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Find a 3x3 Magic Square where each row, column and both diagonals result in positive number 9

You are allowed any of the 4 math operations : add, subtract, multiply or divide. NO OTHER OPERATIONS ALLOWED

You must use numbers 1 thru 9 (positive integers only, no negative signs within the number boxes). All 9 numbers must be used.

You can use a Square root or Cube root or a Factorial function WITHIN the number box only. No other function allowed. For example the number box can say 3! which would be a value of 6 but you have used the number 3 only.( you cannot use 3 somewhere else). If you use 3! then you must use 6 somewhere else. The square must show all 9 positive integers.

Mathematical operations must be in the direction of the arrows. No programming.

MAGIC SQUARE 9

I found one solution but there may be more.

As a broader question can one program the square to get any number (not just 9) with multiple math operations using 1 to 9??

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  • $\begingroup$ I am confused. What original source? I have created this with the idea that instead of the Original Magic Square of 3X3 for getting 15 by just adding all rows and columns annd diagonals I could get another number by adding other math operations. Please tell me what original source you are talking about. $\endgroup$ – DEEM Jun 5 '18 at 19:19
  • $\begingroup$ Whoops. This looks pretty much exactly like what puzzle copy-pasters do, so apologies for jumping to that conclusion. I've removed the hold. Nothing to see here, move along ... ;) $\endgroup$ – Rubio Jun 5 '18 at 19:26
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magic square­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­­

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  • $\begingroup$ Way to go noedne! I thought there would be more than one answers. ( Mine is different). Out of curiosity how did you get there? Any system? If yes can this be done for ANY numbers (just not nine) $\endgroup$ – DEEM Jun 5 '18 at 21:24
  • $\begingroup$ @DEEM I started with 7 because it seems finicky, but beyond that it was just trial and error. This surely does not work for any "number," for any reasonable definition of number, because it obviously lacks the flexibility to support many large integers. $\endgroup$ – noedne Jun 5 '18 at 21:35

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