Grid 3x3

Question

Grid problem 3x3

 

Answer 1

Let:

A B C

D E F 

G H I

represent the initial 3×3 grid, where the letters represent a configuration of the numbers from 1 to 9. The products form two (algebraic) sets:

Z={ABC DEF GHI} (horizontal)

V={ADG BEH CFI} (vertical)

If Z=V then, for example, ABC=ADG or BEH or CFI. In each case, one of the numbers cancels out, so we are left with equating the product of two pairs of numbers. So we only have to consider which products of a pair of numbers are the same.

(a) 1×6=2×3, (b) 1×8=2×4, (c) 2×6=3×4, (d) 3×6=2×9, (e) 4×6=3×8.

If, for example, ABC=ADG, then BC=DG, if DEF=BEH, then DF=BH, and if GHI=CFI, then GH=CF. In this example A, E and I have cancelled out. These form a diagonal in the grid. The numbers missing from the pairs are 5 and 7. But another number is required for the diagonal—set X.

There are 10 ways (combos) to select 3 sets out of 5:

abc, abd, abe, acd, ace, ade, bcd, bce, bde, cde.

In the products we are using B, C, D, F, G, H, that is, 6 out of 9 numbers:

BC=DG, DF=BH, GH=CF. Each letter appears twice. We need at least one combo in which each number appears twice, and only one number is omitted. This omitted number can then be added to the diagonal set. We can capture those combos satisfying the constraints: abe is the only one:

1×6=2×3, 1×8=2×4, 4×6=3×8. The number 9 is missing so it can be included in set X, X={5 7 9}.

Now let’s prove the configuration:

B=1, C=6, D=2, G=3, F=4, H=8, A=5, E=7, I=9:

5	1	6	30
2	7	4	56
3	8	9	216
30	56	216

It works! There are 6 ways of arranging the numbers along the diagonal, and each one will produce a grid that meets the requirements without changing the positions of any other numbers. So straight away we have 6 possible grids! For example, swapping 9 and 5, we get 54 as the product of the first row and column; and 120 for the last row and column.

 

More to follow...

Comments

...answer continued:

Applying the method to an n×n grid filled with integers from 1 to n², we can generalise:

List all pairs of composites up to n⁴-n² with n-1 different factors, including 1 (that is, 1 and n-2 different factors). There must be no fewer than n pairs. 

Each pair will contain 2(n-1) numbers. 

There are n numbers making up the diagonals.

If p is the number of pairs of factors, there must be sufficient pairs so that 2p(n-1)+n≥n². In particular, if r is the actual number of pairs used out of a possible p pairs, then ᵖCᵣis the number of combinations that need to be examined to look for a set of pairs to create the required grid.

 

Still more to follow...