# This Pied Piper of Hamelin

"The Pied Piper of Hamelin'' is a story you may have heard or read. This man, who is often dressed in very bright colours, drives the many rats out of town by his pipe playing - and the children follow his tune.

Suppose that there were $100$ children and $100$ rats. Supposing they all have the usual number of legs, there will be $600$ legs in the town belonging to people and rats.

But now, what if you were only told that there were $600$ legs belonging to people and rats but you did not know how many children/rats there were?

The challenge is to * investigate how many children/rats there could be if the number of legs was $600$.* To start you off, it is not too hard to see that you could have $100$ children and $100$ rats;

*you could have had $250$ children and $25$ rats. See what other numbers you can come up with.*

**or**Remember that you have to have $600$ legs altogether and rats will have $4$ legs and children will have $2$ legs.

When it's time to have a look at all the results that you have got and see what things you notice you might write something like this:

a) $100$ Children and $100$ Rats - the same number of both,

b) $150$ Children and $75$ Rats - twice as many children as rats,

c) $250$ Children and $25$ Rats - ten times as many children as rats.

This seems as if it could be worth looking at more deeply. I guess there are other things which will "pop up'', to explore.

Then there is the chance to think about the usual question, "I wonder what would happen if ...?''

How are you keeping track of what you have done?

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 $5”¨$ pupils from St Ambrose Catholic Primary School said;

”¨

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:

Children | Rats | Legs |

$300$ | $0$ | $600 + 0$ |

$298$ | $1$ | $596+ 4$ |

$296$ | $2$ | $592 + 8$ |

$294$ | $3$ | $588 + 12$ |

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.

**Why do this problem?**

### Possible approach

### Key questions

How many legs do your rats have?

### Possible extension

Setting different target numbers of legs offers the chance to explore multiples of 2 and 4 and how they are related. Each target number will have a range of possible solutions. Encourage the children to generalise about how the numbers of rats and people are related.

Another avenue for extension woul be to look at animals with other numbers of legs and perhaps three types of different-legged animals at the same time - eg. birds, spiders and pigs. This option links with Noah.

### Possible support

Some children may find the large numbers being considered in the presentation of the problem too high to make sense of so start them off with lower targets such as 20 or 30 legs. Noah is a similar problem involving fewer legs. Some toys or pictures representing the different animals may help some pupils to get started. Modelling clay bodies with straw legs can also be very helpful. Children could be given 20 lengths of straw and work on sharing them between people and rats as a way in to dealing with the larger numbers in a more abstract way.