We had just a few submissions to this challenge.

Brodie & Tully from  St Patrick's School Macksville NSW in Australia sent in the following;

Highest

(4-3+5)x6=36

Lowest

6/3+4-5=1

(4-3+5)/6

Highest-1, 2, 3, 4, 5, 6

(3+4)/(2-1)x5x6=210

Lowest

(4-3)x2/1+(5-6)=1

M, T and G (Y3) from  Monteney Primary School, Sheffield wrote;

M, T and G say that to make the greatest possible total you need to "multiply all the numbers together but not one"  Multiplying all of the numbers 2, 3, 4, 5 and 6 together generate a total of 720.

2 x 3 = 6

6 x 4  = 24

24 x 5 = 120

120 x 6 = 720

If you add one to your total you get a total of 721.

720 + 1 = 721

M, T and G said that you add one to make the biggest total, because if you multiplied your total by one then your answer of 720 wouldn't change.

Ashkan from Gorsefield Primary wrote;

The highest I found was:

3 - 4 + (5 x 6) = 29.

I did 5 x 6 which = 30. Then I did 3 - 4 which is -1. Finally I did -1 + 30 = 29.

The lowest I found was:

3 - 4 + (5 / 6) = 0.167

I did 5 / 6 = 0.833. Then I did 3 - 4 = -1. Finally I did -1 + 0.833 = 0.167.

Kestrel class at Churchfields, the Village School in Wiltshire, wrote:

We decided that we had to keep the numbers in order.

At first we could use the operations as many times as we wanted.

Our highest score was 360

3 x 4 x 5 x 6 = 360 (I wonder why you didn't use the 2 as well...)

Our lowest score was -117.

3 – (4 x 5 x 6) = -117

At first we experimented with division before we realised that negative numbers were lower.

Then we decided to see what would happen if we could only use each symbol once (but we kept the numbers in order).

Our highest score was 34.5.

(3 ÷ 4 + 5) x 6 = 34.5

Our lowest score was -51.

3 – ((4 + 5) x 6) = - 51

Thank you, Kestrel class. What a good idea to decide on particular 'rules', like having the numbers in order, or using each operation once.

Thank you for all these different solutions. You may like to think about the ideas in each solution and whether you can improve on them.

We had a good number of solutions sent in for this challenge, here are a few to think about.

Jack from Whangaparaoa School in  New Zealand

3 x 1 = 3   52 - 3   = 49 divided by 7 = 7

3 x 8 = 24  52 - 24  = 28 divided by 7 = 4

3 x 15 = 45 52 - 45  = 7 divided by 7 = 1

Shivoham from Robin Hood Junior School sent in the following:

 

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From Hamilton South Public in Australia we had a few submitted solutions:

Henry:

Yes there are different ways to make 52, what I did was I had 7 Zepts and 1 Zios which makes 52 legs. But you can also make 52 legs with 4 x 7 Zepts = 28 legs and 3 x 8 = 24 so 28 + 24 = 52. So this shows there is more than 1 way to answer this question.

Caylum:

52 Legs would make 4 Zepts and 8 Zios

Finn:

In solving this problem of zepts and zois, there can be numerous answers to this problem. Some of the solutions include- 1 lot of zois and 7 lots of zeptsI worked this out by the  olution- 52/7= 7 with 3 left over  there for on of the solutions is 1 lot of zois and 7 lots of  Zepts. Another solution I came up with was- 15 lots of zois and 1 lot of zepts  I worked this solution out by swapping the numbers around from solution one- 52/3=15 with 7 left over there for another one of the solutions is 15 lots of zois and 1 lot of zepts. As you can see there are more than one answer I have demonstrated only one though.

Matilda.

I know my 3 times table found that 3 times 15 = 45 and 45 is 7 less that 52.

7 Times 7 is 49 which is 3 less than 52.

Jett :

I added up the legs by doing the Zepts legs and the Zios legs at the same time so

7+3=10 10+3+7=20 20+7=3=30 30+7+3=40 40+3+3+3+3=52

My other way was 7+7+7+7+7+7+7=49,  49+3=52

First answer 8 Zios 4 Zepts.

Second answer 1 Zios 7 Zepts. So there are 2 answers to the Zios and Zepts question.

Advika from Wimbledon Chase Primary School in England

There are 4 Zepts and 8 Zios

At first I tried to do it in ratio but it didn't work.I would write out the timetables of three and seven and then find out numbers that give a sum of fifty-two. Ratio does not work because (seven is to three )-the ratio- are not equal in one portion.My words may be rather. You could also approximate fifty two and use times tables to help you work

the main answer out. I could not find out any corresponding ways to get fifty two and would be grateful to get another answer that doesn't match mine. Below,I have uploaded a file that you may want to consider seeing as it should help you if you get stuck.I hope it is of help to you. Please consider my working out as I could not think of any other way to

come across the answer and would be happy to come across the method if

there is any. I hope this has come to use of you if any use. 

 

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As a Ratio it can't work 7:3 is the ratio for Zepts to Zios. When you add them together it equals to 10 and 52 divided by 10 equals (5.2).Th is method does not work because each portion does not equal the same .This is because of the fact that Zepts equal to seven and Zios equals to three.

Amaan from Christ Church School England wrote:

7x4=28  4 Zepts - 3x8=24   8 Zios

I got this answer by trying to go up systematically in the 7 times table. First of I tried finding out how many Zios there were.Firstly,I took away 14 away from 52 because of the description in the clue. The answer to this sum was 38.From the answer of 38 I tried dividing 38 from 3.But there was no full answer. Next I tried 21,52-21=31. So once again my answer was not in the 3 times table. The next number in the 7 times table was 28. So once again took took my answer 28 away from 52.52-28=24.I then found out 8x3=24.So that was the answer to how many Zios there were and because 28 is the answer to 7x4 that told me the answer to how many Zepts there were.

Anaya from GEMS Willington Academy-AlKhail in the UAE:

As mentioned one zios has 3 legs and one zepts has 7 legs and  the scientist found total of 52 legs. We need to find the number of zios and zepts in the planet.

HERE IS THE SOLUTION

1 Zios    =   3 legs

8 zios  have   =  3 x 8 = 24 legs

1 zepts  = 7 legs

4 zepts have =  7x4=28 legs

So total  no of legs =24+28=52

So the answer is There are 8 NUMBER of zios and 4 NUMBER of zepts in the planet.

Great! You have worked hard on getting solutions and there are some good explanations as to how you went about finding solutions. 

 

We had a good number of solutions sent in, thank you.  Here we will feature those of you who've looked at many possibilities. From George at Linton Heights Junior School”¨ we had the following:

a) $0$ rats and $300$ children

b) $1$ rat and $298$ children

c) $5$ rats and $290$ children

d) $10$ rats and $280$ children

e) $25$ rats and $250$ children

f) $50$ rats and $200$ children

g) $100$ rats and $100$ children

h) $125$ rats and $50$ children

i) $150$ rats and $0$ children 

From Patrick at Manorcroft Primary School we had this good explanation, well done:”¨

There are $148$ different combinations of child and rat because I figured out that you could replace $1$ rat with $2$ children (because one rat has twice as many legs as a child) so the maximum possible children to rats are $298$ children to $1$ rat and the maximum possible rats to children is $149$ rats to $2$ children. If you take $1$ from $149$ to get $148$ possiblities.

Year $5Ӭ$ pupils from St Ambrose Catholic Primary School said;

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There are many possible answers to this question, to find out how many children, you can start with the number of rats. You can go from $1$ rat to $149$ rats to work out how many children.  The rule is: multiply by $4$, take away from $600$, divide by $2$. 

For example, $1$ rat= $298$ children because $1$x$4=4 600-4=596, 596$ divided by $2=298$.

If we start with the number of children the rule is: multiply by $2$, take away from $600$, divide by $4$.

For example, $2$ children $ = 149$ rats because $2$x$2=4$, $600-4= 596, 596$ divided by $4= 149$.

There are some patterns that we noticed, such as: If you take $1$ away from the amount of rats, it adds $2$ to the amount of children. For example: $86$ children, $107$ rats; $106$ rats, $88$ children.

Matthew from Calcot Junior School also discovered this general rule. He said:

There is a formula which is however many rats you have, double that number and take it off 300. Your result is how many children there are.

The total number of legs is 600. Each child has 2 legs, each rat has 4. So the total number of legs can be written as:

Legs = Children x 2 + Rats x 4

That's very helpful, Matthew. By 'children', I think Matthew means the number of children, and 'rats' means the number of rats.

Because we know the number of legs, we can write:

600 = Children x 2 + Rats x 4

Then if we know either children or rats, we can use this to work out the missing one. For example, if there are 100 Children:

600 = 100 x 2 + Rats x 4

600 = 200 + Rats x 4

400 = Rats x 4

Rats = 100

So Rats = (Legs - 2 x Children) $\div$ 4

If you know legs and rats, to work out children:

Children = (Legs - 4xRats) $\div$ 2

This is extremely well explained, thank you Matthew. Matthew calculated all the possible answers, which you can see here. Thank you for sharing this with us, Matthew.

James”¨, who calls his school  "BG"”¨ sent in the following good explanation:

When working out the pattern for children/rats, I thought it would be a good idea to start with either everything as children or everything as rats. I decided to start with everything as children, which would be $300$ children and $0$ rats, since $600$ halved is $300$.

Then, I knew if I wanted to get all possible solutions, I'd need to come up with a pattern. My pattern was adding some of the children's legs, to the rats each time. But of course, since rats have twice as many legs as children, that wouldn't work, so I took away two children for every rat I added, as shown in the pattern below:

I did that on and on, until I was certain, that every time, the legs would total up to $600$. By looking at the number of children, and noticing that it started at $300$, and got $2$ fewer each time, I divided $300$ by $2$, and then added on the $1$ possibilty, with $0$ children, to work out there would be $151$ possible solutions for this.

These solutions are really good. Well done, keep submitting solutions to other activities.

Thank you for the large number of interesting solutions that came in. Of course your answers depended upon what you decided was 'allowed'.  For the first part, getting an answer of $12$, we had this variety of ideas:

 

From Year $3$ and $4$ Mathematics Extension Programme at Lumen Christi Catholic Primary School in Australia we had this very good solution:

To work out the solution to the problem we first tried using the four operations ($+, -, \times, \div$). This helped us find some of the solutions.

We then realised we would need to think about the order of operations to get different solutions. So we used our knowledge of brackets and BODMAS to find other solutions.

Finally we thought about other mathematical operators we could use to help us find more solutions.

We worked out that $4^2 = 16$; $4^3 = 64$ and $√4 = 2$ (square root $4$).

This information helped us find the remaining solutions except the answer.

Below are the solutions for the Year $3$ and Year $4$ Mathematics Extension Programs. Each group of students worked separately on the tasks, but have found some similar solutions.

Our goal is now to find all the solutions from $0 - 100$.

 

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Rosie Goodleigh C of E Primary SchoolӬ sent in the following which was accompanied by similar solutions from Paige, George, Tom, Eva and Jessica.

$0$. First I added $4+4$ which made $8$. Then $-4$ which made $4$ and then$-4$ again which uses four $4$'s and makes $0$

$1$. First I did $4$ divided by $4$ which equals $1$ so then $+4$ and then $-4$ which makes $1$.

$2$. In brackets do $4$ divided by $4$, do this once more, both of them should

equal $1$ each,add both 1s together which equal $2$.

$3$. First do $4+4+4=12$. Subsequently divide $12$ by $4$ which equals $3$.

$4$. First take $4$ away from $4=0$. $0$x$4=0$,then add $4$ which equals $4$.

$5$. First times $4$ by $4 =16$ $16 + 4 =20$. Divide $20$ by $4=5$.

$6$.First add $4$ and $4 =8$. Divide $8$ by $4$ equals $2$. $2 + 4=6$.

$7$. In brackets divide $4$ by $4=1$. Then add $4$ and $4=8$ $- 1 =7$.

$8$. Add $4$ and $4$ x $4=32$. Divide $32$ by $4= 8$.

$9$. Divide $4$ by $4=1$. Add $4$ and $4=8$ $+ 1 = 9$.

$10$. In brackets there is the square root of $4$ which is $2$. Add $4$ and $4$ and $4

=12$ $-2=10$.

$12$. In brackets do $4$ divided by $4=1$. $4 - 1 = 3$. $3$x$4=12$.

$15$. In brackets do $4$ divided by $4 =1$. Do $4$ times $4$ then minus the $1$ to

equal $15$.

$16$. Add $4$ and $4$ and $4$ and $4 =16$.

$17$. In brackets divide $4$ by $4 =1$. Times $4$ by $4$ then add $1 =17$.

Sam from The Lantern Primary School sent in a really good alternative set of answers as follows.

The first ones I did I found quite easy. I used the four simple operations. 

$1=(4 \div 4) \times (4 \div 4)$

$12=(44+4) \div 4$

$3=(4+4+4) \div 4$

$6=((4+4) \div 4)+4$

$7=(44 \div 4) −4$

$8=(4 \times 4) −4−4$

$9=(4 \div 4)+4+4$

$15=(4 \times 4) − (4 \div 4)$

$16 = 4+4+4+4$

$17 =(4 \times 4)+(4 \div 4)$

Then I came to a problem.  I couldn't do 4, 5 and 10.  Then I realised that I had only used the simple operations, addition, subtraction, multiplication and division.  Then I remembered about square root, factorial and powers.  An example of factorial is:

$4!=4$×$3$×$2$×$1$

  

The first square root I tried was 4, then I realised I needed a few more.  Taking these into account, I was able to do the last few problems.

$4 = \sqrt(4+4+4+4)$

$5 = \sqrt(4!+(4 \div 4)^4)$

$10 = \sqrt(4!+(4 \div 4) \times 4)$

When I checked through my answers, I realised I had forgotten to put brackets in, so I added them in in the appropriate places to make the solutions clearer.

Thank you all for your excellent ideas which I think used so much of your mathematical knowledge and skills.