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 4 hours ago
Jonathan Allan
17.8k14697
asked 11 hours ago
Bernardo Recamán SantosBernardo Recamán Santos
2,4721343
$endgroup$
|
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 4 hours ago
Jonathan Allan
17.8k14697
asked 11 hours ago
Bernardo Recamán SantosBernardo Recamán Santos
2,4721343
$endgroup$
$begingroup$
What is “pairwise different”?
$endgroup$
– Arvasu Kulkarni
11 hours ago
$begingroup$
@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
11 hours ago
$begingroup$
Was this puzzle of your own creation?
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
@Brandon_J: An oldie twisted.
$endgroup$
– Bernardo Recamán Santos
4 hours ago
$begingroup$
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
2 hours ago
|
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 4 hours ago
Jonathan Allan
17.8k14697
asked 11 hours ago
Bernardo Recamán SantosBernardo Recamán Santos
2,4721343
$endgroup$
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 4 hours ago
Jonathan Allan
17.8k14697
asked 11 hours ago
Bernardo Recamán SantosBernardo Recamán Santos
2,4721343
edited 4 hours ago
Jonathan Allan
17.8k14697
asked 11 hours ago
Bernardo Recamán SantosBernardo Recamán Santos
2,4721343
edited 4 hours ago
Jonathan Allan
17.8k14697
edited 4 hours ago
Jonathan Allan
17.8k14697
edited 4 hours ago
Jonathan Allan
17.8k14697
17.8k14697
asked 11 hours ago
Bernardo Recamán SantosBernardo Recamán Santos
2,4721343
asked 11 hours ago
Bernardo Recamán SantosBernardo Recamán Santos
2,4721343
asked 11 hours ago
Bernardo Recamán SantosBernardo Recamán Santos
2,4721343
2,4721343
$begingroup$
What is “pairwise different”?
$endgroup$
– Arvasu Kulkarni
11 hours ago
$begingroup$
@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
11 hours ago
$begingroup$
Was this puzzle of your own creation?
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
@Brandon_J: An oldie twisted.
$endgroup$
– Bernardo Recamán Santos
4 hours ago
$begingroup$
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
2 hours ago
|
show 1 more comment
$begingroup$
What is “pairwise different”?
$endgroup$
– Arvasu Kulkarni
11 hours ago
$begingroup$
@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
11 hours ago
$begingroup$
Was this puzzle of your own creation?
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
@Brandon_J: An oldie twisted.
$endgroup$
– Bernardo Recamán Santos
4 hours ago
$begingroup$
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
2 hours ago
$begingroup$
What is “pairwise different”?
$endgroup$
– Arvasu Kulkarni
11 hours ago
$begingroup$
What is “pairwise different”?
$endgroup$
– Arvasu Kulkarni
11 hours ago
$begingroup$
@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
11 hours ago
$begingroup$
@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
11 hours ago
$begingroup$
Was this puzzle of your own creation?
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
Was this puzzle of your own creation?
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
@Brandon_J: An oldie twisted.
$endgroup$
– Bernardo Recamán Santos
4 hours ago
$begingroup$
@Brandon_J: An oldie twisted.
$endgroup$
– Bernardo Recamán Santos
4 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
|
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 4 hours ago
Jonathan AllanJonathan Allan
17.8k14697
$endgroup$
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
4 hours ago
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
3 hours ago
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 5 hours ago
Brandon_JBrandon_J
1,423127
$endgroup$
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
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
5 hours ago
|
show 1 more 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 2 hours ago
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 4 hours ago
Jonathan AllanJonathan Allan
17.8k14697
$endgroup$
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
4 hours ago
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
3 hours ago
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 4 hours ago
Jonathan AllanJonathan Allan
17.8k14697
$endgroup$
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
4 hours ago
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
3 hours ago
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 4 hours ago
Jonathan AllanJonathan Allan
17.8k14697
$endgroup$
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 4 hours ago
Jonathan AllanJonathan Allan
17.8k14697
edited 2 hours ago
answered 4 hours ago
Jonathan AllanJonathan Allan
17.8k14697
answered 4 hours ago
Jonathan AllanJonathan Allan
17.8k14697
answered 4 hours ago
Jonathan AllanJonathan Allan
17.8k14697
17.8k14697
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
4 hours ago
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
3 hours ago
add a comment |
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
4 hours ago
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
3 hours ago
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
4 hours ago
$begingroup$
markfarrar.co.uk/msq4x4cm.htm does not like your solution, but I can't find the error.
$endgroup$
– Brandon_J
4 hours ago
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
Ah, the diagonals. Thanks!
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
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
3 hours ago
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
Oh dang. Can I give you a +47 or so for finding this answer?
$endgroup$
– Brandon_J
3 hours ago
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
3 hours ago
$begingroup$
Just found better :p
$endgroup$
– Jonathan Allan
3 hours ago
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 5 hours ago
Brandon_JBrandon_J
1,423127
$endgroup$
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
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
5 hours ago
|
show 1 more 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 5 hours ago
Brandon_JBrandon_J
1,423127
$endgroup$
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
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
5 hours ago
|
show 1 more comment
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 5 hours ago
Brandon_JBrandon_J
1,423127
$endgroup$
I've got a solution:
and I admit I found it online.
Here it is:
The largest number in it is
73
answered 5 hours ago
Brandon_JBrandon_J
1,423127
edited 5 hours ago
answered 5 hours ago
Brandon_JBrandon_J
1,423127
answered 5 hours ago
Brandon_JBrandon_J
1,423127
answered 5 hours ago
Brandon_JBrandon_J
1,423127
1,423127
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
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
5 hours ago
|
show 1 more comment
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
Maybe we can make a smaller one by using 1 as one of our numbers?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
I'm not sure. I feel like I found a pretty reliable source @JonathanAllan
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
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– Jonathan Allan
5 hours ago
$begingroup$
I am not doubting Mathworld as a source, but did A. W. Johnson, Jr allow 1 or not?
$endgroup$
– Jonathan Allan
5 hours ago
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
@JonathanAllan ah, I see what you mean. Does the OP allow the number one? I guess I should ask.
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
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
5 hours ago
$begingroup$
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
5 hours ago
|
show 1 more 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 2 hours ago
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 2 hours ago
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 2 hours ago
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 2 hours ago
Freddy BarreraFreddy Barrera
34318
answered 2 hours ago
Freddy BarreraFreddy Barrera
34318
answered 2 hours ago
Freddy BarreraFreddy Barrera
34318
answered 2 hours ago
Freddy BarreraFreddy Barrera
34318
34318
add a comment |
add a comment |
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$begingroup$
What is “pairwise different”?
$endgroup$
– Arvasu Kulkarni
11 hours ago
$begingroup$
@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
11 hours ago
$begingroup$
Was this puzzle of your own creation?
$endgroup$
– Brandon_J
5 hours ago
$begingroup$
@Brandon_J: An oldie twisted.
$endgroup$
– Bernardo Recamán Santos
4 hours ago
$begingroup$
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
2 hours ago