Place digits from 1 to 7 into each row and each column of the grid once. Numbers in the circles give the product of the four surrounding digits

What is the way to resolve the following puzzle?

Link to source. It's from a taiwanese math olympiad held in 2005.

• Hi Grace, and welcome to Puzzling :) Where did you find this puzzle, please? If you can mention its source at the bottom of the post that will help us to credit the original puzzle creator and satisfy our plagiarism policy... Thanks!
– Stiv
Commented Dec 9, 2021 at 14:53
• Is the OP active? Commented Dec 10, 2021 at 3:15

3 Answers

Here's another answer for the sake of detailed solving path.

First to notice is the columns 4-5 and rows 3-4, each of which fully contains three disjoint cages. Note that the product of each row or column is $$7! = 5040 = 2^4 \times 3^2 \times 5 \times 7$$ so two rows or two columns should multiply to its square. For columns 4-5, dividing it by $$168 \times 120 \times 84$$ gives $$15$$, which means R3C4-5 should contain a 3 and a 5. Likewise, R3-4C7 should contain a 3 and a 7. But since R3C4-5 is a naked pair, we can determine that R3C7 is 7 and R4C7 is 3.

The second 7 in R3-4 should go in the 105 cage, but 120 is not a multiple of 7, so there is only one place it can go. And a similar logic applies to 3 in the same cage.

The remaining three cells in the 60 cage have the product of 12, but without including the number 3. So it must be (1, 2, 6). There's only one place in R5 for 7, the 7 in R6 must go inside the 84 cage, and then the 7 in R7 is also decided.

Another logic can be drawn from C1-2. Dividing $$7!^2$$ by 192 and 36 gives $$1 \times 3 \times 5 \times 5 \times 7 \times 7$$, which precisely defines the six numbers outside the two cages. Then all numbers inside 192 or 36 are even except for a single 1 and single 3, none of which can belong to 192. So both go in 36, and the other two numbers are 2 and 6.
And similarly the product of six cells outside 168 and 24 on R1-2 is $$2 \times 2 \times 3 \times 3 \times 5 \times 5 \times 7$$. If the R1-2C1-2 contains (1,3,5,7), then the product of R1-2C3 is 60, which is impossible. So the corner must be (3,5,5,7). This gives a lot of progress with Sudoku logic.

More progress, combined with the fact that additional 3 cannot go into 120.

More progress at the top. Also, the 84 cage cannot contain any 2, so the only possible combination is (1, 3, 4, 7).

The rest of the puzzle is solved easily.

The completed grid will look like this:

Here is what i got (Please forgive the editing I did it in MS Paint from my grandparents' laptop)

Method of solving

Started off with the top right column and calculated the possibilities for each of the circles
(Wavy curve method and Prime Factorisation) and got the values for each of the boxes