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Comments on Where is the large cook promotion?
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Where is the large cook promotion? Question
The time is 8:59 or ...
Where is the large cook promotion?
Note: The correct answer to this puzzle is related to the numbers 3,4,2.
Spoiler! Click here to reveal a hint about filling in the grid.
The phrase “The time is 8:59 or ...” is intended as a hint as to how to fill in the grid.The time 8:59 can be read as 1 minute to 9 (or 1 to 9) meaning that the grid should be filled with all the numbers 1, 2, 3, ..., 9 using each number exactly once.
With this condition, there is only one way to correctly fill in the grid.
Post
This is only a partial answer that solves the grid.
spoiler
A good way to approach solving the grid is to assume that all the cells contain integers. Then we can decompose each of the row and column products into their prime factors: --- --- --- 126: 2 3 3 7
--- --- --- 120: 2 2 2 3 5
--- --- --- 24: 2 2 2 3
36: 72: 140:
2 2 2
2 2 2
3 2 5
3 3 7
3
From this we can immediately see that 7 and 5 are only in one row and column each. Those factors therefore have only one place they can possibly go:
--- --- 7 126: 2 3 3
--- --- 5 120: 2 2 2 3
--- --- --- 24: 2 2 2 3
36: 72: 140:
2 2 2
2 2 2
3 2
3 3
3
Two more 2s are required in the right column. Assuming that none of the squares are intended to contain 1, at least one of those 2s goes in the lower right cell. Since the bottom row needs a total of three 2s, we'll put both the remaining 2s of the right column in the bottom cell, at least for now:
--- --- 7 126: 2 3 3
--- --- 5 120: 2 2 2 3
--- --- 2 2 24: 2 3
36: 72: 140:
2 2
2 2
3 2
3 3
3
We'll consider the right column done for now and see if the remaining factors can be fit into the remaining six cells. There are less 3s to place (4) compared to 2s (5), so let's start with the 3s. Each of the bottom two rows can only hold a single 3. Both the remaining columns need two 3s, so each of the open top cells must have at least one 3 each:
3 3 7 126: 2
--- --- 5 120: 2 2 2 3
--- --- 2 2 24: 2 3
36: 72: 140:
2 2
2 2
3 2
3
The remaining two 3s must go into the corners of the bottom left four cells. The problem is under-constrained, so they can go either way. We'll arbitrarily pick upper left and lower right:
3 3 7 126: 2
3 --- 5 120: 2 2 2
--- 3 2 2 24: 2
36: 72: 140:
2 2
2 2
2
The bottom row can only take a single 2, so we'll stick it in the currently-empty lower left cell:
3 3 7 126: 2
3 --- 5 120: 2 2 2
2 3 2 2 24:
36: 72: 140:
2 2
2
2
The left column and the top row both need a single 2, so we'll stick it in the top left cell:
2 3 3 7 126:
3 --- 5 120: 2 2 2
2 3 2 2 24:
36: 72: 140:
2
2
2
Both the center column and center row need three 2s, so we dump them into the center cell:
2 3 3 7 126:
3 2 2 2 5 120:
2 3 2 2 24:
36: 72: 140:
This is only one of many possible solutions. In any "box" within this matrix that has the same number in two opposite corners, you can switch those numbers to the other opposite corners. As an example, consider the lower right 4 cells as a box. Both the upper left and lower right corners of that box contain 2s. Here is switching one of those 2s to the lower left and upper right corners:
2 3 3 7 126:
3 2 2 2 5 120:
2 2 3 2 24:
36: 72: 140:
Note that all the products still work out. It should therefore be clear that there are quite a few valid integer non-1 solutions to the grid.
The values in each cell are the products of the prime factors we have arranged into them. Here is the full solution as shown in the last grid:
6 3 7 126:
3 4 10 120:
2 6 2 24:
36: 72: 140:
Again, there are many possible solutions to the grid. By working with the prime factors it is easy to rearrange values that still result in the same row and column products.
I realize solving the grid is only part of the overall puzzle, but I'm going to quit here.
You assumed that none of the squares are intended to contain 1 but that is contrary to the subtle hint:The time is 8:59 or ...
There are many solutions to the grid. One way to find them is to start with any one solution, then permute it according to the rule I showed above.
Another way to state the same rule is to find a pair of cells that are not in the same row or column. Any common factor in both those cells can be removed from those cells and placed instead into other two corners of the "box". Less handwavy, find two cells at (X1,Y1) and (X2,Y2) where X1≠X2 and Y1≠Y2. Any common factors in the two cells can be moved to (X1,Y2) and (X2,Y1) from the original cells.
There are 18 such pairs of cells in this 3x3 grid. The next steps would be to go thru them and rearrange factors. Allowing cells to contain only 1 broadens the possible swaps but the mechanics remain the same otherwise.
I expect your intended answer will have some "cute" values. But, since we don't know what we're looking for, and have only a vague hint what your idea of "cute" is, this will be just a bunch of busy-work that's not fun anymore. That's why I quit where I did.
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