Magic square deluxe!
A magic square consists of a 3 × 3 grid filled in with the digits 1-9 such that every column, row, or diagonal adds up to the same constant (typically 15).
Is it possible to fill each box with a distinct integer such that the product of every column, row, or diagonal is the same number?
Note: The integers don't need to be the numbers 1-9.
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11 solutions
But the numbers must be between 1-9
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It says integer in the final question... But I've clarified the question.
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If you read the note, you’d actually understand.
Nice! I believe that the smallest multiplicative magic square "product" is 216, e.g.,
18 .... 1 ..... 12
4 ..... 6 ..... 9
3 ..... 36 .... 2
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Can you prove that claim?
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I think you'd need to use unique factorisation: the product of row/columns can be factorised as p 1 i 1 p 2 i 2 ⋯ p n i n . Each entry of the square is also of this form.
In the case n = 1 you reduce yourself to standard magic squares and get the smallest product to be 2 1 2 = 4 0 9 6 . In the case n = 2 you are looking at Graeco-latin square and the smallest is the one with primes 2 and 3 (i.e. the one given above). In the case n ≥ 3 , you can show that i j ≥ 2 (probably i j ≥ 3 is true) and get a minimal product of at least 300.
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@Antoine G – Why 2^12 and not 2^15=32768, though?
Also, if all powers need to be >=3 (which was basically proven by Yatin above), then minimal value would be 27000. :-B
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@C . – The "traditional" magic squares has integers from 1 to 9. If you add the same number to all numbers in a magic square, you still get a magic square. So let's add -1 to get a magic square with numbers from 0 to 8. 0 is not a positive integer but when you take the powers of 2, 2 0 = 1 .
So basically, since it is known that the smallest magic square is the standard one (minus 1 if you accept 0), the smallest "product magic square" with only one prime has a product of p 1 2 which is much bigger than the actual 216.
Here is the square:
2^1 | 2^6 | 2^5
2^8 | 2^4 | 2^0
2^3 | 2^2 | 2^7
=
2 | 64 | 32
256 | 16 | 1
8 | 4 | 128
So to recap: with one prime it's a 2 1 2 ; with two primes it's a Graeco-Latin square (with smallest elements being a = 2 0 , b = 2 1 , c = 2 2 , α = 3 0 , β = 3 1 , γ = 3 2 ); with three primes [using Yatin's comment which came some days afterwards] you get at least ( 2 ⋅ 3 ⋅ 5 ) 3 = 2 7 0 0 0
Would you accept a brute-force approach, Marcus? :)
The reason this is the smallest, likely has to do with the center number.
This has to be:
- > 4 ( since you need to have 4 positive integers less than it in 4 of the of the other boxes
- have at least 2 prime factors (so that you can have four combinations that pass through the center
6 fits the bill!
It can be easily proved that the product is actually the cube of the number in the central square. Hence, we only need to check cubes; 8,27 and 125 are cubes of primes we cannot have so many distinct factors for them. 64 too doesnt have enough factors required.
As for proof consider:
a b c d e f g h i
As the magic square and K as product.
K=abc=def=ghi=aei=beh=ceg=def
K=K^4/K^3 = (aei)(beh)(ceg)(def)/((abc)(def)(ghi) = e^3.
Brilliant way of viewing it! Bravo!
Wow! Brilliant!
Such simplicity much wow.
Of course (slapping my forehead)! Why didn't I think of that? Very nice!!
I found exactly the same solution ;)
I did not read the note, I really do need to read more
Amazing!!!!
The solution can be generalized with any number set a^n+bc where c is -3,4,-1 on top row; 2, 0 -2 in middle row; and 1, -4, 3 on bottom row. The values a, n, and b can be anything and gives pretty much infinite solutions to the problem.
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Not quite sure I understand... 🤔
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He's talking about the additive square of the exponents.
how though
A general solution could be of the form:
| h f | b | d h |
| d | b h | f |
| b f | h | b d |
With the additional constraint:
bh = df.
Take b,d,f,h to be square numbers satisfying the constraint and you are done!
I think your solution is the best one.
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Apart from b, d, h and f not exactly needing to be square numbers (see the 2^n solution), yes, I agree with you.
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I said square numbers because in order to fill the grid with integers, the easiest way is to have b,d,f,g as squares.
Why did you choose b, d, h and f, may I ask?
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I started off with 9 unknowns - a to i. But after reducing a couple of equations was left with only 4 : b,d,f,g.
The most prefectest answer i've seen
Here's another solution
| 1 2 | 1 | 1 8 |
| 9 | 6 | 4 |
| 2 | 3 6 | 3 |
where all the products equal to 2 1 6
Very nice! As @Brian Charlesworth pointed out to me earlier this could be the one with the smallest integers!
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oh, I didn't see Brian's comment posted earlier
I love it! Hmm. I see that each row, column, diagonal has three factors each of 2 and 3. Do they have to have three? Could I have four factors of 2 and two of 3?? (A common sum of 144?)
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Any magic square of powers of any prime number or factor can be thrown in. For example, throwing in the prime number 5
| 6 0 | 9 | 5 0 |
| 2 5 | 3 0 | 3 6 |
| 1 8 | 1 0 0 | 1 5 |
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And yet three of each - 2 and 3 and 5 - in each row, column, diagonal.
But I’m looking for a lower product than 216.
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@Phillip Wagoner – These numbers are not distinct, however. But you get the idea.
| 4 | 1 | 2 |
| 1 | 2 | 4 |
| 2 | 4 | 1 |
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@Michael Mendrin – I’m looking for distinct numbers. Maybe based on powers of 2. But all I get is
2^1 2^8 2^3 2^6 2^4 2^2 2^5 2^0 2^7
And each product is 2^12. Decidedly not less than 216.
Michael, your use of factors of powers of 2 and 3, I see, is the key. 1,2,3,4,6,9,12,18,36 increases much more slowly than powers of 2 alone: 1,2,4,8,16,32,64,128,256.
A beautiful solution, Michael. Thanks for sharing.
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@Phillip Wagoner – For distinct integers, 216 is the minimum possible product.
@Phillip Wagoner – Nope, i just ran a computer program to check all possible products up to 220. Only 8 solutions exist with that restriction: the rotations and mirror images of what Michael provided.
However, the first in "alphabetical" order would be 2 9 12 / 36 6 1 / 3 4 18 :-B
Not sure you can get to LESS than 3 factors... Looking on the separate power-squares, each of them has a diagonal where the product is formed with 1 factor from each row/column.
The reason this is the smallest, likely has to do with the center number.
This has to be:
- > 4 ( since you need to have 4 positive integers less than it in 4 of the of the other boxes
- have at least 2 prime factors (so that you can have four combinations that pass through the center)
6 fits the bill!
You could replace every number with 5
Numbers need to be distinct.
Don't know how to make boxes in comment, so... Row 1: AD--BE--CF Row 2: CE--AF--BD Row 3: BF--CD--AE Substitute any value for the alphabet, the product will be ABCDEF.
The idea is that say that you have: 35 6 9 5 34 11 10 10 30 The number is 50 (Sorry about the diagonals, plz don’t judge me for that, or the line thing, it was a square when I typed it(if there’s an error at all))
There have been many squares that sum to a constant (50 in your case)... but the problem was asking about MULTIPLYING the numbers to get to a constant product.
Also, you have the number 10 twice in there. All the numbers need to be distinct. :-B
And saying "sorry about the diagonals" doesn't cancel the fact that diagonals not having the same sum WILL make your square not a perfectly magic one, even if it did have all numbers distinct.
The integers need to be distinct.
Ah, my mistake
Assuming 0 is a possible number, out it in the center of the box for a consistent product with any other number.
The problem then becomes the outer rows and columns... 🤔
Just fill each field with the same number. It works with any number.
Convert the multiplication problem into a summation problem bu using the numbers in the given solution as exponents to any fixed integer base where |base| > 1.
That would work, except that the integers all need to be different.
If the digits don't have to be 1 - 9, they could all be identically 1 or 0, so the products are 1 or 0. Found two cases in half a second so I bet clever people more interested could easily find more.
The question specified distinct integers.
You can use the original pattern to create an infinite number of different possibilities. Just add a random constant K to each square and the results of the rows, columns and diagonals will now be 15 + 3K
Since, multiplying two numbers together is like adding exponents, you can take the above square, and replace n with 2 n in each case, giving: