Solve 3x3 Numerical Puzzle: Can it Be Done?

AI Thread Summary
The 3x3 numerical puzzle requires placing the numbers 2, 2, 2, 3, 3, 3, 4, 4, 4 in a way that every row, column, and diagonal sums to 9. Participants initially struggled to find a solution, with one user presenting a configuration that only partially met the criteria. After several attempts, a valid solution was identified: 342, 234, 423, which satisfies the conditions. Discussions highlighted that at least one '3' must be present in every line due to the odd total of 9. Overall, while multiple configurations may exist, the focus remained on finding valid arrangements that meet the puzzle's requirements.
Natasha1
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1. In a 3 x 3 square, place the numbers 2,2,2,3,3,3,4,4,4 in it so that when any line of three numbers is added up in any direction (including diagonally) the total is always 9.

2. I have tried for hours, can anyone tell me if this problem is actually possible?

The best I get is when I do

234
342
423

And only get one diagonal of 4s which you can also do with only 3s or 2s. Any help would be truly welcomed. Thank you!
 
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Start with the center square and think what number can be placed there so that the two diagonals and middle row and middle column can sum to 9 with the same number?
 
Is there only one solution to this problem. Is the centre square 2?
 
I just can't do this... There are no numbers that can go in the middle and satisfy this problem
 
Natasha1 said:
Is there only one solution to this problem. Is the centre square 2?
It took me about 4 tries, but I came up with two solutions, and there are probably four or maybe more. I don't have 2 in the center.

Natasha1 said:
I just can't do this... There are no numbers that can go in the middle and satisfy this problem
Yes there are.
 
Got the answer

342
234
423

Thanks!
 
In my head only 4, 3 and 2 made 9... Ahhh basic! Forgot the 3, 3, 3.

Has anyone got any other combinations by pure interest?
 
Mark44 said:
It took me about 4 tries, but I came up with two solutions, and there are probably four or maybe more. I don't have 2 in the center.
Excluding rotations, I think there's only one. With rotations, four, and allowing reflections makes no difference.

There must be at least one 3 in every line (9 is odd), which forces the diagonal of 3s, and the rest is determined by your next number placement.
 
Gosh how do you guys work all this out! Brainy!
 

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