PK 2016 Q13

Since each of the nine numbers appears exactly once in the three rows, the product of each of the three rows is equal to the product of all nine numbers, that is
1 × 2 × 4 × 5 × 10 × 20 × 25 × 50 × 100. This equals
\(1 \times 2 \times 2^2 \times 5 \times 2 \times 5 \times 2^2 \times 5 \times 5^2 \times 2 \times 5^2 \times 2^2 \times 5^2 = 2^9 \times 5^9\).
Since the products of each of the three rows are equal, these products are equal to \(\sqrt[3]{2^9 \times 5^9} = 2^3 \times 5^3\). So the ‘magic product’ is 1000.
By considering the top row, we can see that the top right cell must contain \(1000 \div (20 \times 1) = 50\).
The only other way to get 1000 as the product of 1 and two of the remaining numbers is 1 × 10 × 100. So 10 and 100 must fill the two spaces below 1. We cannot put the 100 in the centre since the products of the numbers on each diagonal would then be too large. So 10 goes in the centre and 100 below it. From the diagonal we see that the bottom right entry is 5 and so the middle number on the right is 4.
Note: It may interest readers to know that in a 3 × 3 multiplicative magic square the centre number is always the cube root of the magic total.