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help with cross math puzzle?


Due to the 1st Round of this years Puzzle GP hasn't started yet - is there possibly anybody who's willing to help me with a cross math puzzle? I'm on the last page of a Swiss Puzzle Magazine called "Mix Logik Nr. 39", it's the last puzzle (163) I'm fighting with.

I wonder if there's an easier way to solve this kind of puzzle than by method of elimination? I've been able to fill in 3 of 36 integers (1-9) for certain, I have quite a lot possibilities for the others.
Usually, my next step is "trial and error"-method, but in this case that might take an eternity ...

The puzzle is taken from the World Puzzle Championship, unfortunately I'm not sure about the year. I think I bought the magazine this year (2014), so I assume it's 2013 or 2014, but I wasn't able to get the puzzles from those years ...

So, can anybody offer tips and tricks for this kind of puzzles? Is there an easy way I overlooked? I guess I'm not able to post pictures in this forum, but I can email it to those interested in this specific puzzle ... Feel free to message me :)

Thx for the help in advance,
lovely greetings,

Mix Logik is a Magazine that appears every 2nd month in Switzerland. Number 39 appeared in autumn 2013. The puzzle is, what surprize, when you see the puzzle, from 21st WPC in Croatia
Did you read the additional rule, that no 2 numbers can be in the same cell in 2 grids?
A few hints:
Puzzle 2,
- top row: Only 137 are possible, due to other conditions 1 can only be placed in top left cell. So 7 is in the middle and 3 on the right.
- Column three must be 3x(45)+(96)=21
- In 3rd row there can’t be a 6, because you used 3 in top row.-->3x4+9=21
- Row 3 must be 2x6+9=21, 2 is too small for column 2. The rest of puzzle 2 is easy
Puzzle 4
- Top row and left column, you need 789. 123 is impossible. Multiplication in column 3 needs a 7, so 7 is in top right corner. (89)+(65)+7=21
- You have to find a place for one number >3 in the 4 fields left. Column 2 is not possible. Column 3 is fixed with 7x6:2=21
- Top row must be 9+5+7=21 and left column must be 9+4+8=21 because of row 3.
Greetings Markus

Thanks so much for your detailed answer.

I did read the additional rule, but it's not really helpful in the beginning ...

I already filled in the first row of Puzzle 2 correctly, 1 7 3. But from there on, in Column 3, I have 3 x (456)+(369)=21. I just tried to fit in the 6 in Column 3 Row 2, but I could fit in the 2nd Puzzle with my remaining possible numbers ... And why can't it be a 6 in the 3rd row? 3x5+6=21 as well ...

In Puzzle 4 I can restrict the possibilities for the 3rd Column to (679)*(3679):(123)=21, so I know the sum in the first row of the first two summands have to be (12 14 15), but I can only eliminate 1 and 2 for Column 1 Row 1 ... Can you please tell me how you get to your conclusions?

Greetings, Judith