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This magic square question was given to my brothers sixth grader:

-------------------
| 122 |     | 126 |
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| 129 |     |     |
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|     |     |     |
-------------------

This is literally all the information that was given to the kids. So it's not a typical 1-9 magic square, but I don't see how we can guess what range it is. And without knowing the sum, I honestly haven't got the idea how to solve this. If I were to assume that the range of values is 121-129 it's solvable, but is there some reason why we should assume that that I'm not aware of?

Davor
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2 Answers2

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Let $x$ be the element in the lower left. Then the magic sum $S$ is $251+x$. But the $S$ is also $(3/2)(126+x)$ because the central element of a $3×3$ square must be the average of any two elements that are $180°$ apart. So

$251+x=(3/2)(126+x)=189+(3/2)x$

$x=124$

$S=251+124=(3/2)(126+124)=375$

The rest should then follow smoothly.

Oscar Lanzi
  • 39,403
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Let the magic square be represented as $\begin{pmatrix}122 & a&126 \\ 129 &b &c \\ d& e & f\end{pmatrix}$.

From $122+129 + d = 126 + b + d$ we get $b=125$. We find this by searching for a row/column/diagonal were we have a lot of information already. Knowing $b$ we can continue $122+b+f=126+c+f \Longrightarrow 122+b=126+c \Longrightarrow c = 121 $. Know we also now the sum of all colums/rows/diagonals via $129+b+c = 375$. And now everything follows immediately: \begin{align} &a = 375-122-126 = 127 \\ &d = 375-122-129 = 124 \\ &f = 375-126-121 = 128 \\ &e = 375-124-128 = 123 \end{align} And the final result is: \begin{align} \begin{pmatrix} 122 & 127 & 126 \\ 129 & 125 & 121 \\ 124 & 123 & 128 \end{pmatrix} \end{align}

It is even solvable for 6th graders to train solving linear equations but I think it is not an easy question.

includeCMath
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