A 4 x 4 Magic Square with Pairwise Relatively Prime Entries
7
$begingroup$
Find a 4 x 4 magic square of positive integers such that any two of its entries are pairwise different and relatively prime, i.e., have no common divisor greater than 1.
What is the least that the largest number in such a square can be?
mathematics computer-puzzle magic-square
edited Feb 10 at 21:57
Jonathan Allan
17.9k14697
asked Feb 10 at 14:31
Bernardo Recamán SantosBernardo Recamán Santos
2,4991344
$endgroup$
|
show 1 more comment
7
$begingroup$
Find a 4 x 4 magic square of positive integers such that any two of its entries are pairwise different and relatively prime, i.e., have no common divisor greater than 1.
What is the least that the largest number in such a square can be?
mathematics computer-puzzle magic-square
edited Feb 10 at 21:57
Jonathan Allan
17.9k14697
asked Feb 10 at 14:31
Bernardo Recamán SantosBernardo Recamán Santos
2,4991344
$endgroup$
$begingroup$
What is “pairwise different”?
$endgroup$
– Arvasu Kulkarni
Feb 10 at 14:56
$begingroup$
@ArvasuKulkarni: Any two entries are different. They are also relatively prime, that is, they do not have a common divisor greater than 1.
$endgroup$
– Bernardo Recamán Santos
Feb 10 at 14:59
$begingroup$
Was this puzzle of your own creation?
$endgroup$
– Brandon_J
Feb 10 at 20:51
$begingroup$
@Brandon_J: An oldie twisted.
$endgroup$
– Bernardo Recamán Santos
Feb 10 at 21:54
1
$begingroup$
Wow, that's the best puzzle I ever encountered here (and I've been a lurker for more than a year, I think). A question about the definition: can the square contain a 1, i. e. is 1 considered a coprime to any other natural number or not? It commonly is, but not always.
$endgroup$
– kkm
Feb 11 at 0:08
|
show 1 more comment
7
7
7
1
$begingroup$
Find a 4 x 4 magic square of positive integers such that any two of its entries are pairwise different and relatively prime, i.e., have no common divisor greater than 1.
What is the least that the largest number in such a square can be?
mathematics computer-puzzle magic-square
edited Feb 10 at 21:57
Jonathan Allan
17.9k14697
asked Feb 10 at 14:31
Bernardo Recamán SantosBernardo Recamán Santos
2,4991344
$endgroup$
Find a 4 x 4 magic square of positive integers such that any two of its entries are pairwise different and relatively prime, i.e., have no common divisor greater than 1.
What is the least that the largest number in such a square can be?
mathematics computer-puzzle magic-square
mathematics computer-puzzle magic-square
edited Feb 10 at 21:57
Jonathan Allan
17.9k14697
asked Feb 10 at 14:31
Bernardo Recamán SantosBernardo Recamán Santos
2,4991344
edited Feb 10 at 21:57
Jonathan Allan
17.9k14697
asked Feb 10 at 14:31
Bernardo Recamán SantosBernardo Recamán Santos
2,4991344
edited Feb 10 at 21:57
Jonathan Allan
17.9k14697
edited Feb 10 at 21:57
Jonathan Allan
17.9k14697
edited Feb 10 at 21:57
Jonathan Allan
17.9k14697
17.9k14697
asked Feb 10 at 14:31
Bernardo Recamán SantosBernardo Recamán Santos
2,4991344
asked Feb 10 at 14:31
Bernardo Recamán SantosBernardo Recamán Santos
2,4991344
asked Feb 10 at 14:31
Bernardo Recamán SantosBernardo Recamán Santos
2,4991344
2,4991344
$begingroup$
What is “pairwise different”?
$endgroup$
– Arvasu Kulkarni
Feb 10 at 14:56
$begingroup$
@ArvasuKulkarni: Any two entries are different. They are also relatively prime, that is, they do not have a common divisor greater than 1.
$endgroup$
– Bernardo Recamán Santos
Feb 10 at 14:59
$begingroup$
Was this puzzle of your own creation?
$endgroup$
– Brandon_J
Feb 10 at 20:51
$begingroup$
@Brandon_J: An oldie twisted.
$endgroup$
– Bernardo Recamán Santos
Feb 10 at 21:54
1
$begingroup$
Wow, that's the best puzzle I ever encountered here (and I've been a lurker for more than a year, I think). A question about the definition: can the square contain a 1, i. e. is 1 considered a coprime to any other natural number or not? It commonly is, but not always.
$endgroup$
– kkm
Feb 11 at 0:08
|
show 1 more comment
$begingroup$
What is “pairwise different”?
$endgroup$
– Arvasu Kulkarni
Feb 10 at 14:56
$begingroup$
@ArvasuKulkarni: Any two entries are different. They are also relatively prime, that is, they do not have a common divisor greater than 1.
$endgroup$
– Bernardo Recamán Santos
Feb 10 at 14:59
$begingroup$
Was this puzzle of your own creation?
$endgroup$
– Brandon_J
Feb 10 at 20:51
$begingroup$
@Brandon_J: An oldie twisted.
$endgroup$
– Bernardo Recamán Santos
Feb 10 at 21:54
1
$begingroup$
Wow, that's the best puzzle I ever encountered here (and I've been a lurker for more than a year, I think). A question about the definition: can the square contain a 1, i. e. is 1 considered a coprime to any other natural number or not? It commonly is, but not always.
$endgroup$
– kkm
Feb 11 at 0:08
$begingroup$
What is “pairwise different”?
$endgroup$
– Arvasu Kulkarni
Feb 10 at 14:56
$begingroup$
What is “pairwise different”?
$endgroup$
– Arvasu Kulkarni
Feb 10 at 14:56
$begingroup$
@ArvasuKulkarni: Any two entries are different. They are also relatively prime, that is, they do not have a common divisor greater than 1.
$endgroup$
– Bernardo Recamán Santos
Feb 10 at 14:59
$begingroup$
@ArvasuKulkarni: Any two entries are different. They are also relatively prime, that is, they do not have a common divisor greater than 1.
$endgroup$
– Bernardo Recamán Santos
Feb 10 at 14:59
$begingroup$
Was this puzzle of your own creation?
$endgroup$
– Brandon_J
Feb 10 at 20:51
$begingroup$
Was this puzzle of your own creation?
$endgroup$
– Brandon_J
Feb 10 at 20:51
$begingroup$
@Brandon_J: An oldie twisted.
$endgroup$
– Bernardo Recamán Santos
Feb 10 at 21:54
$begingroup$
@Brandon_J: An oldie twisted.
$endgroup$
– Bernardo Recamán Santos
Feb 10 at 21:54
1
1
$begingroup$
Wow, that's the best puzzle I ever encountered here (and I've been a lurker for more than a year, I think). A question about the definition: can the square contain a 1, i. e. is 1 considered a coprime to any other natural number or not? It commonly is, but not always.
$endgroup$
– kkm
Feb 11 at 0:08
$begingroup$
Wow, that's the best puzzle I ever encountered here (and I've been a lurker for more than a year, I think). A question about the definition: can the square contain a 1, i. e. is 1 considered a coprime to any other natural number or not? It commonly is, but not always.
$endgroup$
– kkm
Feb 11 at 0:08
|
show 1 more comment
3 Answers
3
active
oldest
votes
3
$begingroup$
I believe this is optimal (unless I've missed a trick):
1 13 47 53
29 59 7 19
41 11 37 25
43 31 23 17
Which has
a maximal value of 59 (and a sum of 114)
Note: all values are prime except for 1 and the composite number 25
I also found these two with the same maximum value:
...using primes, 1, 9 and 49 (with a sum of 114);1 17 37 59
53 29 23 9
47 19 43 5
13 49 11 41
and
...using primes, 1, 39 and 49 (with a sum of 126)1 29 47 49
43 41 37 5
59 17 31 19
23 39 11 53
First I found these two:
and1 11 41 61
47 31 17 19
43 13 53 5
23 59 3 29
both of which have a maximal value of 61 (and a sum of 114)1 13 47 53
29 59 7 19
61 31 17 5
23 11 43 37
For these I restricted myself to fifteen odd primes less than 73 and added the number one as the sixteenth value. These two have the smallest maximal value given this additional constraint.
answered Feb 10 at 22:11
Jonathan AllanJonathan Allan
17.9k14697
$endgroup$
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
Feb 10 at 22:22
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
Feb 10 at 22:25
$begingroup$
The diagonals work, but not whatever else this considers when deciding if it is "perfect" - looking at the results when you give a magic total it seems to want squares and offset 2-by-1s or 1-by-2s and non-main diagonals.
$endgroup$
– Jonathan Allan
Feb 10 at 22:35
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
Feb 10 at 22:35
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
Feb 10 at 22:40
add a comment |
3
$begingroup$
I've got a solution:
and I admit I found it online.
Here it is:
The largest number in it is
73
answered Feb 10 at 20:30
Brandon_JBrandon_J
1,653228
$endgroup$
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
Feb 10 at 20:56
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
Feb 10 at 20:56
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
$endgroup$
– Jonathan Allan
Feb 10 at 21:00
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
Feb 10 at 21:04
$begingroup$
1 is co-prime to all other integers. Note that I asked if we are only allowed positive integers because the same is true of -1.
$endgroup$
– Jonathan Allan
Feb 10 at 21:06
|
show 5 more comments
1
$begingroup$
The following magic square
11 1 53 37
7 47 29 19
71 23 3 5
13 31 17 41
has magic constant
102
and largest number
71
answered Feb 11 at 0:17
Freddy BarreraFreddy Barrera
34318
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
I believe this is optimal (unless I've missed a trick):
1 13 47 53
29 59 7 19
41 11 37 25
43 31 23 17
Which has
a maximal value of 59 (and a sum of 114)
Note: all values are prime except for 1 and the composite number 25
I also found these two with the same maximum value:
...using primes, 1, 9 and 49 (with a sum of 114);1 17 37 59
53 29 23 9
47 19 43 5
13 49 11 41
and
...using primes, 1, 39 and 49 (with a sum of 126)1 29 47 49
43 41 37 5
59 17 31 19
23 39 11 53
First I found these two:
and1 11 41 61
47 31 17 19
43 13 53 5
23 59 3 29
both of which have a maximal value of 61 (and a sum of 114)1 13 47 53
29 59 7 19
61 31 17 5
23 11 43 37
For these I restricted myself to fifteen odd primes less than 73 and added the number one as the sixteenth value. These two have the smallest maximal value given this additional constraint.
answered Feb 10 at 22:11
Jonathan AllanJonathan Allan
17.9k14697
$endgroup$
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
Feb 10 at 22:22
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
Feb 10 at 22:25
$begingroup$
The diagonals work, but not whatever else this considers when deciding if it is "perfect" - looking at the results when you give a magic total it seems to want squares and offset 2-by-1s or 1-by-2s and non-main diagonals.
$endgroup$
– Jonathan Allan
Feb 10 at 22:35
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
Feb 10 at 22:35
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
Feb 10 at 22:40
add a comment |
3
$begingroup$
I believe this is optimal (unless I've missed a trick):
1 13 47 53
29 59 7 19
41 11 37 25
43 31 23 17
Which has
a maximal value of 59 (and a sum of 114)
Note: all values are prime except for 1 and the composite number 25
I also found these two with the same maximum value:
...using primes, 1, 9 and 49 (with a sum of 114);1 17 37 59
53 29 23 9
47 19 43 5
13 49 11 41
and
...using primes, 1, 39 and 49 (with a sum of 126)1 29 47 49
43 41 37 5
59 17 31 19
23 39 11 53
First I found these two:
and1 11 41 61
47 31 17 19
43 13 53 5
23 59 3 29
both of which have a maximal value of 61 (and a sum of 114)1 13 47 53
29 59 7 19
61 31 17 5
23 11 43 37
For these I restricted myself to fifteen odd primes less than 73 and added the number one as the sixteenth value. These two have the smallest maximal value given this additional constraint.
answered Feb 10 at 22:11
Jonathan AllanJonathan Allan
17.9k14697
$endgroup$
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
Feb 10 at 22:22
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
Feb 10 at 22:25
$begingroup$
The diagonals work, but not whatever else this considers when deciding if it is "perfect" - looking at the results when you give a magic total it seems to want squares and offset 2-by-1s or 1-by-2s and non-main diagonals.
$endgroup$
– Jonathan Allan
Feb 10 at 22:35
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
Feb 10 at 22:35
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
Feb 10 at 22:40
add a comment |
3
3
3
$begingroup$
I believe this is optimal (unless I've missed a trick):
1 13 47 53
29 59 7 19
41 11 37 25
43 31 23 17
Which has
a maximal value of 59 (and a sum of 114)
Note: all values are prime except for 1 and the composite number 25
I also found these two with the same maximum value:
...using primes, 1, 9 and 49 (with a sum of 114);1 17 37 59
53 29 23 9
47 19 43 5
13 49 11 41
and
...using primes, 1, 39 and 49 (with a sum of 126)1 29 47 49
43 41 37 5
59 17 31 19
23 39 11 53
First I found these two:
and1 11 41 61
47 31 17 19
43 13 53 5
23 59 3 29
both of which have a maximal value of 61 (and a sum of 114)1 13 47 53
29 59 7 19
61 31 17 5
23 11 43 37
For these I restricted myself to fifteen odd primes less than 73 and added the number one as the sixteenth value. These two have the smallest maximal value given this additional constraint.
answered Feb 10 at 22:11
Jonathan AllanJonathan Allan
17.9k14697
$endgroup$
I believe this is optimal (unless I've missed a trick):
1 13 47 53
29 59 7 19
41 11 37 25
43 31 23 17
Which has
a maximal value of 59 (and a sum of 114)
Note: all values are prime except for 1 and the composite number 25
I also found these two with the same maximum value:
...using primes, 1, 9 and 49 (with a sum of 114);1 17 37 59
53 29 23 9
47 19 43 5
13 49 11 41
and
...using primes, 1, 39 and 49 (with a sum of 126)1 29 47 49
43 41 37 5
59 17 31 19
23 39 11 53
First I found these two:
and1 11 41 61
47 31 17 19
43 13 53 5
23 59 3 29
both of which have a maximal value of 61 (and a sum of 114)1 13 47 53
29 59 7 19
61 31 17 5
23 11 43 37
For these I restricted myself to fifteen odd primes less than 73 and added the number one as the sixteenth value. These two have the smallest maximal value given this additional constraint.
answered Feb 10 at 22:11
Jonathan AllanJonathan Allan
17.9k14697
edited Feb 10 at 23:57
answered Feb 10 at 22:11
Jonathan AllanJonathan Allan
17.9k14697
answered Feb 10 at 22:11
Jonathan AllanJonathan Allan
17.9k14697
answered Feb 10 at 22:11
Jonathan AllanJonathan Allan
17.9k14697
17.9k14697
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
Feb 10 at 22:22
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
Feb 10 at 22:25
$begingroup$
The diagonals work, but not whatever else this considers when deciding if it is "perfect" - looking at the results when you give a magic total it seems to want squares and offset 2-by-1s or 1-by-2s and non-main diagonals.
$endgroup$
– Jonathan Allan
Feb 10 at 22:35
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
Feb 10 at 22:35
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
Feb 10 at 22:40
add a comment |
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
Feb 10 at 22:22
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
Feb 10 at 22:25
$begingroup$
The diagonals work, but not whatever else this considers when deciding if it is "perfect" - looking at the results when you give a magic total it seems to want squares and offset 2-by-1s or 1-by-2s and non-main diagonals.
$endgroup$
– Jonathan Allan
Feb 10 at 22:35
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
Feb 10 at 22:35
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
Feb 10 at 22:40
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
Feb 10 at 22:22
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
Feb 10 at 22:22
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
Feb 10 at 22:25
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
Feb 10 at 22:25
$begingroup$
The diagonals work, but not whatever else this considers when deciding if it is "perfect" - looking at the results when you give a magic total it seems to want squares and offset 2-by-1s or 1-by-2s and non-main diagonals.
$endgroup$
– Jonathan Allan
Feb 10 at 22:35
$begingroup$
The diagonals work, but not whatever else this considers when deciding if it is "perfect" - looking at the results when you give a magic total it seems to want squares and offset 2-by-1s or 1-by-2s and non-main diagonals.
$endgroup$
– Jonathan Allan
Feb 10 at 22:35
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
Feb 10 at 22:35
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
Feb 10 at 22:35
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
Feb 10 at 22:40
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
Feb 10 at 22:40
add a comment |
3
$begingroup$
I've got a solution:
and I admit I found it online.
Here it is:
The largest number in it is
73
answered Feb 10 at 20:30
Brandon_JBrandon_J
1,653228
$endgroup$
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
Feb 10 at 20:56
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
Feb 10 at 20:56
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
$endgroup$
– Jonathan Allan
Feb 10 at 21:00
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
Feb 10 at 21:04
$begingroup$
1 is co-prime to all other integers. Note that I asked if we are only allowed positive integers because the same is true of -1.
$endgroup$
– Jonathan Allan
Feb 10 at 21:06
|
show 5 more comments
3
$begingroup$
I've got a solution:
and I admit I found it online.
Here it is:
The largest number in it is
73
answered Feb 10 at 20:30
Brandon_JBrandon_J
1,653228
$endgroup$
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
Feb 10 at 20:56
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
Feb 10 at 20:56
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
$endgroup$
– Jonathan Allan
Feb 10 at 21:00
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
Feb 10 at 21:04
$begingroup$
1 is co-prime to all other integers. Note that I asked if we are only allowed positive integers because the same is true of -1.
$endgroup$
– Jonathan Allan
Feb 10 at 21:06
|
show 5 more comments
3
3
3
$begingroup$
I've got a solution:
and I admit I found it online.
Here it is:
The largest number in it is
73
answered Feb 10 at 20:30
Brandon_JBrandon_J
1,653228
$endgroup$
I've got a solution:
and I admit I found it online.
Here it is:
The largest number in it is
73
answered Feb 10 at 20:30
Brandon_JBrandon_J
1,653228
edited Feb 10 at 20:49
answered Feb 10 at 20:30
Brandon_JBrandon_J
1,653228
answered Feb 10 at 20:30
Brandon_JBrandon_J
1,653228
answered Feb 10 at 20:30
Brandon_JBrandon_J
1,653228
1,653228
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
Feb 10 at 20:56
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
Feb 10 at 20:56
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
$endgroup$
– Jonathan Allan
Feb 10 at 21:00
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
Feb 10 at 21:04
$begingroup$
1 is co-prime to all other integers. Note that I asked if we are only allowed positive integers because the same is true of -1.
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– Jonathan Allan
Feb 10 at 21:06
|
show 5 more comments
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Maybe we can make a smaller one by using 1 as one of our numbers?
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– Jonathan Allan
Feb 10 at 20:56
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I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
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– Brandon_J
Feb 10 at 20:56
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I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
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– Jonathan Allan
Feb 10 at 21:00
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@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
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– Brandon_J
Feb 10 at 21:04
$begingroup$
1 is co-prime to all other integers. Note that I asked if we are only allowed positive integers because the same is true of -1.
$endgroup$
– Jonathan Allan
Feb 10 at 21:06
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
Feb 10 at 20:56
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
Feb 10 at 20:56
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
Feb 10 at 20:56
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
Feb 10 at 20:56
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
$endgroup$
– Jonathan Allan
Feb 10 at 21:00
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
$endgroup$
– Jonathan Allan
Feb 10 at 21:00
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
Feb 10 at 21:04
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
Feb 10 at 21:04
$begingroup$
1 is co-prime to all other integers. Note that I asked if we are only allowed positive integers because the same is true of -1.
$endgroup$
– Jonathan Allan
Feb 10 at 21:06
$begingroup$
1 is co-prime to all other integers. Note that I asked if we are only allowed positive integers because the same is true of -1.
$endgroup$
– Jonathan Allan
Feb 10 at 21:06
|
show 5 more comments
1
$begingroup$
The following magic square
11 1 53 37
7 47 29 19
71 23 3 5
13 31 17 41
has magic constant
102
and largest number
71
answered Feb 11 at 0:17
Freddy BarreraFreddy Barrera
34318
$endgroup$
add a comment |
1
$begingroup$
The following magic square
11 1 53 37
7 47 29 19
71 23 3 5
13 31 17 41
has magic constant
102
and largest number
71
answered Feb 11 at 0:17
Freddy BarreraFreddy Barrera
34318
$endgroup$
add a comment |
1
1
1
$begingroup$
The following magic square
11 1 53 37
7 47 29 19
71 23 3 5
13 31 17 41
has magic constant
102
and largest number
71
answered Feb 11 at 0:17
Freddy BarreraFreddy Barrera
34318
$endgroup$
The following magic square
11 1 53 37
7 47 29 19
71 23 3 5
13 31 17 41
has magic constant
102
and largest number
71
answered Feb 11 at 0:17
Freddy BarreraFreddy Barrera
34318
answered Feb 11 at 0:17
Freddy BarreraFreddy Barrera
34318
answered Feb 11 at 0:17
Freddy BarreraFreddy Barrera
34318
answered Feb 11 at 0:17
Freddy BarreraFreddy Barrera
34318
34318
add a comment |
add a comment |
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$begingroup$
What is “pairwise different”?
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– Arvasu Kulkarni
Feb 10 at 14:56
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@ArvasuKulkarni: Any two entries are different. They are also relatively prime, that is, they do not have a common divisor greater than 1.
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– Bernardo Recamán Santos
Feb 10 at 14:59
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Was this puzzle of your own creation?
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– Brandon_J
Feb 10 at 20:51
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@Brandon_J: An oldie twisted.
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– Bernardo Recamán Santos
Feb 10 at 21:54
1
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Wow, that's the best puzzle I ever encountered here (and I've been a lurker for more than a year, I think). A question about the definition: can the square contain a 1, i. e. is 1 considered a coprime to any other natural number or not? It commonly is, but not always.
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– kkm
Feb 11 at 0:08