Scheduling based problem Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraDeducing Two Numbers based on their Difference and RatioThe 2 million, er, 20 dollar problemAdd a number to each vertex of a triangle such that each edge adds to a perfect squareHexagonal sum fillingEight distinct numbers in the tableIncreasing rows and columnsA simple grid puzzle123456789=1 problemFilling a chessboard with -1, 0 and 1Input/Output Problem #7
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Scheduling based problem
Scheduling based problem
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraDeducing Two Numbers based on their Difference and RatioThe 2 million, er, 20 dollar problemAdd a number to each vertex of a triangle such that each edge adds to a perfect squareHexagonal sum fillingEight distinct numbers in the tableIncreasing rows and columnsA simple grid puzzle123456789=1 problemFilling a chessboard with -1, 0 and 1Input/Output Problem #7
$begingroup$
Can you place these numbers into 5 rows of 4 such that each row totals 20?
1, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 7, 8, 8, 8
mathematics packing
$endgroup$
add a comment |
$begingroup$
Can you place these numbers into 5 rows of 4 such that each row totals 20?
1, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 7, 8, 8, 8
mathematics packing
$endgroup$
add a comment |
$begingroup$
Can you place these numbers into 5 rows of 4 such that each row totals 20?
1, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 7, 8, 8, 8
mathematics packing
$endgroup$
Can you place these numbers into 5 rows of 4 such that each row totals 20?
1, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 7, 8, 8, 8
mathematics packing
mathematics packing
asked 3 hours ago
JonMark PerryJonMark Perry
20.9k64199
20.9k64199
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Sure.
1,5,6,8
2,4,6,8
3,4,5,8
4,4,5,7
5,5,5,5
Method:
Started from the bigger numbers, and partitioned into 5 parts of 20: 8,8,4, 8,7,5 and so on.
Then swapped a big number with two smaller ones (with the same sum) on another row until I had 4 numbers on each row.
With some fiddling, it's also possible to get all the columns to add up to 25:
1,8,6,5
5,4,7,4
5,5,5,5
6,4,2,8
8,4,5,3
And here's a magic square (with duplicates, unavoidably) followed by a row of fives:
1,6,8,5
5,7,4,4
8,5,4,3
6,2,4,8
5,5,5,5
And finally:
4 4 8 4
8 5 4 3
1 5 6 8
7 6 2 5
5 5 5 5
This has
* 20 on all 5 rows
* 20 on all 4 long diagonals
* 25 in all 4 columns
* a magic square on the first 4 rows
$endgroup$
$begingroup$
I'm late ! But I have differents rows :D
$endgroup$
– Narlore
3 hours ago
$begingroup$
I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
$endgroup$
– 5202456
2 hours ago
$begingroup$
@5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
$endgroup$
– Bass
59 mins ago
add a comment |
$begingroup$
This one works
7 3 5 5
8 2 5 5
6 4 5 5
8 6 1 5
8 4 4 4
Method :
Write randomly the numbers on a piece of paper during 5 min until the solution appears magically.
$endgroup$
add a comment |
$begingroup$
A bit late to the party and cannot beat the excellent answer from @Bass.
I worked out the number of distinct solutions bearing in mind they can be further permuted by ordering each row and the row sequence. I found
16 distinct solutions
I did this by first finding all the sets of four digits which sum to 20.
There are 17 sets of digits
1 3 8 8
1 4 7 8
1 5 6 8
1 6 6 7
2 3 7 8
2 4 6 8
2 5 5 8
2 5 6 7
3 4 5 8
3 4 6 7
3 5 5 7
3 5 6 6
4 4 4 8
4 4 5 7
4 4 6 6
4 5 5 6
5 5 5 5
I then permuted them for each row so that each digit is used the right number of times.
These are the 16 solutions
1 3 8 8 1 3 8 8 1 3 8 8 1 3 8 8
2 4 6 8 2 5 5 8 2 5 5 8 2 5 6 7
4 4 5 7 4 4 5 7 4 4 5 7 4 4 4 8
4 5 5 6 4 4 6 6 4 5 5 6 4 5 5 6
5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5
1 4 7 8 1 4 7 8 1 4 7 8 1 4 7 8
2 4 6 8 2 5 5 8 2 5 5 8 2 5 5 8
3 4 5 8 3 4 5 8 3 4 5 8 3 5 6 6
4 5 5 6 4 4 6 6 4 5 5 6 4 4 4 8
5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5
1 5 6 8 1 5 6 8 1 5 6 8 1 5 6 8
2 3 7 8 2 4 6 8 2 4 6 8 2 5 5 8
4 4 4 8 3 4 5 8 3 5 5 7 3 4 5 8
4 5 5 6 4 4 5 7 4 4 4 8 4 4 5 7
5 5 5 5 5 5 5 5 5 5 5 5 4 5 5 6
1 5 6 8 1 5 6 8 1 5 6 8 1 6 6 7
2 5 5 8 2 5 5 8 2 5 6 7 2 5 5 8
3 4 6 7 3 5 5 7 3 4 5 8 3 4 5 8
4 4 4 8 4 4 4 8 4 4 4 8 4 4 4 8
5 5 5 5 4 5 5 6 5 5 5 5 5 5 5 5
Method:
A computer program written in C.
$endgroup$
$begingroup$
Nice! FYI I generated the puzzle from row 2 column 4.
$endgroup$
– JonMark Perry
39 mins ago
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
$begingroup$
Sure.
1,5,6,8
2,4,6,8
3,4,5,8
4,4,5,7
5,5,5,5
Method:
Started from the bigger numbers, and partitioned into 5 parts of 20: 8,8,4, 8,7,5 and so on.
Then swapped a big number with two smaller ones (with the same sum) on another row until I had 4 numbers on each row.
With some fiddling, it's also possible to get all the columns to add up to 25:
1,8,6,5
5,4,7,4
5,5,5,5
6,4,2,8
8,4,5,3
And here's a magic square (with duplicates, unavoidably) followed by a row of fives:
1,6,8,5
5,7,4,4
8,5,4,3
6,2,4,8
5,5,5,5
And finally:
4 4 8 4
8 5 4 3
1 5 6 8
7 6 2 5
5 5 5 5
This has
* 20 on all 5 rows
* 20 on all 4 long diagonals
* 25 in all 4 columns
* a magic square on the first 4 rows
$endgroup$
$begingroup$
I'm late ! But I have differents rows :D
$endgroup$
– Narlore
3 hours ago
$begingroup$
I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
$endgroup$
– 5202456
2 hours ago
$begingroup$
@5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
$endgroup$
– Bass
59 mins ago
add a comment |
$begingroup$
Sure.
1,5,6,8
2,4,6,8
3,4,5,8
4,4,5,7
5,5,5,5
Method:
Started from the bigger numbers, and partitioned into 5 parts of 20: 8,8,4, 8,7,5 and so on.
Then swapped a big number with two smaller ones (with the same sum) on another row until I had 4 numbers on each row.
With some fiddling, it's also possible to get all the columns to add up to 25:
1,8,6,5
5,4,7,4
5,5,5,5
6,4,2,8
8,4,5,3
And here's a magic square (with duplicates, unavoidably) followed by a row of fives:
1,6,8,5
5,7,4,4
8,5,4,3
6,2,4,8
5,5,5,5
And finally:
4 4 8 4
8 5 4 3
1 5 6 8
7 6 2 5
5 5 5 5
This has
* 20 on all 5 rows
* 20 on all 4 long diagonals
* 25 in all 4 columns
* a magic square on the first 4 rows
$endgroup$
$begingroup$
I'm late ! But I have differents rows :D
$endgroup$
– Narlore
3 hours ago
$begingroup$
I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
$endgroup$
– 5202456
2 hours ago
$begingroup$
@5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
$endgroup$
– Bass
59 mins ago
add a comment |
$begingroup$
Sure.
1,5,6,8
2,4,6,8
3,4,5,8
4,4,5,7
5,5,5,5
Method:
Started from the bigger numbers, and partitioned into 5 parts of 20: 8,8,4, 8,7,5 and so on.
Then swapped a big number with two smaller ones (with the same sum) on another row until I had 4 numbers on each row.
With some fiddling, it's also possible to get all the columns to add up to 25:
1,8,6,5
5,4,7,4
5,5,5,5
6,4,2,8
8,4,5,3
And here's a magic square (with duplicates, unavoidably) followed by a row of fives:
1,6,8,5
5,7,4,4
8,5,4,3
6,2,4,8
5,5,5,5
And finally:
4 4 8 4
8 5 4 3
1 5 6 8
7 6 2 5
5 5 5 5
This has
* 20 on all 5 rows
* 20 on all 4 long diagonals
* 25 in all 4 columns
* a magic square on the first 4 rows
$endgroup$
Sure.
1,5,6,8
2,4,6,8
3,4,5,8
4,4,5,7
5,5,5,5
Method:
Started from the bigger numbers, and partitioned into 5 parts of 20: 8,8,4, 8,7,5 and so on.
Then swapped a big number with two smaller ones (with the same sum) on another row until I had 4 numbers on each row.
With some fiddling, it's also possible to get all the columns to add up to 25:
1,8,6,5
5,4,7,4
5,5,5,5
6,4,2,8
8,4,5,3
And here's a magic square (with duplicates, unavoidably) followed by a row of fives:
1,6,8,5
5,7,4,4
8,5,4,3
6,2,4,8
5,5,5,5
And finally:
4 4 8 4
8 5 4 3
1 5 6 8
7 6 2 5
5 5 5 5
This has
* 20 on all 5 rows
* 20 on all 4 long diagonals
* 25 in all 4 columns
* a magic square on the first 4 rows
edited 1 hour ago
answered 3 hours ago
BassBass
31.5k474192
31.5k474192
$begingroup$
I'm late ! But I have differents rows :D
$endgroup$
– Narlore
3 hours ago
$begingroup$
I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
$endgroup$
– 5202456
2 hours ago
$begingroup$
@5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
$endgroup$
– Bass
59 mins ago
add a comment |
$begingroup$
I'm late ! But I have differents rows :D
$endgroup$
– Narlore
3 hours ago
$begingroup$
I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
$endgroup$
– 5202456
2 hours ago
$begingroup$
@5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
$endgroup$
– Bass
59 mins ago
$begingroup$
I'm late ! But I have differents rows :D
$endgroup$
– Narlore
3 hours ago
$begingroup$
I'm late ! But I have differents rows :D
$endgroup$
– Narlore
3 hours ago
$begingroup$
I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
$endgroup$
– 5202456
2 hours ago
$begingroup$
I like the way you went on to challenge yourself by finding solutions to your own proposed questions.
$endgroup$
– 5202456
2 hours ago
$begingroup$
@5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
$endgroup$
– Bass
59 mins ago
$begingroup$
@5202456 Thanks! The original solution seemed to leave an awful lot of "wiggle room" in the pattern, so I wanted to see what I could do with it. I added one more "extra magical" solution after your comment; hopefully I didn't make any mistakes, as the numbers are starting to bounce around in my eyes :-)
$endgroup$
– Bass
59 mins ago
add a comment |
$begingroup$
This one works
7 3 5 5
8 2 5 5
6 4 5 5
8 6 1 5
8 4 4 4
Method :
Write randomly the numbers on a piece of paper during 5 min until the solution appears magically.
$endgroup$
add a comment |
$begingroup$
This one works
7 3 5 5
8 2 5 5
6 4 5 5
8 6 1 5
8 4 4 4
Method :
Write randomly the numbers on a piece of paper during 5 min until the solution appears magically.
$endgroup$
add a comment |
$begingroup$
This one works
7 3 5 5
8 2 5 5
6 4 5 5
8 6 1 5
8 4 4 4
Method :
Write randomly the numbers on a piece of paper during 5 min until the solution appears magically.
$endgroup$
This one works
7 3 5 5
8 2 5 5
6 4 5 5
8 6 1 5
8 4 4 4
Method :
Write randomly the numbers on a piece of paper during 5 min until the solution appears magically.
edited 3 hours ago
answered 3 hours ago
NarloreNarlore
35615
35615
add a comment |
add a comment |
$begingroup$
A bit late to the party and cannot beat the excellent answer from @Bass.
I worked out the number of distinct solutions bearing in mind they can be further permuted by ordering each row and the row sequence. I found
16 distinct solutions
I did this by first finding all the sets of four digits which sum to 20.
There are 17 sets of digits
1 3 8 8
1 4 7 8
1 5 6 8
1 6 6 7
2 3 7 8
2 4 6 8
2 5 5 8
2 5 6 7
3 4 5 8
3 4 6 7
3 5 5 7
3 5 6 6
4 4 4 8
4 4 5 7
4 4 6 6
4 5 5 6
5 5 5 5
I then permuted them for each row so that each digit is used the right number of times.
These are the 16 solutions
1 3 8 8 1 3 8 8 1 3 8 8 1 3 8 8
2 4 6 8 2 5 5 8 2 5 5 8 2 5 6 7
4 4 5 7 4 4 5 7 4 4 5 7 4 4 4 8
4 5 5 6 4 4 6 6 4 5 5 6 4 5 5 6
5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5
1 4 7 8 1 4 7 8 1 4 7 8 1 4 7 8
2 4 6 8 2 5 5 8 2 5 5 8 2 5 5 8
3 4 5 8 3 4 5 8 3 4 5 8 3 5 6 6
4 5 5 6 4 4 6 6 4 5 5 6 4 4 4 8
5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5
1 5 6 8 1 5 6 8 1 5 6 8 1 5 6 8
2 3 7 8 2 4 6 8 2 4 6 8 2 5 5 8
4 4 4 8 3 4 5 8 3 5 5 7 3 4 5 8
4 5 5 6 4 4 5 7 4 4 4 8 4 4 5 7
5 5 5 5 5 5 5 5 5 5 5 5 4 5 5 6
1 5 6 8 1 5 6 8 1 5 6 8 1 6 6 7
2 5 5 8 2 5 5 8 2 5 6 7 2 5 5 8
3 4 6 7 3 5 5 7 3 4 5 8 3 4 5 8
4 4 4 8 4 4 4 8 4 4 4 8 4 4 4 8
5 5 5 5 4 5 5 6 5 5 5 5 5 5 5 5
Method:
A computer program written in C.
$endgroup$
$begingroup$
Nice! FYI I generated the puzzle from row 2 column 4.
$endgroup$
– JonMark Perry
39 mins ago
add a comment |
$begingroup$
A bit late to the party and cannot beat the excellent answer from @Bass.
I worked out the number of distinct solutions bearing in mind they can be further permuted by ordering each row and the row sequence. I found
16 distinct solutions
I did this by first finding all the sets of four digits which sum to 20.
There are 17 sets of digits
1 3 8 8
1 4 7 8
1 5 6 8
1 6 6 7
2 3 7 8
2 4 6 8
2 5 5 8
2 5 6 7
3 4 5 8
3 4 6 7
3 5 5 7
3 5 6 6
4 4 4 8
4 4 5 7
4 4 6 6
4 5 5 6
5 5 5 5
I then permuted them for each row so that each digit is used the right number of times.
These are the 16 solutions
1 3 8 8 1 3 8 8 1 3 8 8 1 3 8 8
2 4 6 8 2 5 5 8 2 5 5 8 2 5 6 7
4 4 5 7 4 4 5 7 4 4 5 7 4 4 4 8
4 5 5 6 4 4 6 6 4 5 5 6 4 5 5 6
5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5
1 4 7 8 1 4 7 8 1 4 7 8 1 4 7 8
2 4 6 8 2 5 5 8 2 5 5 8 2 5 5 8
3 4 5 8 3 4 5 8 3 4 5 8 3 5 6 6
4 5 5 6 4 4 6 6 4 5 5 6 4 4 4 8
5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5
1 5 6 8 1 5 6 8 1 5 6 8 1 5 6 8
2 3 7 8 2 4 6 8 2 4 6 8 2 5 5 8
4 4 4 8 3 4 5 8 3 5 5 7 3 4 5 8
4 5 5 6 4 4 5 7 4 4 4 8 4 4 5 7
5 5 5 5 5 5 5 5 5 5 5 5 4 5 5 6
1 5 6 8 1 5 6 8 1 5 6 8 1 6 6 7
2 5 5 8 2 5 5 8 2 5 6 7 2 5 5 8
3 4 6 7 3 5 5 7 3 4 5 8 3 4 5 8
4 4 4 8 4 4 4 8 4 4 4 8 4 4 4 8
5 5 5 5 4 5 5 6 5 5 5 5 5 5 5 5
Method:
A computer program written in C.
$endgroup$
$begingroup$
Nice! FYI I generated the puzzle from row 2 column 4.
$endgroup$
– JonMark Perry
39 mins ago
add a comment |
$begingroup$
A bit late to the party and cannot beat the excellent answer from @Bass.
I worked out the number of distinct solutions bearing in mind they can be further permuted by ordering each row and the row sequence. I found
16 distinct solutions
I did this by first finding all the sets of four digits which sum to 20.
There are 17 sets of digits
1 3 8 8
1 4 7 8
1 5 6 8
1 6 6 7
2 3 7 8
2 4 6 8
2 5 5 8
2 5 6 7
3 4 5 8
3 4 6 7
3 5 5 7
3 5 6 6
4 4 4 8
4 4 5 7
4 4 6 6
4 5 5 6
5 5 5 5
I then permuted them for each row so that each digit is used the right number of times.
These are the 16 solutions
1 3 8 8 1 3 8 8 1 3 8 8 1 3 8 8
2 4 6 8 2 5 5 8 2 5 5 8 2 5 6 7
4 4 5 7 4 4 5 7 4 4 5 7 4 4 4 8
4 5 5 6 4 4 6 6 4 5 5 6 4 5 5 6
5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5
1 4 7 8 1 4 7 8 1 4 7 8 1 4 7 8
2 4 6 8 2 5 5 8 2 5 5 8 2 5 5 8
3 4 5 8 3 4 5 8 3 4 5 8 3 5 6 6
4 5 5 6 4 4 6 6 4 5 5 6 4 4 4 8
5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5
1 5 6 8 1 5 6 8 1 5 6 8 1 5 6 8
2 3 7 8 2 4 6 8 2 4 6 8 2 5 5 8
4 4 4 8 3 4 5 8 3 5 5 7 3 4 5 8
4 5 5 6 4 4 5 7 4 4 4 8 4 4 5 7
5 5 5 5 5 5 5 5 5 5 5 5 4 5 5 6
1 5 6 8 1 5 6 8 1 5 6 8 1 6 6 7
2 5 5 8 2 5 5 8 2 5 6 7 2 5 5 8
3 4 6 7 3 5 5 7 3 4 5 8 3 4 5 8
4 4 4 8 4 4 4 8 4 4 4 8 4 4 4 8
5 5 5 5 4 5 5 6 5 5 5 5 5 5 5 5
Method:
A computer program written in C.
$endgroup$
A bit late to the party and cannot beat the excellent answer from @Bass.
I worked out the number of distinct solutions bearing in mind they can be further permuted by ordering each row and the row sequence. I found
16 distinct solutions
I did this by first finding all the sets of four digits which sum to 20.
There are 17 sets of digits
1 3 8 8
1 4 7 8
1 5 6 8
1 6 6 7
2 3 7 8
2 4 6 8
2 5 5 8
2 5 6 7
3 4 5 8
3 4 6 7
3 5 5 7
3 5 6 6
4 4 4 8
4 4 5 7
4 4 6 6
4 5 5 6
5 5 5 5
I then permuted them for each row so that each digit is used the right number of times.
These are the 16 solutions
1 3 8 8 1 3 8 8 1 3 8 8 1 3 8 8
2 4 6 8 2 5 5 8 2 5 5 8 2 5 6 7
4 4 5 7 4 4 5 7 4 4 5 7 4 4 4 8
4 5 5 6 4 4 6 6 4 5 5 6 4 5 5 6
5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5
1 4 7 8 1 4 7 8 1 4 7 8 1 4 7 8
2 4 6 8 2 5 5 8 2 5 5 8 2 5 5 8
3 4 5 8 3 4 5 8 3 4 5 8 3 5 6 6
4 5 5 6 4 4 6 6 4 5 5 6 4 4 4 8
5 5 5 5 5 5 5 5 4 5 5 6 5 5 5 5
1 5 6 8 1 5 6 8 1 5 6 8 1 5 6 8
2 3 7 8 2 4 6 8 2 4 6 8 2 5 5 8
4 4 4 8 3 4 5 8 3 5 5 7 3 4 5 8
4 5 5 6 4 4 5 7 4 4 4 8 4 4 5 7
5 5 5 5 5 5 5 5 5 5 5 5 4 5 5 6
1 5 6 8 1 5 6 8 1 5 6 8 1 6 6 7
2 5 5 8 2 5 5 8 2 5 6 7 2 5 5 8
3 4 6 7 3 5 5 7 3 4 5 8 3 4 5 8
4 4 4 8 4 4 4 8 4 4 4 8 4 4 4 8
5 5 5 5 4 5 5 6 5 5 5 5 5 5 5 5
Method:
A computer program written in C.
answered 48 mins ago
Weather VaneWeather Vane
2,623114
2,623114
$begingroup$
Nice! FYI I generated the puzzle from row 2 column 4.
$endgroup$
– JonMark Perry
39 mins ago
add a comment |
$begingroup$
Nice! FYI I generated the puzzle from row 2 column 4.
$endgroup$
– JonMark Perry
39 mins ago
$begingroup$
Nice! FYI I generated the puzzle from row 2 column 4.
$endgroup$
– JonMark Perry
39 mins ago
$begingroup$
Nice! FYI I generated the puzzle from row 2 column 4.
$endgroup$
– JonMark Perry
39 mins ago
add a comment |
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