Determine all integers m for which a square of length m can be dissected into five rectangles, the side lengths of which are the integers 1,2,3,...,10 in some order.
The previous post should be ignored. It would neither fully upload nor delete. :-(
There are many candidate solutions where the ariel sum of 5 pieces add to a square: 14 candidates for 121; 9 for 144; and 12 for 169. Of these, only two for area 121 and two for area 169 can be fit together into a square. Note, any one map is one of eight rotations and/or reflections. The candidates and the 4 solution maps are shown below. The program is shown here.
area = 121 index = 1 ( 1, 6) ( 2, 9) ( 3,10) ( 4, 8) ( 5, 7)
area = 121 index = 2 ( 1, 6) ( 2,10) ( 3, 8) ( 4, 9) ( 5, 7)
*area = 121 index = 3 ( 1, 6) ( 2,10) ( 3, 9) ( 4, 7) ( 5, 8)
area = 121 index = 4 ( 1, 7) ( 2,10) ( 3, 6) ( 4, 9) ( 5, 8)
area = 121 index = 5 ( 1, 8) ( 2, 6) ( 3,10) ( 4, 9) ( 5, 7)
area = 121 index = 6 ( 1, 8) ( 2, 7) ( 3,10) ( 4, 6) ( 5, 9)
area = 121 index = 7 ( 1, 8) ( 2, 9) ( 3, 7) ( 4, 6) ( 5,10)
area = 121 index = 8 ( 1, 8) ( 2,10) ( 3, 5) ( 4, 9) ( 6, 7)
area = 121 index = 9 ( 1, 9) ( 2, 7) ( 3, 6) ( 4,10) ( 5, 8)
area = 121 index = 10 ( 1, 9) ( 2, 7) ( 3, 8) ( 4, 6) ( 5,10)
area = 121 index = 11 ( 1, 9) ( 2, 7) ( 3,10) ( 4, 5) ( 6, 8)
*area = 121 index = 12 ( 1, 9) ( 2, 8) ( 3, 6) ( 4, 7) ( 5,10)
area = 121 index = 13 ( 1,10) ( 2, 5) ( 3, 9) ( 4, 8) ( 6, 7)
area = 121 index = 14 ( 1,10) ( 2, 8) ( 3, 7) ( 4, 5) ( 6, 9)
area = 144 index = 1 ( 1, 3) ( 2, 9) ( 4,10) ( 5, 7) ( 6, 8)
area = 144 index = 2 ( 1, 3) ( 2,10) ( 4, 7) ( 5, 9) ( 6, 8)
area = 144 index = 3 ( 1, 3) ( 2,10) ( 4, 8) ( 5, 7) ( 6, 9)
area = 144 index = 4 ( 1, 5) ( 2, 6) ( 3, 8) ( 4,10) ( 7, 9)
area = 144 index = 5 ( 1, 7) ( 2, 4) ( 3, 8) ( 5, 9) ( 6,10)
area = 144 index = 6 ( 1, 7) ( 2, 9) ( 3, 5) ( 4, 6) ( 8,10)
area = 144 index = 7 ( 1, 9) ( 2, 5) ( 3, 7) ( 4, 6) ( 8,10)
area = 144 index = 8 ( 1, 9) ( 2, 6) ( 3, 5) ( 4, 7) ( 8,10)
area = 144 index = 9 ( 1,10) ( 2, 4) ( 3, 5) ( 6, 8) ( 7, 9)
area = 169 index = 1 ( 1, 2) ( 3, 5) ( 4, 9) ( 6,10) ( 7, 8)
*area = 169 index = 2 ( 1, 2) ( 3, 7) ( 4, 6) ( 5,10) ( 8, 9)
area = 169 index = 3 ( 1, 2) ( 3, 7) ( 4, 9) ( 5, 6) ( 8,10)
*area = 169 index = 4 ( 1, 2) ( 3, 8) ( 4, 5) ( 6,10) ( 7, 9)
area = 169 index = 5 ( 1, 3) ( 2, 5) ( 4, 8) ( 6, 9) ( 7,10)
area = 169 index = 6 ( 1, 3) ( 2, 7) ( 4, 5) ( 6,10) ( 8, 9)
area = 169 index = 7 ( 1, 3) ( 2, 7) ( 4, 8) ( 5, 6) ( 9,10)
area = 169 index = 8 ( 1, 4) ( 2, 5) ( 3, 7) ( 6, 9) ( 8,10)
area = 169 index = 9 ( 1, 5) ( 2, 3) ( 4, 9) ( 6, 7) ( 8,10)
area = 169 index = 10 ( 1, 5) ( 2, 4) ( 3, 8) ( 6, 7) ( 9,10)
area = 169 index = 11 ( 1, 5) ( 2, 7) ( 3, 4) ( 6, 8) ( 9,10)
area = 169 index = 12 ( 1, 6) ( 2, 3) ( 4, 8) ( 5, 7) ( 9,10)
* = solutions
121 combo 3 of 14, 5-order = 3 of 120 orient = 19 of 32
1:( 1, 6) 2:(10, 2) 3:( 7, 4) 4:( 3, 9) 5:( 8, 5)
12222222222
12222222222
13333333444
13333333444
13333333444
13333333444
55555555444
55555555444
55555555444
55555555444
55555555444
121 combo 12 of 14, 5-order = 6 of 120 orient = 15 of 32
1:( 9, 1) 2:( 2, 8) 3:( 5,10) 4:( 4, 7) 5:( 6, 3)
11111111122
33333444422
33333444422
33333444422
33333444422
33333444422
33333444422
33333444422
33333555555
33333555555
33333555555
169 combo 2 of 12, 5-order = 33 of 120 orient = 6 of 32
1:( 7, 3) 2:( 6, 4) 3:( 5,10) 4:( 2, 1) 5:( 8, 9)
1111111222222
1111111222222
1111111222222
3333344222222
3333355555555
3333355555555
3333355555555
3333355555555
3333355555555
3333355555555
3333355555555
3333355555555
3333355555555
169 combo 4 of 12, 5-order = 33 of 120 orient = 6 of 32
1:( 8, 3) 2:( 5, 4) 3:( 6,10) 4:( 2, 1) 5:( 7, 9)
1111111122222
1111111122222
1111111122222
3333334422222
3333335555555
3333335555555
3333335555555
3333335555555
3333335555555
3333335555555
3333335555555
3333335555555
3333335555555
lord@rabbit-3 pid12125 %
Note in passing - I modified this code to solve the similar problem of rectangles of sides (1,2,3,...,8) and also (1,2,3,...,12). While there were many candidates for m=8,9,10 and m=14,15,16,17 respectively, there were no solutions.
Edited on October 4, 2020, 2:35 pm
solution (take two)
