This Pied Piper of Hamelin

8315 /www/nrich/html/content/id/8315/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div style="text-align: center;"> <mdo:image src="icon2.png"></mdo:image></div> "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; or you could have had $250$ children and $25$ rats. See what other numbers you can come up with. 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 put the usual question "I wonder what would happen if ...?'' </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">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. 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: <hr></hr> <table style="width: 500px;border-spacing:1px;" border="1"> <tbody> <tr> <td>Children</td> <td>Rats</td> <td>Legs</td> </tr> <tr> <td>$300$</td> <td>$0$</td> <td>$600 + 0$</td> </tr> <tr> <td>$298$</td> <td>$1$</td> <td>$596+ 4$</td> </tr> <tr> <td>$296$</td> <td>$2$</td> <td>$592 + 8$</td> </tr> <tr> <td>$294$</td> <td>$3$</td> <td>$588 + 12$</td> </tr> </tbody> </table> 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. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div class="embed"> <h2>This Pied Piper of Hamelin</h2> <div style="text-align: center;"> <mdo:image alt="image" height="196" src="images.jpg" width="177"></mdo:image></div> "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; or you could have had $250$ children and $25$ rats. See what other numbers you can come up with. 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 $25$ rates - 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 put the usual question "I wonder what would happen if ...?'' </div> <h3>Why do this problem?</h3> <div>This <a href="http://nrich.maths.org/public/viewer.php?obj_id=1996&part=">activity,</a> based on the well-known story, opens the door to opportunities for doing mathematical calculations that can be explored with or without a spreadsheet. The story scenario is motivating and gives the children a meaningful context in which to make sense of these calculations. It can be extended by allowing pupils to create further questions to answer.</div> <h3>Possible approach</h3> <div>Reading a version of The Pied Piper of Hamlin with the children so that they are familiar with the story before starting this investigation is a good way to start.</div> <div>Then you could use the story to talk about the number of legs at particular times. You could also pose some theoretical questions, such as asking the children to imagine you've opened the book at a page which had 10 legs on it in total. How many people and how many rats could there have been? Learners could work on this in pairs using mini-whiteboards and then you can talk about the possiblities as a whole group. This will lead into general conversations about the number of animals/people and how the number of each affects the other.</div> <div>You might also want to spend some time sharing ways of recording what the children are doing. Some might be drawing pictures or symbols for the rats/people, others might be recording numbers only. It is worth talking about the different ways and the advantages/disadvantages of each. You may find that after some discussion, a few children adopt a different way of recording to the one they started with.</div> <h3>Key questions</h3> <div>How many legs do your rats have?</div> <div>What could you replace a rat with?</div> <div>Can you tell me about the way you are working out so many possibilities?</div> <div>(And for the pupils who have gone much further)</div> <div>What have you noticed about all your results so far?</div> <div>Can you explain why . . . . . has happened?</div> <h3>Possible extension</h3> <div>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 <a href="http://nrich.maths.org/136">Noah</a>.</div> <h3>Possible support</h3> <div>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. <a href="http://nrich.maths.org/136">Noah</a> 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.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">If you had one fewer rat, what could you replace it with to keep the number of legs the same? How are you keeping track of what you have done? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Here is just a taste of the results with 720 qne 600 legs showing the ratios of rats to humans nad humans to rats. <div style="text-align: center;"><mdo:image alt="pic 1" height="423" src="Picture%201.jpg" width="372"></mdo:image></div> Lara, John-Anthony, Harry and Richard of Loretto Junior say: If you start with 100 rats and 100 humans and then do the following: Humans 98 and rats 101 Humans 96 and rats 102 And so on -- then you get a lot of combinations!! Yes - you're right, well done. That's a great way to start. So, everytime you have one more rat, you need to lose two humans so that the total number of legs stays at 600. Is it possible to have more than 100 humans then and fewer than 100 rats? We'd love to hear from you if you've investigated this problem further.Please don't worry that your solution is not "complete" - we'd like to hear about anything you have tried. Thanks to all who sent in solutions. Here are some from Oakley School and St. George's school. This is what Emily sent us; If you TAKE 1 rat you have to ADD 2 people E.G 2 people + 196 rats 98 people + 101 rats 4 people + 195 rats 96 people + 102 rats 6 people + 194 rats 8 people + 193 rats 10 people + 192 rats etc. Hope this helps! Ben, Hannah & Scott sent in these thoughts; We figured that if you add 1 rat and take 2 people. For example 108 rats and 84 people, 109 rats and 82 people and 110 rats and 80 people. Otis, Jason & Sophis sent their solutions for 600 legs; our solutions are: 298 people, 1 rat 298x2 = 596 legs 1 rat = 4 legs 298 people(596 legs) + 1 rat(4 legs) 150x2 = 300 legs 75x4 = 300 legs 150 people(300 legs) + 75 rats(300 legs) </mdoxml> 2 4 0 1 0 0 0 This Pied Piper of Hamelin This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether! Using, Applying and Reasoning about Mathematics Working systematically Admin Upper primary mapping document