Create a magic square of 4-digit numbersMagic square with the position of 8 fixedMagic square with the position of 8 fixedMagic Matrices?The magic square with a holeUnsolved Mysteries: Magic Square of SquaresNo ordinary magic squaremodify a magic square - part IIA challenging Magic SquareCreate a 3x3 Magic Square that uses integers from -10 to -2Albrecht Durer Inspired Magic Square
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Create a magic square of 4-digit numbers
Magic square with the position of 8 fixedMagic square with the position of 8 fixedMagic Matrices?The magic square with a holeUnsolved Mysteries: Magic Square of SquaresNo ordinary magic squaremodify a magic square - part IIA challenging Magic SquareCreate a 3x3 Magic Square that uses integers from -10 to -2Albrecht Durer Inspired Magic Square
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;
$begingroup$
Example:
4567 4567 4567
4567 4567 4567
4567 4567 4567
what is magic square? if you add up each diagonal, row and column of
above matrix it will sum upto 13701.
Above is a 3*3 matrix where each entry is the same number. You need to replace the "4567"s with 9 different 4-digit numbers to create a perfect magic square.
Remember that the full square must contain nine of each digit 1, 2, 3, 4, and that all nine entries must be four-digit integers.
You need to use only the four digits 1, 2, 3, 4 to solve the problem so that we won't end up with multiple solutions.
You can use numbers like 1234, 4321, 2211, 2121 and so on. But if you used 4321 once in any of the 9 cells you cannot use it again.
Similar puzzle link.
magic-square
$endgroup$
add a comment
|
$begingroup$
Example:
4567 4567 4567
4567 4567 4567
4567 4567 4567
what is magic square? if you add up each diagonal, row and column of
above matrix it will sum upto 13701.
Above is a 3*3 matrix where each entry is the same number. You need to replace the "4567"s with 9 different 4-digit numbers to create a perfect magic square.
Remember that the full square must contain nine of each digit 1, 2, 3, 4, and that all nine entries must be four-digit integers.
You need to use only the four digits 1, 2, 3, 4 to solve the problem so that we won't end up with multiple solutions.
You can use numbers like 1234, 4321, 2211, 2121 and so on. But if you used 4321 once in any of the 9 cells you cannot use it again.
Similar puzzle link.
magic-square
$endgroup$
5
$begingroup$
I've just made an edit, attempting to make your question more clear/coherent/comprehensible. Please let me know if the question as it's now written is what you intended.
$endgroup$
– Rand al'Thor
Sep 18 at 10:29
add a comment
|
$begingroup$
Example:
4567 4567 4567
4567 4567 4567
4567 4567 4567
what is magic square? if you add up each diagonal, row and column of
above matrix it will sum upto 13701.
Above is a 3*3 matrix where each entry is the same number. You need to replace the "4567"s with 9 different 4-digit numbers to create a perfect magic square.
Remember that the full square must contain nine of each digit 1, 2, 3, 4, and that all nine entries must be four-digit integers.
You need to use only the four digits 1, 2, 3, 4 to solve the problem so that we won't end up with multiple solutions.
You can use numbers like 1234, 4321, 2211, 2121 and so on. But if you used 4321 once in any of the 9 cells you cannot use it again.
Similar puzzle link.
magic-square
$endgroup$
Example:
4567 4567 4567
4567 4567 4567
4567 4567 4567
what is magic square? if you add up each diagonal, row and column of
above matrix it will sum upto 13701.
Above is a 3*3 matrix where each entry is the same number. You need to replace the "4567"s with 9 different 4-digit numbers to create a perfect magic square.
Remember that the full square must contain nine of each digit 1, 2, 3, 4, and that all nine entries must be four-digit integers.
You need to use only the four digits 1, 2, 3, 4 to solve the problem so that we won't end up with multiple solutions.
You can use numbers like 1234, 4321, 2211, 2121 and so on. But if you used 4321 once in any of the 9 cells you cannot use it again.
Similar puzzle link.
magic-square
magic-square
edited Sep 19 at 6:15
Sayed Mohd Ali
asked Sep 18 at 10:10
Sayed Mohd AliSayed Mohd Ali
1,2261 silver badge20 bronze badges
1,2261 silver badge20 bronze badges
5
$begingroup$
I've just made an edit, attempting to make your question more clear/coherent/comprehensible. Please let me know if the question as it's now written is what you intended.
$endgroup$
– Rand al'Thor
Sep 18 at 10:29
add a comment
|
5
$begingroup$
I've just made an edit, attempting to make your question more clear/coherent/comprehensible. Please let me know if the question as it's now written is what you intended.
$endgroup$
– Rand al'Thor
Sep 18 at 10:29
5
5
$begingroup$
I've just made an edit, attempting to make your question more clear/coherent/comprehensible. Please let me know if the question as it's now written is what you intended.
$endgroup$
– Rand al'Thor
Sep 18 at 10:29
$begingroup$
I've just made an edit, attempting to make your question more clear/coherent/comprehensible. Please let me know if the question as it's now written is what you intended.
$endgroup$
– Rand al'Thor
Sep 18 at 10:29
add a comment
|
3 Answers
3
active
oldest
votes
$begingroup$
Building on the strategy of Omega Krypton, this is one possibility which also gets the diagonals to sum to the magic total
1214 3134 2324
3334 2224 1114
2124 1314 3234
To clarify, the sum of the numbers in each row, each column and along each diagonal is 6672 (the magic total) and each of the digits 1,2,3,4 appears nine times.
First of all, construct four single digit magic squares...
132
321
213
213
321
132
132
321
213
444
444
444
Then concatenate them to get a 4-digit magic square!
$endgroup$
$begingroup$
That is the cleverest way to arrive at a solution that I've yet seen.
$endgroup$
– Brandon_J
Sep 18 at 16:32
1
$begingroup$
Wow, the most elegant solution I've seen recently! :)
$endgroup$
– user47134
Sep 18 at 17:16
3
$begingroup$
@SayedMohdAli I rather think an explanation of what a magic square is would be up to you as the puzzle-poser. Granted, I don't think it would hurt for hexomino to include the final magic square in his answer.
$endgroup$
– Brandon_J
Sep 18 at 19:44
add a comment
|
$begingroup$
Here is another one
2243 1341 3142
3141 2242 1343
1342 3143 2241
All rows, columns and diagonal sums 6,726 and there is only 9 of each 1, 2, 3, 4
I will edit the explaination later.
$endgroup$
add a comment
|
$begingroup$
I think this is the answer where each number consisting of 4 digits
with only 1,2,3,4 number and calculation of this 3*3 matrix will be
equals from each side maybe this the combination of digits which can
be considered as a magic number.
$endgroup$
8
$begingroup$
Remember that the full square must contain nine of each digit 1, 2, 3, 4
$endgroup$
– Omega Krypton
Sep 18 at 13:09
5
$begingroup$
... and the diagonals don't make the same sum.
$endgroup$
– Weather Vane
Sep 18 at 13:11
add a comment
|
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Building on the strategy of Omega Krypton, this is one possibility which also gets the diagonals to sum to the magic total
1214 3134 2324
3334 2224 1114
2124 1314 3234
To clarify, the sum of the numbers in each row, each column and along each diagonal is 6672 (the magic total) and each of the digits 1,2,3,4 appears nine times.
First of all, construct four single digit magic squares...
132
321
213
213
321
132
132
321
213
444
444
444
Then concatenate them to get a 4-digit magic square!
$endgroup$
$begingroup$
That is the cleverest way to arrive at a solution that I've yet seen.
$endgroup$
– Brandon_J
Sep 18 at 16:32
1
$begingroup$
Wow, the most elegant solution I've seen recently! :)
$endgroup$
– user47134
Sep 18 at 17:16
3
$begingroup$
@SayedMohdAli I rather think an explanation of what a magic square is would be up to you as the puzzle-poser. Granted, I don't think it would hurt for hexomino to include the final magic square in his answer.
$endgroup$
– Brandon_J
Sep 18 at 19:44
add a comment
|
$begingroup$
Building on the strategy of Omega Krypton, this is one possibility which also gets the diagonals to sum to the magic total
1214 3134 2324
3334 2224 1114
2124 1314 3234
To clarify, the sum of the numbers in each row, each column and along each diagonal is 6672 (the magic total) and each of the digits 1,2,3,4 appears nine times.
First of all, construct four single digit magic squares...
132
321
213
213
321
132
132
321
213
444
444
444
Then concatenate them to get a 4-digit magic square!
$endgroup$
$begingroup$
That is the cleverest way to arrive at a solution that I've yet seen.
$endgroup$
– Brandon_J
Sep 18 at 16:32
1
$begingroup$
Wow, the most elegant solution I've seen recently! :)
$endgroup$
– user47134
Sep 18 at 17:16
3
$begingroup$
@SayedMohdAli I rather think an explanation of what a magic square is would be up to you as the puzzle-poser. Granted, I don't think it would hurt for hexomino to include the final magic square in his answer.
$endgroup$
– Brandon_J
Sep 18 at 19:44
add a comment
|
$begingroup$
Building on the strategy of Omega Krypton, this is one possibility which also gets the diagonals to sum to the magic total
1214 3134 2324
3334 2224 1114
2124 1314 3234
To clarify, the sum of the numbers in each row, each column and along each diagonal is 6672 (the magic total) and each of the digits 1,2,3,4 appears nine times.
First of all, construct four single digit magic squares...
132
321
213
213
321
132
132
321
213
444
444
444
Then concatenate them to get a 4-digit magic square!
$endgroup$
Building on the strategy of Omega Krypton, this is one possibility which also gets the diagonals to sum to the magic total
1214 3134 2324
3334 2224 1114
2124 1314 3234
To clarify, the sum of the numbers in each row, each column and along each diagonal is 6672 (the magic total) and each of the digits 1,2,3,4 appears nine times.
First of all, construct four single digit magic squares...
132
321
213
213
321
132
132
321
213
444
444
444
Then concatenate them to get a 4-digit magic square!
edited Sep 18 at 20:09
answered Sep 18 at 16:13
hexominohexomino
69.6k6 gold badges194 silver badges300 bronze badges
69.6k6 gold badges194 silver badges300 bronze badges
$begingroup$
That is the cleverest way to arrive at a solution that I've yet seen.
$endgroup$
– Brandon_J
Sep 18 at 16:32
1
$begingroup$
Wow, the most elegant solution I've seen recently! :)
$endgroup$
– user47134
Sep 18 at 17:16
3
$begingroup$
@SayedMohdAli I rather think an explanation of what a magic square is would be up to you as the puzzle-poser. Granted, I don't think it would hurt for hexomino to include the final magic square in his answer.
$endgroup$
– Brandon_J
Sep 18 at 19:44
add a comment
|
$begingroup$
That is the cleverest way to arrive at a solution that I've yet seen.
$endgroup$
– Brandon_J
Sep 18 at 16:32
1
$begingroup$
Wow, the most elegant solution I've seen recently! :)
$endgroup$
– user47134
Sep 18 at 17:16
3
$begingroup$
@SayedMohdAli I rather think an explanation of what a magic square is would be up to you as the puzzle-poser. Granted, I don't think it would hurt for hexomino to include the final magic square in his answer.
$endgroup$
– Brandon_J
Sep 18 at 19:44
$begingroup$
That is the cleverest way to arrive at a solution that I've yet seen.
$endgroup$
– Brandon_J
Sep 18 at 16:32
$begingroup$
That is the cleverest way to arrive at a solution that I've yet seen.
$endgroup$
– Brandon_J
Sep 18 at 16:32
1
1
$begingroup$
Wow, the most elegant solution I've seen recently! :)
$endgroup$
– user47134
Sep 18 at 17:16
$begingroup$
Wow, the most elegant solution I've seen recently! :)
$endgroup$
– user47134
Sep 18 at 17:16
3
3
$begingroup$
@SayedMohdAli I rather think an explanation of what a magic square is would be up to you as the puzzle-poser. Granted, I don't think it would hurt for hexomino to include the final magic square in his answer.
$endgroup$
– Brandon_J
Sep 18 at 19:44
$begingroup$
@SayedMohdAli I rather think an explanation of what a magic square is would be up to you as the puzzle-poser. Granted, I don't think it would hurt for hexomino to include the final magic square in his answer.
$endgroup$
– Brandon_J
Sep 18 at 19:44
add a comment
|
$begingroup$
Here is another one
2243 1341 3142
3141 2242 1343
1342 3143 2241
All rows, columns and diagonal sums 6,726 and there is only 9 of each 1, 2, 3, 4
I will edit the explaination later.
$endgroup$
add a comment
|
$begingroup$
Here is another one
2243 1341 3142
3141 2242 1343
1342 3143 2241
All rows, columns and diagonal sums 6,726 and there is only 9 of each 1, 2, 3, 4
I will edit the explaination later.
$endgroup$
add a comment
|
$begingroup$
Here is another one
2243 1341 3142
3141 2242 1343
1342 3143 2241
All rows, columns and diagonal sums 6,726 and there is only 9 of each 1, 2, 3, 4
I will edit the explaination later.
$endgroup$
Here is another one
2243 1341 3142
3141 2242 1343
1342 3143 2241
All rows, columns and diagonal sums 6,726 and there is only 9 of each 1, 2, 3, 4
I will edit the explaination later.
answered Sep 19 at 8:08
Pʀıncess AnayaPʀıncess Anaya
4507 bronze badges
4507 bronze badges
add a comment
|
add a comment
|
$begingroup$
I think this is the answer where each number consisting of 4 digits
with only 1,2,3,4 number and calculation of this 3*3 matrix will be
equals from each side maybe this the combination of digits which can
be considered as a magic number.
$endgroup$
8
$begingroup$
Remember that the full square must contain nine of each digit 1, 2, 3, 4
$endgroup$
– Omega Krypton
Sep 18 at 13:09
5
$begingroup$
... and the diagonals don't make the same sum.
$endgroup$
– Weather Vane
Sep 18 at 13:11
add a comment
|
$begingroup$
I think this is the answer where each number consisting of 4 digits
with only 1,2,3,4 number and calculation of this 3*3 matrix will be
equals from each side maybe this the combination of digits which can
be considered as a magic number.
$endgroup$
8
$begingroup$
Remember that the full square must contain nine of each digit 1, 2, 3, 4
$endgroup$
– Omega Krypton
Sep 18 at 13:09
5
$begingroup$
... and the diagonals don't make the same sum.
$endgroup$
– Weather Vane
Sep 18 at 13:11
add a comment
|
$begingroup$
I think this is the answer where each number consisting of 4 digits
with only 1,2,3,4 number and calculation of this 3*3 matrix will be
equals from each side maybe this the combination of digits which can
be considered as a magic number.
$endgroup$
I think this is the answer where each number consisting of 4 digits
with only 1,2,3,4 number and calculation of this 3*3 matrix will be
equals from each side maybe this the combination of digits which can
be considered as a magic number.
edited Sep 18 at 16:31
Brandon_J
8,1921 gold badge9 silver badges60 bronze badges
8,1921 gold badge9 silver badges60 bronze badges
answered Sep 18 at 13:08
ankitkanojiaankitkanojia
1172 bronze badges
1172 bronze badges
8
$begingroup$
Remember that the full square must contain nine of each digit 1, 2, 3, 4
$endgroup$
– Omega Krypton
Sep 18 at 13:09
5
$begingroup$
... and the diagonals don't make the same sum.
$endgroup$
– Weather Vane
Sep 18 at 13:11
add a comment
|
8
$begingroup$
Remember that the full square must contain nine of each digit 1, 2, 3, 4
$endgroup$
– Omega Krypton
Sep 18 at 13:09
5
$begingroup$
... and the diagonals don't make the same sum.
$endgroup$
– Weather Vane
Sep 18 at 13:11
8
8
$begingroup$
Remember that the full square must contain nine of each digit 1, 2, 3, 4
$endgroup$
– Omega Krypton
Sep 18 at 13:09
$begingroup$
Remember that the full square must contain nine of each digit 1, 2, 3, 4
$endgroup$
– Omega Krypton
Sep 18 at 13:09
5
5
$begingroup$
... and the diagonals don't make the same sum.
$endgroup$
– Weather Vane
Sep 18 at 13:11
$begingroup$
... and the diagonals don't make the same sum.
$endgroup$
– Weather Vane
Sep 18 at 13:11
add a comment
|
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$begingroup$
I've just made an edit, attempting to make your question more clear/coherent/comprehensible. Please let me know if the question as it's now written is what you intended.
$endgroup$
– Rand al'Thor
Sep 18 at 10:29