how to fill in 9 digit numbers in nine squares where the addition of any 3 numbers is 18?

Answer 1

First divide, the sum with 3,
18/3 = 6 place it in the central place.
place, (6+3) = 9 in the top right corner,& (6-3) = 3 in bottom left corner.
place, (6-1)= 5 in the top left corner & ( 6+1)= 7 in the bottom right corner, now fill up the blanck areas so that you get a total of 18, counting rowwise,columnwise or along diagonals.
In this way you can construct any such magic square.

In this case, it is

5....4....9
10..6....2
3....8....7

Note that you will always have to place 6 at the central place, So if you want to place 1, one row or column or diagonal has the elements 1 & 6, so the other element will be 18-6-1 = 11; so you can not construct such a square using only the numbers, 1 to 9.

Answer 2

Well i think there is a glitch in question.It should be how to fill 9 digits in nine squares so that .......is 18.But u may be right too.I hav solution but problem is i hav completed 93% of answer so there is possible solution.Also, if i m right the answer u want definitely includes som sort of diagrammatic solution n i hav it too using 9 circles.If u want the soluion pls mail me question n i will reply the answer.

Answer 3

I f you mean nine, 1-digit numbers in nine squares, which adds up to 18 in all directions, then it is impossible. The triplet in which zero occurs, MUST have two 9s to add up to 18. So without repeating numbers, you can't do it.

If numbers can be repeated, then the simple solution is to fill all squares with 6.

Answer 4

I think the question is, how do you put the numbers 1 - 9 onto a tic-tac-toe board so that along any row of three, the numbers add up to 18.
I don't think it can be done.

Answer 5

I have no idea what you're saying. Can the 9-digit numbers be negative?

OK, so I do get what you mean. The answer is:

uhh... I don't think there is one...