Zios and Zepts

1005 /www/nrich/html/content/00/06/penta2/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> On the planet Vuv there are two sorts of creatures. The Zios have $3$ legs and the Zepts have $7$ legs. <mdo:image alt="" height="247" src="Zio.gif" width="285"></mdo:image> The great planetary explorer Nico, who first discovered the planet, saw a crowd of Zios and Zepts. He managed to see that there was more than one of each kind of creature before they saw him. Suddenly they all rolled over onto their backs and put their legs in the air. He counted $52$ legs. How many Zios and how many Zepts were there? Do you think there are any different answers? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div class="embed"> <h2>Zios and Zepts</h2> On the planet Vuv there are two sorts of creatures. The Zios have $3$ legs and the Zepts have $7$ legs. <mdo:image alt="" height="247" src="Zio.gif" width="285"></mdo:image> The great planetary explorer Nico, who first discovered the planet, saw a crowd of Zios and Zepts. He managed to see that there was more than one of each kind of creature before they saw him. Suddenly they all rolled over onto their backs and put their legs in the air. He counted $52$ legs. How many Zios and how many Zepts were there? Do you think there are any different answers to this that woud work? </div> <h3>Why do this problem?</h3> This problem focuses on numbers that are multiples of $3$ and $7$ and offers a motivating context in which to develop instant recall of them. It also provides an opportunity for learners to discuss alternative strategies and to consider the merits of each. <h3>Possible approach</h3> <div>Children enjoy considering strange forms of life in imaginary planets and these weird creatures with odd numbers of legs should appeal to them. You could begin by telling the story in more detail and introducing a simpler version of the problem, for example, if Nico saw $6$ legs what creatures might he have seen? And how about $14$ legs? Or $13$ legs? Ask them to work in pairs, perhaps on mini-whiteboards, to think about these first challenges. As they work, observe different ways of recording and thinking about the problem which you can share with the whole group.</div> <div>After this learners could work in pairs on the actual problem which you could present orally with the image. Again, listening to learners as they work can be an interesting insight to their thinking. Some children may like to use practical resources to help them with their thinking such as modelling clay and headless matches.</div> <div>At the end of the session learners could share not only their solutions but also their methods. Some may have used a trial and improvement approach, either with the materials provided or using pictures; some may have written lists of multiples of $3$ and $7$ in a very systematic way and then made totals. Children could consider whether some strategies were more effective than others, although of course a strategy that is effective for one child may be very confusing for another. Having the freedom to approach this problem in any way is key here, but in talking to others, some pupils might change the way they work, which is interesting in itself.</div> <h3>Key questions</h3> <div>What have you tried so far?</div> <div>What happens if you replace this seven-legged creature with some three-legged creatures?</div> <div>What happens if you replace these three-legged creature with some seven-legged creatures?</div> <h3>Possible extension</h3> Learners could investigate other possible numbers of legs of a group of Zios and Zepts. Perhaps some can be done in more than one way. How many ways can be found for each target number? The investigations can go on and on! <h3>Possible support</h3> Starting with lower totals and using practical equipment to support them will help some children. They may like to write down all the multiples of $3$ and $7$ as an aid. Some children might benefit from trying the <a href="http://nrich.maths.org/public/viewer.php?obj_id=182&part=index">Spiders and Flies</a> problem first which involves lower totals. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Drawing some pictures or using sticks/counters for legs might help. Do you have any idea approximately how many Zios and Zepts there might be to make $52$ legs? You could try choosing a certain number of Zios and a certain number of Zepts, then count their legs. Do you need more legs or fewer to have a total of $52$? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Well everyone agreed that there must be 8 Zios and 4 Zepts, but not everyone worked it out the same way! Daniel from Anglo-Chinese School (Primary) drew a table: <table > <tbody> <tr> <th>Zios(3)</th> <th>Zepts(7)</th> <th>Legs</th> <th><mdo:image alt="" src="cross.gif"></mdo:image> or <mdo:image alt="" src="tick.gif"></mdo:image> or</th> </tr> <tr> <td>2</td> <td>2</td> <td>2</td> <td>20 <mdo:image alt="" src="cross.gif"></mdo:image></td> </tr> <tr> <td>3</td> <td>3</td> <td>30</td> <td><mdo:image alt="" src="cross.gif"></mdo:image></td> </tr> <tr> <td>10</td> <td>3</td> <td>51</td> <td><mdo:image alt="" src="cross.gif"></mdo:image></td> </tr> <tr> <td>8</td> <td>4</td> <td>52</td> <td><mdo:image alt="" src="tick.gif"></mdo:image></td> </tr> </tbody> </table> Here's what the children from Wesley College Prahan Prep School, Melbourne, Australia did: Kev : I got that answer by 7x 4 + 3 x 8 = 52, the total number of legs he saw. Dougall : I started counting backwards from 52 and I counted the Zios first and then I counted the Zepts second. Nick : I drew circles with 3 legs and circles with 7 legs Mark and Stuart from Cummersdale Lower Juniors used calculation and tables facts to help them work it out: "I wrote down both times tables and added the numbers together and got, 4 Zepts and 8 Zios", and "2 zios and 2 zepts times 4 is 40 add 4 zios is 52, 4 zept and 8 zios". This "using the times tables method" was explained in detail by Clementine and Laura from The Mount School, York, who began with listing multiples of Zios (3) then checking the number of Zepts (7) needed to make the total of 52 legs. Charlotte, Sarah and Kate from The Mount School, York, and Christina from Malborough Primary started the other way around - first the multiples of Zepts (7) then the Zios (3). Emma and Eleanor also from The Mount School, York, and Elizabeth from Stamford High School, Lincolnshire realised that the number of Zio legs (3) plus the number of Zept legs (7) make 10 legs. So the worked with pairs of Zios and Zepts - that is, multiples of 10. <blockquote> <pre> 3 + 7 = 10 10 x 4 = 40 and 52 - 40 = 12, which is 3 x 4 </pre></blockquote> so there are 4 Zepts and 8 Zios Lucy and Sarah from Stamford High School, Lincolnshire each found a way of starting with the total number of legs (52), then taking it apart into numbers that were multiples of 3 or 7. They thought about division, rather than multiplication. Sarah says: "I found this by taking away 3 and then dividing by 7 and if it divides into a whole number that was the answer. If it did not divide into a whole number I took away 6 and so on until I found the answer. Then I took the number of legs of the Zepts away from 52 and divided it to find the number of Zios." Lucy says: "You have to find two numbers which add up to 52 and one has to divide by three and the other by seven. <blockquote> <pre> 28 + 24 = 52 28 Ã?· 7 = 4 24 Ã?· 3 = 8 </pre></blockquote> We had a very large number of solutions sent in showing various ways of approaching the problem. Ramesh sent in the following solution: Ramesh 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 Rowena</mdoxml> 2 3 0 1 0 0 0 Zios and Zepts On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there? Using, Applying and Reasoning about Mathematics Trial and improvement Numbers and the Number System Factors and multiples Calculations and Numerical Methods Multiplication & division Calculations and Numerical Methods Addition & subtraction Admin Upper primary mapping document