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# Multiplication Arithmagons

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.

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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.

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