### Brain Teasers

# Tic Tac Toe

Math
Math brain teasers require computations to solve.

Arrange the numbers 1 through 9 on a tic tac toe board such that the numbers in each row, column, and diagonal add up to 15.

### Answer

4, 3, 8,9, 5, 1,

2, 7, 6,

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## Comments

Nice one... many possible scramblings of the answer too ;P

rather simple.. just have 2,4,6 and 8 anywhere at the corners and 5 in the middle and u can easily solve it

This one was really fun because there are so many ways to solve it, and many ways to get the "aha!" - you can check for the center (same number, 4 combos - only 5 works), or do the big numbers (987 need to be on different rows/columns and not on a diag) - on and on. very good fun.

I got that in like 5 minutes. Is that good? Dang. I sound like a girl. Oh well....there's nothing wrong with being in touch with my feminine side is there?

a little easy for my tastes because we do that stuff in math all of the time

this was pretty boring...sorry but it was. i've had to do this i don't know how many times in school so far for EXTRA CREDIT...pretty lame.

GOOD 1! -luvupurple

I also have to say it was pretty easy but a good teaser

Takes prettly long to work out, but time consuming and fun, good job

ive done this before, you can even do boxes of four by four but i forget what they have to add up to...

Hmm.. Answer maybe should have indicated that there are several other answers, and maybe some reasoning behind it. The people that said it is easy say that because they got lucky in their trial-and-error. No one mentioned any logic, like 1 can't go with 2 or 3, 7 can't go with 4, 9 can't go with 3 or 6-8, and that 5 is the only number w/o any restrictions and thusly must go in the middle, since every other square forms a sum somehow with the middle square.

I was told of the solution to this problem before when I was a child. But it only works if the box is an odd number, meaning its 3x3, 5x5, 7x7, etc...

Firstly, you can draw a grid, say 5x5 grid.

Then place the 1 in the right-most column, middle row.

Next step is, go 1 box down, 1 box to right, then you place the next number (2) in that box.

But if you're on the right-most column, and you need to go 1 box to the right, you'll have to go 4 boxes to left instead (or go to the left-most box in the same row).

And if you need to go 1 box below and you're in the bottom row, you'll have to go 4 boxes above instead (or go to the top-most box in the same column).

You'll notice that if you're already in the number 5, if you go 1 box down and 1 box to right, that box is already occupied by number 1. So instead, go 1 box to left and place the next number (6) there. Then repeat the process.

Here is the solution for a 5x5 grid.

11 10 4 23 17

18 12 6 5 24

25 19 13 7 1

2 21 20 14 8

9 3 22 16 15

The sums of the columns, rows and diagonal are all 65.

You can try it in a 7x7 grid, the sums should be 175!

Firstly, you can draw a grid, say 5x5 grid.

Then place the 1 in the right-most column, middle row.

Next step is, go 1 box down, 1 box to right, then you place the next number (2) in that box.

But if you're on the right-most column, and you need to go 1 box to the right, you'll have to go 4 boxes to left instead (or go to the left-most box in the same row).

And if you need to go 1 box below and you're in the bottom row, you'll have to go 4 boxes above instead (or go to the top-most box in the same column).

You'll notice that if you're already in the number 5, if you go 1 box down and 1 box to right, that box is already occupied by number 1. So instead, go 1 box to left and place the next number (6) there. Then repeat the process.

Here is the solution for a 5x5 grid.

11 10 4 23 17

18 12 6 5 24

25 19 13 7 1

2 21 20 14 8

9 3 22 16 15

The sums of the columns, rows and diagonal are all 65.

You can try it in a 7x7 grid, the sums should be 175!

11_10__4_23_17

18_12__6__5_24

25_19_13__7__1

_2_21_20_14__8

_9__3_22_16_15

18_12__6__5_24

25_19_13__7__1

_2_21_20_14__8

_9__3_22_16_15

Very nice solution pating. I think I saw that before, but had forgotten about it.

The sum for each row/column/diagonal in an NxN magic square is (NxN+1) x N/2.

The sum for each row/column/diagonal in an NxN magic square is (NxN+1) x N/2.

The middle number requires 4 sets of pairs with equal sum. Hence, 5 is the only possible middle number. From there, it's casework with 9-1 being a sum in a diagonal or non-diagonal. Use the 9 to figure out where 8 can go, and you eventually arrive at the only case that works.

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