The following problem was posed in the March 2025 issue of Crux Mathematicorum.
Solution to MA311 by Srikanth, Mudhitha Maths Academy, Chennai.
Show Solution
a) Condition on the Center Cell
In the given square, the sum of numbers horizontally, vertically, and diagonally must be divisible by 9. So:
\[ A + E + I,\quad B + E + H,\quad C + E + G,\quad F + E + D \]
are all multiples of 9. Adding these gives:
\[ S = 3E + A + B + C + D + E + F + G + H + I \]
Since each number from 1 to 9 appears exactly once, the total sum is 45. Thus:
\[ S = 3E + 45 \]
So \( 3E \) must be a multiple of 9, which means \( E \) is a multiple of 3.
b) Example Grid:
\[ \begin{bmatrix} 1 & 5 & 3 \\ 8 & 6 & 4 \\ 9 & 7 & 2 \\ \end{bmatrix} \]
Remarks
We’ve shown a stronger claim: If the four main lines (horizontal, vertical, diagonals) each sum to a multiple of 9, then the center value must be divisible by 3.
However, note that this doesn’t guarantee a full magic square. For example:
\[ \begin{bmatrix} 1 & 9 & 7 \\ 2 & 3 & 4 \\ 8 & 6 & 5 \\ \end{bmatrix} \]
satisfies the key lines but not all row and column sums.
Authors:
Srikanth is the instructor at Mudhitha Maths Academy, Chennai. He is very fond of problem solving and puzzle games. He is an assistant professor of mathematics by the week, enjoys designing software and loves teaching mathematics. Srikanth likes to play role playing or strategy video games, enjoys reading fantasy novels and collect fountain pens. His favorite subjects are geometry and category theory.