Counting Cogs

6966 /www/nrich/html/content/id/6966/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This problem has been designed to be worked on in a group. For more details about how you might go about doing this, please read the <a href="http://nrich.maths.org/6966&part=note">Teachers' Notes</a>. Here are nine different cogs: <mdo:image width="500" height="412" src="cogs.gif" alt=""></mdo:image> If you coloured one tooth on each cog, which pairs of cogs let the coloured tooth go into every 'gap' on the other cog? Which pairs do not let this happen? Why? Can you explain how to determine which pairs will work, and why? You could cut out the cogs from <a href="/content/id/6966/CogsToPrint.pdf">these sheets</a> or you could use the <a href="http://nrich.maths.org/4775&part=">cogs interactive environment</a> to try out your ideas. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This was a tricky problem but we received some detailed solutions - well done. Frank, Noah and Ben from Stanford Junior School said: If the number of teeth on one cog is the multiple of teeth on the other cog it will not work. So if the teeth on one cog is 5 and on the other cog is ten teeth it will not work. Also the number you times the teeth to make the number of teeth on the other cog that is the number of spaces it goes into. cog $1$: $5$ teeth cog $2$: $10$ teeth spaces gone into = $2$ $5 \times 2 =10$ Sean, Ben, Lane and Jonathan from Meridian Primary School told us about some of the things they noticed: First of all, all consecutive numbers will work because the smaller will go round once but won't make it to the start point by one so eventually it will make it the whole way round (e.g. $4$, $5$). Two even numbers will never ever go together because they are multiples of two so they will keep going in a rut. If two odd numbers are factors of two consecutive numbers they will work because they'll still act as consecutive numbers would e.g. $5$ and $7$: $5 \times 3 = 15$ and $7 \times 2 = 14$ $11$ works with all numbers. Pranjal from Garden International School, Kuala Lumpur took these ideas a bit further and wrote: There are three different ways of determining which pairs of cogs work and which don't: 1. Find consecutive numbers from $4$ to $12$ and pair them up. The solutions that are in this category are as follows: $4$-$5$, $5$-$6$, $6$-$7$, $7$-$8$, $8$-$9$, $9$-$10$, $10$-$11$ and $11$-$12$ 2. Make pairs of odd numbers. The pairs are: $5$-$7$, $5$-$9$, $5$-$11$, $7$-$9$, $7$-$11$ and $9$-$11$ 3. The last, and the category that most numbers fall into, is the HCF (highest common factor) category. This category also includes the pairs in the previous category. The rule is that the numbers have to have a HCF of $1$. All the numbers that work are in this category too. ANSWER The pairs that go together are: $4$-$5$, $5$-$6$, $6$-$7$, $7$-$8$, $8$-$9$, $9$-$10$, $10$-$11$, $11$-$12$, $5$-$7$, $5$-$9$, $5$-$11$, $7$-$9$, $7$-$11$, $6$-$11$, $7$-$10$, $5$-$8$, $9$-$11$, $4$-$7$, $4$-$9$, $4$-$11$, $5$-$12$, $6$-$11$, $8$-$11$ Note: Even-Even numbers can never go together, no matter what the case. This is because their one factor is definitely $2$, if not the highest. Some particular odd-even numbers can't go together either. Examples are $5$-$10$ and $9$-$12$.The rules are that there cannot be a pair which consists of numbers which are both divisible by a certain number other than one or that one number is divisible by the other number in the pair. QUESTION We had to ask ourselves a question as we progressed in the solution. The question was that which pairs work and why, and is there a pattern between the pairs that work. Jong Woong, Jayme, Denise and Mariana, also from Garden International School, have reasoned through this problem very carefully. They told us how they worked to solve the problem. First, they said they wrote down all the pairs possible using the $4$, $5$, $6$, $7$, $8$, $10$, $11$ and $12$ cogs: ($4$, $5$) ($4$, $6$) ($4$, $7$) ($4$, $8$) ($4$, $10$) ($4$, $11$) ($4$, $12$) ($5$, $6$) ($5$, $7$) ($5$, $8$) ($5$, $10$) ($5$, $11$) ($5$, $12$) ($6$, $7$) ($6$, $8$) ($6$, $10$) ($6$, $11$) ($6$, $12$) ($7$, $8$) ($7$, $10$) ($7$, $11$) ($7$, $12$) ($8$, $10$) ($8$, $11$) ($8$, $12$) ($10$, $11$) ($10$, $12$) ($11$, $12$) Then they tried random cogs to see if they worked and tried to find a pattern among them. Pairs that work ($4$, $5$) ($4$, $7$) ($4$, $11$) ($5$, $6$) ($5$, $7$) ($5$, $11$) ($5$, $12$) ($6$, $7$) ($6$, $11$) ($7$, $8$) ($7$, $10$) ($7$, $11$) ($7$, $12$) ($8$, $11$) ($10$, $11$) ($11$, $12$) Pairs that don't work ($4$, $6$) ($4$, $8$) ($4$, $10$) ($4$, $12$) ($5$, $8$) ($5$, $10$) ($6$, $8$) ($6$, $10$) ($6$, $12$) ($8$, $10$) ($8$, $12$) ($10$, $12$) They asked themselves questions: Why does this happen? Is there a pattern? They then wrote: We found out that that the pairs that have a HCF (highest common factor) of $1$ work. Reason This only happens if the highest common factor is one, because the cogs get filled in by the highest common factor. For example, the pair ($4$, $6$) doesn't work because the HCM is $2$ which means that the space will be filled in by '$2$'. Well done to all of you who sent a solution to this problem. I wonder whether anyone else can find a way to explain why cogs will only work if they have a HCF of $1$? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div class="embed"> <h2>Counting Cogs</h2> This problem has been designed to be worked on in a group. For more details about how you might go about doing this, please read the <a href="http://nrich.maths.org/6966&part=note">Teachers' Notes</a>. Here are nine different cogs: <mdo:image alt="" height="412" src="cogs.gif" width="500"></mdo:image> If you coloured one tooth on each cog, which pairs of cogs let the coloured tooth go into every 'gap' on the other cog? Which pairs do not let this happen? Why? Can you explain how to determine which pairs will work, and why? You could cut out the cogs from <a href="/content/id/6966/CogsToPrint.pdf">these sheets</a> or you could use the <a href="http://nrich.maths.org/4775&part=">cogs interactive environment</a> to try out your ideas.</div> <h3>Why do this problem?</h3> <div><a href="http://nrich.maths.org/6966&part=">This problem</a> requires children to think about factors and multiples and, in particular, common factors, but it is not necessary for them to have met this term prior to having a go at the task. It offers opportunities for pupils to ask their own questions, find examples, make conjectures and begin to generalise.</div> <div>The problem lends itself to collaborative working, both for children who are inexperienced at working in a group and children who are used to working in this way. By working together on this problem, the task is shared and therefore becomes more manageable than if working alone.</div> Many NRICH tasks have been designed with group work in mind. <a href="http://nrich.maths.org/7011&part=">Here</a> we have gathered together a collection of short articles that outline the merits of collaborative work, together with examples of teachers' classroom practice. <h3>Possible approach</h3> <div>This is an ideal problem for learners to tackle in groups of four. Allocating these clear roles (<a href="/content/id/6966/Roles.doc">Word</a>, <a href="/content/id/6966/Roles.pdf">pdf</a>) can help the group to work in a purposeful way - success on this task could be measured by how effectively members of the group work together as well as by the solutions they reach.</div> <div>Click <a href="http://nrich.maths.org/7014&part=">here</a> to see a couple of video clips of two classes organised into groups to work on this task.</div> <div>Introduce the four group roles to the class. It may be appropriate, if this is the first time the class has worked in this way, to allocate particular roles to particular children. If the class works in roles over a series of lessons, it is desirable to make sure everyone experiences each role over time.</div> <div>For suggestions of team-building maths tasks for use with classes unfamiliar with group work, take a look at <a href="http://nrich.maths.org/6933&part=">this article</a> and the accompanying resources.</div> Give each group a copy of <a href="/content/id/6966/CogsGroupSheet.doc">this sheet</a>, which outlines the task. The idea is for them to read it together to find out what to do. Cut out a set of cogs for each group using <a href="/content/id/6966/CogsToPrint.pdf">this sheet</a> and give them out so each person in a group has two or three cogs. Alternatively, it might be possible for children to use <a href="http://nrich.maths.org/4775&part=">the interactivity</a> to try out their cogs. Children should begin by working individually, investigating several pairs of cogs, then they will pool their findings as a group so that they have worked on all combinations of cogs. Explain that each group will be expected to report back at the end of the session, showing the patterns they noticed, at least one conjecture they have and at least one question. Exploring the full potential of this task is likely to take more than one lesson, allowing time in each lesson for children to feed back ideas and share their thoughts and questions. Ask each group to record their reasoning, conjectures, explanations and any generalisations on a large sheet of paper (for example flipchart paper) in preparation for reporting back. There are many ways that groups can report back. Here are just a few suggestions: <ul> <li>Every group is given a couple of minutes to report back to the whole class. Learners can seek clarification and ask questions. After each presentation, children are invited to offer positive feedback. Finally, pupils can suggest how the group could have improved their work on the task.</li> <li>Everyone's posters are put on display at the front of the room, but only a couple of groups are selected to report back to the whole class. Feedback and suggestions can be given in the same way as above. Additionally, children from the groups which don't present can be invited to share at the end anything they did differently.</li> <li>Two children from each group move to join an adjacent group. The two "hosts" explain their findings to the two "visitors". The "visitors" act as critical friends, requiring clear mathematical explanations and justifications. The "visitors" then comment on anything they did differently in their own group.</li> </ul> <h3>Key questions</h3> <div>Which cogs have you found that work so far?</div> <div>Which pairs didn't work? Can you explain why?</div> <div>How could you predict whether a pair will work before you try them?</div> <div>What questions would you like to ask?</div> <h3>Possible extension</h3> <div>Children could begin to work on a question that they have, or a question posed by another group.</div> <h3>Possible support</h3> <div>By working in groups with clearly assigned roles we are encouraging students to take responsibility for ensuring that everyone understands before the group moves on.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Which pairs of cogs have you found that work? Which pairs didn't work? Do you get a sense of why? What do you notice about the numbers of teeth in each case? </mdoxml> 2 4 0 1 1 0 0 Counting Cogs Which pairs of cogs let the coloured tooth touch every tooth on the other cog? Which pairs do not let this happen? Why? Numbers and the Number System Factors and multiples Numbers and the Number System Common factors Using, Applying and Reasoning about Mathematics Group worthy Secondary Mapping Document Factors, multiples and primes Secondary Mapping Document DisplayCabinet Admin Upper primary mapping document