# Strategies for solving Magic Squares

E11/38.

By an exhaustive process of elimination I can work this out as 39, but there must be a quicker strategy for solving these kind of questions. Advice please.

Hint: The sum of all $9$ numbers is $43+51+59+39+47+55+63+67+71 = 495$, so each row must sum to $495/3 = 165$.
EDIT: Alternatively, you can apply the linear transformation $x \to \dfrac{x-35}{4}$ to all of the numbers. Then, the numbers already filled in are $2$, $4$, and $6$ and the missing numbers are $1$, $3$, $5$, $7$, $8$, and $9$. So, if you know what a $3 \times 3$ magic square looks like, you can easily fill in the missing numbers. Finally, undo the transformation using $x \to 4x+35$.