that you cannot repeat a number, I got this:




      6                 10

4 12 36

8 14



Every number shares a divisor of 2, which is more than 1.



So my final answer is:




4, 6, 8, 10, 12, 14, 36.







share|improve this answer

















  • 1




    12 and 36 are not relatively prime, so should be connected. Also 4 and 12. 6 and 10. 6 and 10, etc etc
    – Dr Xorile
    Dec 1 at 17:52
















-2














Seems like there are a lot of possible solutions - but assuming the two connected numbers common divisor cannot be one of the two numbers, and that you cannot repeat a number, I got this:




      6                 10

4 12 36

8 14



Every number shares a divisor of 2, which is more than 1.



So my final answer is:




4, 6, 8, 10, 12, 14, 36.







share|improve this answer

















  • 1




    12 and 36 are not relatively prime, so should be connected. Also 4 and 12. 6 and 10. 6 and 10, etc etc
    – Dr Xorile
    Dec 1 at 17:52














-2












-2








-2






Seems like there are a lot of possible solutions - but assuming the two connected numbers common divisor cannot be one of the two numbers, and that you cannot repeat a number, I got this:




      6                 10

4 12 36

8 14



Every number shares a divisor of 2, which is more than 1.



So my final answer is:




4, 6, 8, 10, 12, 14, 36.







share|improve this answer












Seems like there are a lot of possible solutions - but assuming the two connected numbers common divisor cannot be one of the two numbers, and that you cannot repeat a number, I got this:




      6                 10

4 12 36

8 14



Every number shares a divisor of 2, which is more than 1.



So my final answer is:




4, 6, 8, 10, 12, 14, 36.








share|improve this answer












share|improve this answer



share|improve this answer










answered Dec 1 at 17:04









a guy

42714




42714








  • 1




    12 and 36 are not relatively prime, so should be connected. Also 4 and 12. 6 and 10. 6 and 10, etc etc
    – Dr Xorile
    Dec 1 at 17:52














  • 1




    12 and 36 are not relatively prime, so should be connected. Also 4 and 12. 6 and 10. 6 and 10, etc etc
    – Dr Xorile
    Dec 1 at 17:52








1




1




12 and 36 are not relatively prime, so should be connected. Also 4 and 12. 6 and 10. 6 and 10, etc etc
– Dr Xorile
Dec 1 at 17:52




12 and 36 are not relatively prime, so should be connected. Also 4 and 12. 6 and 10. 6 and 10, etc etc
– Dr Xorile
Dec 1 at 17:52


















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