# Making a multiplying magic square

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Here is a simple magic square, the three columns, three rows, and two diagonals adding up 72:


+------+------+------+
|      |      |      |
|  27  |  20  |  25  |
|      |      |      |
+------+------+------+
|      |      |      |
|  22  |  24  |  26  |
|      |      |      |
+------+------+------+
|      |      |      |
|  23  |  28  |  21  |
|      |      |      |
+------+------+------+



The puzzle is to convert it into a multiplying magic square, in which the numbers in all the eight lines if multiplied together give the same product in every case. You are not allowed to change, or add to, any of the figures in a cell or use any arithmetical sign whatever! But you may shift the two figures within a cell. Thus, you may write 27 as 72, if you like. These simple conditions make the puzzle absurdly easy, if you once hit on the idea; if you miss it, it will appear to be an utter impossibility.

Clarification: shift (verb): Move or cause to move from one place to another, especially over a small distance.

"Shift" the ones digit up a bit to be an exponent!

Since the original is an addition magic square, and all the tens digits are 2, it follows that the ones digits alone form a magic square. Thus the 2^(ones digit) formed by the shift makes a multiplicative magic square.

• +1 Welcome to PSE (Puzzling Stack Exchange)! Commented Jun 2 at 21:25

Alright, note that when I refer to "swap", it's about swapping digits, or figures (please correct me if I am wrong in the comments).

This definitely has something to do with factors of numbers. Brute force shows that there are roughly 256 possible combinations to test, but that's no fun.

Right off the bat, it would be interesting to note that

Since 23 is prime, that means that if it is part of the multiplying magic square, all 8 lines must also have it as a prime factor. Yet, that is false for 5 of the lines. Therefore, 23's digits must be swapped.

In addition, an even more interesting insight is that

22 is the same whether or not it is swapped. Therefore, the first column (from left to right) only has one uncertain number (to swap or not to swap): 27.

From there,

If we maintain the digits of 27, that results in a product of 27 * 32 * 22 = 19008. We can observe that the top-right digit is 25, which is not divisible by 19008. Therefore, its digits must be swapped. However, by doing so, there is no integer $$a$$ such that 27 * 52 * $$a$$ = 19008. That leaves us with an actual magic product of 72 * 22 * 32 = 50688.

Therefore, we have this so far:

Note that with the same logic from a while ago, we can deduce that the top-right cell is in fact 52.
72 ? 52
22 ? ?
32 ? ?

Alright, so what's funny is that this still leads to no right answer.

I might have really misinterpreted this

• Please alert me ASAP as I may have just misinterpreted the question. I really want to actually solve this problem. Thank you. Commented Jun 2 at 5:07
• Swapping is only one type of shifting. Commented Jun 2 at 5:23
• What are the other types of shifting? Please demonstrate ;-; Commented Jun 2 at 13:57
• A direct answer would pretty much give it away. Here's a hopefully less obvious hint: How would you construct a 3x3 multiplying magic square where the nine numbers could be any nine distinct positive integers? Commented Jun 2 at 17:58
• I think the operation of "shifting" should be clarified as I really do not understand the conditions of the puzzle; e.g. can I use "shift" to make a "2" become "@"? @EdMurphy Commented Jun 2 at 18:04

The trick is to do bit-shifting. That is, replace every number $$n$$ in the grid by $$2^n$$.

• I think you are absurdly close to the intended solution but still missed one detail. Commented Jun 2 at 21:04