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We had a very large number of solutions sent in showing various ways of approaching the problem.

Ramesh sent in the following solution:

Natalie and Emma wrote the following:

Our target was $52$ and this was our strategy: We guessed, checked and improved. We took a number from the $3$x table for the zios and we took a number from the $7$x table for the zepts and we put them together and saw what number it came to. If it was too high we lowered a number and if it was too low we highered a number.

Once we lowered it because our first answer was too high we managed to get $3\times8=24$ and $4\times7=28$ and our answer was $8$ zios and $4$ zepts and that was our solution.

Tenisha explained it this way:

When Nico went to the planet of Vuv and saw $52$ legs put up by the zios and zepts I figured out that that meant there were $4$ zepts and $8$ zios. I figured that because $7\times4=28$ and $8\times3=24$ and together they equal $52$. Also,there is no way you can have the product $52$ with different factors to answer this question correctly. This is my conclusion.

Jack, Jack, Eleanor, Rose, Adam, Gregor, Adela and Emily from Brennands Endowed Primary School got together and wrote:

We used multilink to help us see the problem.

We counted out $52$ pieces of multilink and put them into groups of $3$ (Zios)

This left $1$ piece over.

We used $2\times3$ zios and the one left over to make $1$ Zept ($7$)

We then used the groups of $3$ to make more Zepts until we ended up with

$4$ Zepts ($4\times7 = 28$)

$8$ Zios ($8\times3 = 24$)

$28 + 24 = 52$ legs

Quite a number of you used 'guess, check and improve'.

Megan and Sarah wrote this account:

To start our solution to this problem we would like to inform you we chose to use the theory, guess, check and improve. To start our workings out we worked on the number of Zios, we first tried $2$ then we worked our way up until we came to $8$ Zios which we calculated was the correct amount for reasons that will be divulged later in our answer.To mathematically calculate the number of Zepts our first calculation was to do $52-24$ which equals $28$ remaining. Our next sum was to do $28\div7$ which equals $4$. So using these numbers we were finally able to conclude that there were $8$ Zios and $4$ Zept.

So thank you for taking the time to read our answer.

Rowena explained her working and drew a table:

To start with I guessed. I went for $10$ Zios, which had $3\times10 = 30$ legs. I then counted up in $7$s. $7$, $14$, $21$ but found that none gave the right answer when I added them to $30$.

I tried $3\times10 + 3\times7 = 30 + 21 = 51$. No good!

I thought about times tables and wrote out the $3$ times table up to $51$ (no point in going any higher than that). This would give me the number of Zios. In the next column I took that number away from $52$ and was hoping for a multiple of $7$ which would give me the number of Zepts. Three answers gave multiples of $7$! (See below.)

The possible answers were $3 + 49$ ($1$ Zio and $7$ Zepts), $24 + 28$ ($8$ Zios and $4$ Zepts) and $45 + 7$ ($15$ Zios and $1$ Zept).

Then I looked back at the question - there had to be more than one of each kind of creature, so the only possible answer is:

$8$ Zios and $4$ Zepts