The goal of the puzzle is to fill in the cells with numbers from 1-9 so that all cells multiplied, in each row and column, equal the product given at the end of each row and column.
First, let’s start with the lowest product, 27, found in the third column.
By the rule, 1 can only be used once, and 27 only has two possible dividers: 1 and 3. The only possible solution is thus 3 x 3 x 3 x 1 = 27.
Now if look at the products in intersecting rows: 72, 42 and 81 are divisible by 3, while 50 is not. Therefore, 1 goes into the yellow cell in image above, while three other third column numbers and all 3s.
Next, let’s observe fourth column, where the numbers multiply to give 30. Number 30 has the following numbers for possible dividers: 1, 2, 3, 5 and 6. Among the given, 5 is an essential divider – no combination of other numbers can replace it. Using the same technique as in the first step, comparing it to the intersecting rows, we can see that the only row product that is divisible by 5 is the third row, so 5 has to go into the yellow cell in the image below.
Additionally, 6 is not in the column 4 as it would require two 1s to complete the equation, so the rest of the cells are 1, 2 and 3. 2 can’t go into fourth row, as 81 is not divisible by 2.
For the third step, focus on the third row, with the product of 50 and two known multiplicators, 1 and 5. The remaining two numbers need to give us 10, so they can only be 2 and 5. First column (112) is divisible by 2, and second column (135) by 5, so we complete that row.
Next, do the same thing for second row, where the product is 42.
We already have one number in the row, so the remaining cells need to multiply to 42/3=14. The only possible solution is 1 x 2 x 7, and since only the first column result (112) is divisible by 7, that number goes into the yellow cell in the image above. Using the same logic, 2 can now only go into fourth column, and finally 1 goes into second column.
Next, let’s solve the first column. 112/14=8, so the remaining two numbers are either 1 and 8, or 2 and 4. Since 81 (fourth row) is not divisible by either 2, 4 or 8, it has to be 8 in the first row and 1 in the fourth row.
Since we’re missing 1 and 3 in the fourth column and 1s cannot be repeated, 1 can’t go into the yellow cell in the image below. We can now solve the fourth column, and subsequently, the remaining rows that only miss one cell apiece.
