H6. The figure shows seven circles joined by three straight lines.
The numbers $9,12,18,24,36,48$ and 96 are to be placed into the circles, one in each, so that the product of the three numbers on each of the three lines is the same.
Which of the numbers could go in the centre?
Solution
Since we are interested in the products of various numbers, it seems natural to find the prime factorisations of the seven numbers we are given:
Let the six numbers around the edge of the figure be $a, b, c, d, e$ and $f$, and the central number $x$ (as in the diagram below).
Then we need $a \times x \times d=b \times x \times e=c \times x \times f$. Since $x \neq 0$, we can divide by $x$, giving $a d=b e=c f$.
Now consider the number $a b c d e f$. This must be a cube (since it is equal to, for example, $\left.(a d)^{3}\right)$. This number can also be written as $\frac{a b c d e f x}{x}$, which is useful since $a b c d e f x$ is just the product of the seven numbers we have been given. Hence $a b c d e f=\frac{2^{17} \times 3^{10}}{x}$, and, if we write $x$ as $2^{m} \times 3^{n}$, we have abcdef $=2^{17-m} \times 3^{10-n}$.
Note that, since $x$ must be one of the given numbers, there are only a few possible values for $m$ and $n$; in particular $m \leqslant 5$ and $n \leqslant 2$.
Since abcdef is a cube, $2^{17-m}$ and $3^{10-n}$ must both be cubes too, which means that both $17-m$ and $10-n$ must be multiples of 3 . Hence $m$ could be equal to 2 or 5 , and $n$ must be equal to 1 .
Hence $(m, n)=(2,1)$ which gives $x=12$, or $(m, n)=(5,1)$ which gives $x=96$. As mentioned in the comment above, we must now check that each of these cases enables the shape to be filled as required by the question:
When $x=12$, we have abcdef $=2^{15} \times 3^{9}$, so each pair of outside numbers must multiply to $2^{5} \times 3^{3}$.

When $x=96$, we have abcdef $=2^{12} \times 3^{9}$, so each pair of outside numbers must multiply to $2^{4} \times 3^{3}$.
In each case, a small amount of experimentation quickly leads to arrangements that work in each case, such as:
Hence both cases are possible and there are two numbers that could go in the centre space; namely 12 and 96 .