I've found this problem in an older book which goes by the name of Logical aptitude circa 2019. It doesn't list any other markings. The thing is no matter how I attempt to look into it, I'm trapped in circles as I have no idea on how to approach this problem.

The problem is as follows:

The figure from below shows a set of different domino pieces. By using only 4 of these pieces you can make a 4x4 magic square. Find the sum of the dots situated in the diagonal of the magic square.

I've went to the source again and it doesn't give any other details or conditions for this problem. However from looking other similar problems which I had solved, the intended approach is that some squares will be left blank and its only allowed to use the pieces once.

Sketch of the problem

The alternatives given in the sheet are as follows:

  1. 6
  2. 5
  3. 4
  4. 8

Logically what I attempted to do was trial and error. But after several attempts I couldn't come to a method to make the requested magic square.

Recalling an older classes on this particular subject I remembered that Bachet de Méziriac's diamond method proposed a way to fill in the gaps of a magic square with the identity that the square in the center is the half of the sum of the two opposing vertex of the square.

By the way, to illustrate what I meant with the method. It goes as this:

Sketch of the method

As mentioned above it seems to work using an empty 3x3 table to make a magic square but that's it. How can this puzzle be solved?. It would help a lot if an answer would contain some sort of picture or a detailed step by step analysis on how to approach this sort of question in a systematic and logical manner so I can fill in the blanks without just plugin numbers randomly.

  • $\begingroup$ Can I assume: the magic square is a 3x3, dominoes can't overlap, you can't use the same domino $\endgroup$
    – Alto
    Oct 26, 2020 at 2:04
  • 3
    $\begingroup$ The original question is not quite clear: it doesn't specify what a "magic square" means in this context. How large is the grid? Should diagonal sums be equal to row sums and column sums (some puzzles exclude diagonals)? How are we supposed to place the dominoes (4 dominoes cannot form a full square of any size)? Definitely it isn't a standard magic square because we can't use 1-9 to fill a 3x3. $\endgroup$
    – Bubbler
    Oct 26, 2020 at 2:05
  • $\begingroup$ @Bubbler I'm very sorry. I forgot to include the statement that they want to make a $4 x 4$ magic square. I'm adding it right now. $\endgroup$ Oct 26, 2020 at 3:05
  • $\begingroup$ @Alto I'm sorry, I missed to include a statement from the puzzle, the intended magic square is by arranging the domino pieces to make a $4 \times 4$ magic square, however it is not possible to use the method which I proposed because it only works with $3 \times 3$ magic square. Hence I'm stuck. $\endgroup$ Oct 26, 2020 at 3:08
  • 2
    $\begingroup$ Using that interpretation, there is no solution: the maximum sum of four pieces shown is 3+3+4+5 = 15, but they must sum to at least 16 if the diagonal sum is to be at least 4 (assuming the diagonal sum is equal to the row or column sum). $\endgroup$
    – Bubbler
    Oct 26, 2020 at 3:50

2 Answers 2


This puzzle creator must have some very unusual definition of "magic square":

The diagonals can't possibly have the same sum as the rows and columns: if each column had 4 dots, that would come up to a total of 16 dots, and any 4-tile combination has 14 dots or fewer. So the common sum cannot be 4 or more, which covers all the answer options.

Turns out, by similar logic, that it's impossible to construct any other kind of square with the same sum on every row and column:

If the common sum were 3,

the 2-2 and 2-3 tiles couldn't be used, they would put too many dots in whichever row or column they are in. The rest of the dominoes don't have 12 dots among them.

If the common sum were 2,

again, we couldn't use any domino with more than 2 dots total, and the remaining dominoes would have fewer than 8 dots altogether.

If the sum were 1 (or 0),

There would be fewer than 4 dominoes we can use.

In conclusion, the puzzle is unsolvable under these rules.

  • $\begingroup$ Sorry!, I have to mention that the paragraph included below the question is my interpretation of what is the intended way to approach this question based on other problems, but it is not specifically indicated. There isn't any condition indicating that some pieces cannot be used more than once.Yes I agree there's an unusual definition of a magic square. If you read my question, I know there's a method to fill out a 3 times 3 magic square but there isn't one for a 4 times 4 square which I could remember (maybe there is) but given this conditions I don't think it will be applicable. $\endgroup$ Oct 27, 2020 at 2:20
  • $\begingroup$ If we assume that some pieces may be used more than once can this problem be solved. Can you include an answer given this scenario? and what sort of logic should be used to reach an answer by following that condition?. Can you help me with that?. $\endgroup$ Oct 27, 2020 at 2:21
  • $\begingroup$ @ChrisSteinbeckBell there are still some "moving parts", do we have to use each of the four tiles at least once? Or maybe we should use each exactly twice? The way I would approach this is to start with some common sum. This gives the total number of dots in the pattern, which helps with choosing a possible tile combination. As an oversimplified example, if I chose "1" as the common sum, my tiles could only be 4 x 0-0 and 4 x 0-1, and from those tiles it's easy to check if a magic square can be made. $\endgroup$
    – Bass
    Oct 27, 2020 at 6:25

From Bass's answer it seems we have to interpret the question differently. Maybe in this puzzle you can use only 4 pieces but you can use them multiple times.

This would allow for the following magic square.

enter image description here

I don't know if there are other solutions.

If the aim is a proper magic square where we don't repeat values in a row or column and make the diagonals sums equal the row and column sums, then the diagonal sum must be 6 regardless of how and if you can build it with these dominoes. It is more a question about basic properties of magic squares.

So I don't know whether this is the intended solution. At least it is a magic square.

  • $\begingroup$ Interesting approach. As I mentioned there isn't a statement which indicated to use all the pieces. It seems that in order to solve this you may use one some of them which would yield a magic square. However what steps have you used to get that magic square?. Was it just trial and error?. What I attempted to do at first was to sum all dots. By doing this I obtained $20$ and this can be split in two halves of 10 and $10$ I assumed that the sum of the diagonal to be $5$. But it isn't the case it was $6$. $\endgroup$ Oct 27, 2020 at 2:17
  • $\begingroup$ Again Can you include which sorts of steps you used to get that combination?. Was it just trial and error?. Or does it exist some sort of thing which I should attempt to began to look for first?. $\endgroup$ Oct 27, 2020 at 2:18
  • $\begingroup$ What I did is I first created a magic square with values 0 to 3. Then I split it into dominoes. My first version used only vertical tiles, resulting in only 2 tiles ([0,1] and [2,3]) being used 4 times each. So I replaced some vertical pairs by horizontals. Then it required a [1,3] tile which is not available, so I switched the 0's and 1's. So yes, trial and error. $\endgroup$
    – Florian F
    Oct 27, 2020 at 9:03

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