7 24 20 18 8
13 22 16 12 11
17 14 10 19 2
5 23 6 21 4
25 9 15 3 1
The first step in solving was to divide 25! = 15511210043330985984000000 by all of the first four row products to get the last row product as 10125, and to divide 25! by all of the four column products to get the last column product, 704.
Then factor each of the row and column products into primes:
483840=29*33*5*7
604032=27*3*112*13
90440=23*5*7*17*19
57960=23*32*5*7*23
10125=34*53
193375=53*7*13*17
1530144=25*33*7*11*23
288000=28*32*53
258552=23*35*7*19
704=26*11
The primes 17, 19 and 23 are the easiest to place, at the intersections of the rows and columns containing those factors.
The factor 11 can appear only twice: once as 11 and once in 22, so both appearances must be in row two: one in column 2 and one in column 5.
Once the factor 11 in the final column is used for either 11 or 22, what remains of 704 is 26. If one of the six factors of 2 in the final column were used to combine with the 11 to produce 22, there wouldn't be enough factors of 2 remaining to make four different other numbers for the other rows in that column, so the 22 is in column 2 and the 11 in column 5, while the remaining numbers in column 5 are all powers of 2, the powers being 0,1,2,3, making the numbers in that column 1, 2, 4, 8 besides the 11 already mentioned.
Row 2 now has 13, 11 and 22, leaving 26*3. This can only be divided up as 16*12 as any other factorization would leave a power of two that's larger than 25 or one that's already been assigned to column 5, so 16 and 12 are the remaining numbers in row 2, and only column 3 has a high enough power of 2 to allow 16, so that's the column that gets the 16 and the 12 goes into column 4.
Similar reasoning fills out the remainder of the array, such as recognizing that 5, 7 and 25 must be in the first column, and starting with the placement of 1, 15 and 25 in their respective spots on the last row, noticing that 14 must be in the middle row, and 9 and 3 in the bottom row, etc. When I did it, 24 and 18 were among the last four numbers to be placed, by looking for places they'd fit.
Based on Enigma No. 1625, "Diagonal diversion", by Gwyn Owen, New Scientist, 11 December 2010, page 28. |
