## Multiplying Magic Square

#### Problem

By carefully placing the numbers 1, 2, 4, 8, 16, 32, 64, 128, and 256 in a $3 \times 3$ square grid show how it is possible to ensure that the product of each row, column, and diagonal can give the same value.

#### Solution

By considering the numbers as powers of 2: $1 = 2^0, 2 = 2^1, 4 = 2^2, ... 256 = 2^8,$, it is possible to complete this problem like a standard adding 3x3 Magic Square; instead of solving for the digits 1 through 9 we could solve for the digits 0 through 8, which would correspond to the exponents. However, we shall tackle this in a slightly different way...

Although we do not know the value of each cell we do know that each of the values $2^0, 2^1, ... , 2^8$ will be assigned in some order. If we let the "magic product" for each row, column, and diagonal be $P$ then it can be seen that multiplying the product of each row is equivalent to $P^{ 3}$, but it will also be equivalent to multiplying every value in the grid.

$\therefore 2^0 \times 2^1 \times ... \times 2^8 = 2^{36} = P^{ 3} \Rightarrow P = 2^{12} ( = 4096)$

If we now consider the three values that will be found along the same row, column, diagonal as the largest value: $P = 2^8 \times 2^a \times 2^b \Rightarrow a + b = 4$. So we deduce that $a + b$ can be $0 + 4$ or $1 + 3$, but it cannot be $2 + 2$, as $a$ and $b$ must be different values. Because $2^8$ belongs to a triplet which has two solutions it must be placed in one of the edge squares (a corner would have three solutions and the centre square would have four solutions).

Now we consider the smallest value in the same way: $P = 2^0 \times 2^c \times 2^d \Rightarrow c + d = 12$. Therefore $c + d$ can be $8 + 4$ or $7 + 5$ which means that is must also be placed on an edge square, and it will be found opposite $2^8$.

So without loss of generality we can fill in the values for $2^8, 2^1, 2^3, 2^4$ and $2^0$ (values in black in the diagram below). The red values can be completed by first ensuring that the sum of the indices in the two diagonals add to twelve, then the first and last columns can be completed.

Hence we obtain the following unique solution (considering rotations and reflections to be equivalent).

By considering different sets of nine distinct positive integers prove that $P = 216$ is a minimum and find a solution.

Problem ID: 374 (16 Aug 2010)     Difficulty: 3 Star

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