Copyright © University of Cambridge. All rights reserved.

## 'Multiplication Arithmagons' printed from http://nrich.maths.org/

Alexander from Wilson's School described
his method for finding the missing numbers:
There is a method to solve multiplication arithmagons that works
every time. You first multiply all of the numbers in the red
bordered boxes together, and then find the square root. This will
find the product of all the numbers in the purple bordered boxes.
This is because all of the numbers in the purple bordered boxes are
multiplied together at some point, but they are multiplied two
times. So you find the square root.

To find the numbers in the purple bordered boxes you divide one of
the red-bordered boxes from the number that you got in the previous
operation. This will find the number in the purple bordered box
that is not connected to the red one. You can then find the rest of
the numbers by dividing.

Brandon and Fran from Coombe Dean School,
Alex and Rhys from Llandovery College, and William
from Barnton Community Primary School found the same
method.
Fionn from Thomas Hardye School explained
how to find the numbers by solving simultaneous
equations:
Let $x$, $y$ and $z$ be the vertex numbers and $A$, $B$ and $C$ be
the edge numbers.

$A$, $B$, and $C$ are the product of $xy$, $xz$ and $yz$
respectively.

$A=xy, B=xz, C=yz$.

Then $y=\frac{A}{x}$ and $z=\frac{B}{x}$

We can rearrange to get everything in terms of $X$:

$yz=\frac{AB}{x^2}$

$\Rightarrow x^2yz=AB$

$\Rightarrow Cx^2=AB$

$\Rightarrow x^2=\frac{AB}{C}$

$\Rightarrow x=\sqrt{\frac{AB}{C}}$

From this, we can easily work out:

$y=\sqrt{\frac{AC}{B}}$

and

$z=\sqrt{\frac{BC}{A}}$

Francesco from Lecce in Italy worked in a
similar way to Fionn but also pointed out that the values could
each be multiplied by -1 to give two possible solutions to each
arithmagon.
Oliver from Wilson's School pointed
out:
It's quite obvious that the product of all of the edges is equal to
all of the vertices multiplied together then squared as the edges
are equivalent to $xy, yz, xz$: $xy\times yz\times xz=x^2y^2z^2$
and $\sqrt{x^2y^2z^2}=xyz$

Morgan from Llandovery College sent
us this solution which
considered what happens when you scale the numbers on the
edges.
Finally, Niharika from Leicester High
School for Girls sent us this
solution, and Rajeev from Haberdashers'
Aske's Boys' School sent us this
solution.