jasen 4 Report post Posted July 16, 2016 Put number 2,3,4,5,6,9,10,12,15,18,20,25,30,45,50 and 75 to a 4x4 square table, so the product of 4 squares at every horizontal, vertical and diagonal line are same. Share this post Link to post Share on other sites

0 Thalia 61 Report post Posted July 19, 2016 (edited) I started with making a 4x4 using the factor distribution numbers DejMar mentioned. After a combination of the above and a lot of guessing. . . Spoiler 10 3 20 45 18 15 25 4 75 12 6 5 2 50 9 30 Edited July 19, 2016 by Thalia Share this post Link to post Share on other sites

1 DejMar 9 Report post Posted July 19, 2016 (edited) There are more solutions than the one given by thalia: Spoiler 12 25 3 30 2 9 20 75 15 25 6 12 75 9 10 4 15 30 5 12 18 30 5 10 2 6 50 45 50 4 45 3 2 9 20 75 15 20 18 5 18 25 6 10 50 4 45 3 Other solutions may (and I believe) exist. Edited July 19, 2016 by DejMar correction of poster Share this post Link to post Share on other sites

0 Buddyboy3000 4 Report post Posted July 17, 2016 I was able to find out one part of this question Spoiler The product of the 4 squares has to be 27000. I find this out by first multiplying all of the numbers together, which is 531441000000000000. Next, because there are 16 numbers, I found the 16th root of it, which is about 12.82. Then, because you multiply four of the numbers together, I put that to the fourth power. This got me the answer of 27000. This can be simplified to the product of all of the numbers, 531441000000000000, to the fourth root to make 27000. However, I have not been able to find the arrangement of the numbers in the four by four grid. Share this post Link to post Share on other sites

0 DejMar 9 Report post Posted July 18, 2016 On 7/17/2016 at 6:43 PM, Buddyboy3000 said: I was able to find out one part of this question Hide contents The product of the 4 squares has to be 27000. I find this out by first multiplying all of the numbers together, which is 531441000000000000. Next, because there are 16 numbers, I found the 16th root of it, which is about 12.82. Then, because you multiply four of the numbers together, I put that to the fourth power. This got me the answer of 27000. This can be simplified to the product of all of the numbers, 531441000000000000, to the fourth root to make 27000. However, I have not been able to find the arrangement of the numbers in the four by four grid. Buddyboy3000, Spoiler Your are correct that the product of each row, column, and main diagonals will each total 27000. Each set of four numbers will have the factors 2^{3}×3^{3}×5^{3}. You can determine this by factoring each of the sixteen given numbers and discover that there are twelve each of the primes 2, 3, and 5. Dividing by 4 this gives three each of the distinct primes for each individual row and column, and, for the "magic square", the main diagonals. No square will comprise of more than three factors, and as there are nine factors to be distributed between four squares in a row, column, or diagonal, the distribution of factors must be {3,3,2,1} or {3,2,2,2}. Given there are three single-digit factors, three of each of the rows and columns will be {3,3,2,1}, and no column, row, or main diagonal will contain any two of the single digit primes 2, 3, and 5. However, as of yet, I, too, have not found the arrangement of the numbers in the four by four grid. Share this post Link to post Share on other sites

0 araver 10 Report post Posted July 18, 2016 (edited) I believe this is Spoiler impossible. tl;dr An exhaustive search showed me that it's impossible. But I don't have a complete proof of contradiction as I tried to manually construct such a solution. A few notes though: All numbers are have as prime divisors 2, 3, 5. My approach is to split the dimensions. I lack a very clear mathematical notation for the argument. Or a picture. Oh well, a thousand words should do the trick. Leave some powers aside and concentrate only on 2s (the argument applies to the other prime powers as well). Assuming there is such an arrangement: a11 ... a14 a21 ... a24 a31 ... a34 a41 ... a44 The number of powers of 2 from the product of all numbers is 12. Since the product on each line is equal it is divisible with 2^3=8 and not divisible by 2^4=16. That means that the following square (bij)1<=i,j<=4 where bij is the highest power of two in the decomposition of aij, also forms a regular magic square. There are 16 numbers, 3 of them are divisible by 4 (4,12,20), 6 are divisible by 2 (2, 6, 10, 18, 30,50) and 7 of them odd. That means the sequence of bij is a permutation of [2,2,2,1,1,1,1,1,1,0,0,0,0,0,0,0], with each row, column and diagonals summing up as 3. It is easy to see that the only arrangement for the 4 rows is 2+1+0+0=2+1+0+0=2+1+0+0=1+1+1+0=3. Similarly for columns. The diagonals are themselves either 2+1+0+0 or 1+1+1+0 since there is no other way to make a sum of 3 out of twos, ones and zeroes. Case 1) Assume one diagonal (symmetric, so let's take the main diagonal) is 2+1+0+0 and the other has the same pattern permuted. 2 x x y x 1 x x x x 0 x y x x 0 If the other diagonal has 2 it can't be in the same row/column as the 2 on the main diagonal. Again, the choice is symmetric along the main diagonal, so let's pick one of the remaining 2 spots and finish the row. 2 x x y 0 1 2 0 x x 0 x y x x 0 Here's the catch, column 1 and column 3 have twos and column 4 already has 2 zeros. So the column with 3 ones can only be the second column. Hence there is a 2 on the 4th column, can't be y, so we place 2 between the zeros. 1 follows in the north east corner, then zeroes fill the first row naturally: 2 0 0 1 0 1 2 0 x x 0 2 y x x 0 Now since the second column is the one having the 3 ones and 1 zero: 2 0 0 1 0 1 2 0 x 1 0 2 y 1 x 0 And we have just obtained 1+2+1+y on the other diagonal. Contradiction. Case 2) Going back, assume one diagonal (symmetric, so let's take the main diagonal) is 2+1+0+0 and the other one has 3 ones and 1 zero 2 x x x x 1 y x x y 0 x x x x 0 Obviously at least one of the y positions has a 1 in it which creates a line or a row with the other 1 from the main diagonal. For symmetry reasons, we can assume that 1 is above the first 0. 2 x x x x 1 1 x x x 0 y x y y 0 Now there are 2 twos to place in the square and since these cannot be placed on the same row or on the same column or on the same row/column with the other 2 or on the secondary diagonal or on the row with 2 ones only 3 possibilities remain marked with y in the above. 2 x x x x 1 1 x x x 0 2 x y y 0 After placing it, we run into the problem of placing the remaining 4 ones. We know exactly 2 are on the secondary diagonal out of the 3 remaining x's. Exactly 1 other is on the second row to complement the 2 ones already there. Easier to see if we start to explore the two possibilities for 2: Case 2.1 2 0 x x x 1 1 x x 0 0 2 x 2 y 0 If we choose the first y in the last row, there is one way to complete column 2, with zeroes. Which gives another unique way to complete the secondary diagonal with ones. 2 0 x 1 x 1 1 x x 0 0 2 1 2 y 0 which leads (b/c of restrictions on first and last row) to: 2 0 0 1 x 1 1 x x 0 0 2 1 2 0 0 That is hardly magical since column 3 has a sum of 1. Case 2.2 2 x 0 x x 1 1 x x x 0 2 x y 2 0 If we choose the first y in the last row, there is one way to complete column 3, with zeroes. That almost seems plausible given that we have 4 ones to place and there must be exactly 1 on every row. Assuming the southwestern bit is either 0 or 1 we get two solutions: Case 2.2.1 2 1 0 0 0 1 1 1 0 1 0 2 1 0 2 0 and Case 2.2.2 2 0 0 1 1 1 1 0 0 1 0 2 0 1 2 0 Case 3) Going back assume both diagonals are 1+1+1+0. One cannot have more than one row with 2 ones on it since the 4 rows are 2+1+0+0=2+1+0+0=2+1+0+0=1+1+1+0=3. However following the pigeonhole principle the 3 ones in the two diagonals will collide on at least two rows (sample below): 0 x x 1 x 1 0 x x 1 1 x 1 x x 1 Exhaustive combinations show that (leaving symmetry aside) there are solutions only with a diagonal (main or secondary) of ones and zeros. Coming back, we have as divisors powers of 2, 3 and 5. Looking at the three square matrix projects for 2, 3 and 5, the above argument and the Dirichlet/pigeonhole principle states that there is a diagonal where the pattern 1+1+1+0 can be found for two of those powers. As we saw before one cannot have 2 permutations of 1,1,1,0 in top of one another without 2 ones overlapping. E.g. that gives me starting from case 2.2.1 the following square (where x,y,z denotes 2^x*2^y*2^z) 2,x,x 1,x,x 0,x,x 0,x,x 0,x,x 1,x,x 1,1,x 1,x,x 0,x,x 1,1,x 0,x,x 2,x,x 1,0,x 0,x,x 2,x,x 0,x,x The two center pieces on the secondary diagonal are overlapping ones in the first and second dimension. The lowest southwest corner cannot be 1,1,x since there are only 1,1,0 and 1,1,1 among the starting numbers. I do not see how one could continue to branch the possibilities though, without a lot of sub-cases. Edited July 18, 2016 by araver Share this post Link to post Share on other sites

0 jasen 4 Report post Posted July 19, 2016 (edited) To attack the problem, first write the numbers into its prime factors 2 = 2, 3 = 3, 4 = 2.2, 5 = 5 6 = 2.3, 9 = 3.3, 10 = 2.5, 12 = 2.2.3 15 = 3.5, 18 = 2.3.3, 20 = 2.2.5, 25 = 5.5 30 = 2.3.5, 45 = 3.3.5, 50 = 2.5.5 75 = 3.5.5 So all numbers are consist of 3 prime seed 2,3 and 5 Now replace each seed with 3 symbols, A,B and C So what we get now is A, B,C,AA,BB,CC,AB,AC,BC,ABB,ACC,BAA,BCC,CAA,CBB, and ABC There are 12A,12B, and 12C Total symbols, so each line must have 3A,3B and 3C What we have to do now is Put A,B,C,AA,BB,CC,AB,AC,BC,ABB,ACC,BAA,BCC,CAA,CBB, and ABC to a 4x4 square table, so every horizontal, vertical and diagonal line have 3A,3B, and 3C. Edited July 19, 2016 by jasen Share this post Link to post Share on other sites

0 jasen 4 Report post Posted July 19, 2016 2 hours ago, Thalia said: I started with making a 4x4 using the factor distribution numbers DejMar mentioned. After a combination of the above and a lot of guessing. . . Hide contents 10 3 20 45 18 15 25 4 75 12 6 5 2 50 9 30 If we change Thalia answer with ABC solution, than replacing ABC with permutation [2,3,5], we can get 6 diffrent answers, one of it is as picture bellow Spoiler Share this post Link to post Share on other sites

0 Logophobic 9 Report post Posted July 23, 2016 A note on the distribution of factors and the number of distinct solutions: Spoiler There are six distinct distributions of factors: /---------+---------+---------\ | A | B | C | | 1 2 3 3 | 1 3 2 3 | 1 3 2 3 | | 2 3 1 3 | 3 3 1 2 | 3 3 2 1 | | 3 2 3 1 | 2 2 3 2 | 3 2 2 2 | | 3 2 2 2 | 3 1 3 2 | 2 1 3 3 | +---------+---------+---------+ | D | E | F | | 1 2 3 3 | 2 1 3 3 | 2 2 3 2 | | 3 2 2 2 | 3 2 1 3 | 3 1 2 3 | | 3 2 3 1 | 2 3 3 1 | 2 3 3 1 | | 2 3 1 3 | 2 3 2 2 | 2 3 1 3 | \---------+---------+---------/ All solutions will fit one of these six patterns through rotation and/or reflection. As jasen pointed out, each solution is one of a group of six permutations of a single base solution. Assuming a single solution for each of the above patterns, there only six base solutions and a total of 36 distinct solutions. The solutions given so far fit patterns A, B and C. Thalia's solution and DejMar's first solution (pattern C) are not distinct through permutation. It is still unknown if there are multiple solutions for any of the patterns, or if there are any solutions to patterns D, E and F. Share this post Link to post Share on other sites

Put number 2,3,4,5,6,9,10,12,15,18,20,25,30,45,50 and 75 to a 4x4 square table,

so the product of 4 squares at every horizontal, vertical and diagonal line are same.

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