Break it up!

2284 /www/nrich/html/content/id/2284/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> You have a stick of $7$ interlocking cubes. You cannot change the order of the cubes. <mdo:image width="260" height="48" align="top" bgcolor="" src="7-stick.gif" alt="stick of seven differently coloured cubes"></mdo:image> You break off a bit of it leaving it in two pieces. Here are $3$ of the ways in which you can do it: <mdo:image width="288" height="49" align="top" bgcolor="" src="2-5-stick.gif" alt="2 interlocking cubes and 5 interlocking cubes"></mdo:image> <mdo:image width="288" height="57" align="top" bgcolor="" src="3-4-stick.gif" alt="3 interlocking cubes and 4 interlocking cubes"></mdo:image> <mdo:image width="288" height="50" align="top" bgcolor="" src="5-2-stick.gif" alt="5 interlocking cubes and 2 interlocking cubes"></mdo:image> In how many different ways can it be done? Now try with a stick of $8$ cubes and a stick of $6$ cubes: <mdo:image width="288" height="50" align="top" bgcolor="" src="8-stick.gif" alt="eight interlocking cubes of different colours"></mdo:image> <mdo:image width="221" height="49" align="top" bgcolor="" src="6-stick.gif" alt="six interlocking cubes of different colours"></mdo:image> Make a table of your results like this: <table border="0" style="border-style: solid; border-width: 0pt; width: 80%;"> <tbody> <tr> <td valign="top" style="border-style: solid; width: 50%;"> Number of cubes </td> <td valign="top" style="border-style: solid; width: 50%;"> Number of ways </td> </tr> <tr> <td valign="top" style="border-style: solid; width: 50%;"> $6$ cubes </td> <td valign="top" style="border-style: solid; width: 50%;"> ? </td> </tr> <tr> <td valign="top" style="border-style: solid; width: 50%;"> $7$ cubes </td> <td valign="top" style="border-style: solid; width: 50%;"> ? </td> </tr> <tr> <td valign="top" style="border-style: solid; width: 50%;"> $8$ cubes </td> <td valign="top" style="border-style: solid; width: 50%;"> ? </td> </tr> </tbody> </table> Now predict how many ways there will be with $5$ cubes. Were you right? How many ways with $20$ cubes? $50$ cubes? $100$ cubes? ANY number of cubes? * * * * * * * * * * * * * * * * * * * * If all the cubes are the same colour, a split of $4$ and $2$ will look the same as a split of $2$ and $4$. How many ways are there of splitting $6$ cubes now? Can you predict how may ways there will be with any number of cubes? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Uygor from FMV Erenkoy Isik School in Turkey sent us a very clearly reasoned solution. He says: I can break the first stick (that consists of $7$ blocks) as: $1$ block + $6$ blocks $2$ blocks + $5$ blocks $3$ blocks + $4$ blocks $4$ blocks + $3$ blocks $5$ blocks + $2$ blocks $6$ blocks + $1$ block I found $6$ different styles for a $7$-block stick. Then I tried a $6$-block stick. I found the following breaking styles: $1$ block + $5$ blocks $2$ blocks + $4$ blocks $3$ blocks + $3$ blocks $4$ blocks + $2$ blocks $5$ blocks + $1$ block So I found $5$ different styles for a $6$-block stick. I predict I can find $7$ styles for an $8$-block stick. The number of ways is $1$ less than the number of blocks of the stick. So for $5$ cubes, I should find $4$ ways by doing it like this: $1$ block + $4$ blocks $2$ blocks + $3$ blocks $3$ blocks + $2$ blocks $4$ blocks + $1$ block I saw that my prediction is right! For $20$ cubes, I can find $19$ ways For $50$ cubes, I can find $49$ ways For $100$ cubes, I can find $99$ ways On the other hand, if the cubes are the same colour: For $6$ cubes, we have $3$ ways For $7$ cubes, we have $3$ ways For $8$ cubes , we have $4$ ways For $9$ cubes, we have $4$ ways ... In this case, we can find the number of ways by dividing the number of cubes by $2$ then we only look at the quotient. (By this Uygor means to ignore any remainder.) This is very well explained, thank you. Thomas from New York looked at the last part of the problem in a slightly different way. If n is the number of cubes then: To find out how many ways any number of cubes of the same color can be split, the formula is $(n-1)/2$ if $n$ is odd, and $n/2$ if $n$ is even. Well done both of you. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div class="embed"> <h2>Break it Up!</h2> You have a stick of $7$ interlocking cubes. You cannot change the order of the cubes. <mdo:image align="top" alt="stick of seven differently coloured cubes" bgcolor="" height="48" src="7-stick.gif" width="260"></mdo:image> You break off a bit of it leaving it in two pieces. Here are $3$ of the ways in which you can do it: <mdo:image align="top" alt="2 interlocking cubes and 5 interlocking cubes" bgcolor="" height="49" src="2-5-stick.gif" width="288"></mdo:image> <mdo:image align="top" alt="3 interlocking cubes and 4 interlocking cubes" bgcolor="" height="57" src="3-4-stick.gif" width="288"></mdo:image> <mdo:image align="top" alt="5 interlocking cubes and 2 interlocking cubes" bgcolor="" height="50" src="5-2-stick.gif" width="288"></mdo:image> In how many different ways can it be done? Now try with a stick of $8$ cubes and a stick of $6$ cubes: <mdo:image align="top" alt="eight interlocking cubes of different colours" bgcolor="" height="50" src="8-stick.gif" width="288"></mdo:image> <mdo:image align="top" alt="six interlocking cubes of different colours" bgcolor="" height="49" src="6-stick.gif" width="221"></mdo:image> Make a table of your results like this: <table style="border-width: 0pt; border-style: solid; width: 80%;" border=""> <tbody> <tr> <td style="border-style: solid; width: 50%;" valign="top"> Number of cubes </td> <td style="border-style: solid; width: 50%;" valign="top"> Number of ways </td> </tr> <tr> <td style="border-style: solid; width: 50%;" valign="top"> $6$ cubes </td> <td style="border-style: solid; width: 50%;" valign="top"> ? </td> </tr> <tr> <td style="border-style: solid; width: 50%;" valign="top"> $7$ cubes </td> <td style="border-style: solid; width: 50%;" valign="top"> ? </td> </tr> <tr> <td style="border-style: solid; width: 50%;" valign="top"> $8$ cubes </td> <td style="border-style: solid; width: 50%;" valign="top"> ? </td> </tr> </tbody> </table> Now predict how many ways there will be with $5$ cubes. Were you right? How many ways with $20$ cubes? $50$ cubes? $100$ cubes? ANY number of cubes? * * * * * * * * * * * * * * * * * * * * If all the cubes are the same colour, a split of $4$ and $2$ will look the same as a split of $2$ and $4$. How many ways are there of splitting $6$ cubes now? Can you predict how may ways there will be with any number of cubes? </div> <h3>Why do this problem?</h3> <a bitly="BITLY_PROCESSED" href="http://nrich.maths.org/2284&part=">This problem</a> makes a very good introduction to algebra, however, it also begins very simply so everyone can make a reasonable start. <h3>Possible approach</h3> <div>You could start by giving all the group some interlocking cubes and asking them to make a stick of seven. Then tell them to break their stick into two pieces and hold one piece in each hand. Ask how they have done this and write the various ways on the board and then enquire if there are any other ways that it could have been done. Is there a way that the results can be organised better? This may be the time to show the group how to make a simple table to record their results.</div> <div>Some learners may claim that $2 + 5$ is the same as $5 + 2$. "Not always!" is the answer to this. Of course, numerically they are identical, but the context is also important. A good example for when this is not so is this. If five people are sitting down to a meal and two more turn up, it is quite possible there will be enough food for them. If however, two are starting their meal it is very unlikely that there will be enough food for five more! In this case, is $5 + 2$ the same as $2 + 5$? Just because the answer is the same, it does not mean that the question is the same!</div> <div>After the introduction, learners could work in pairs on finding the number of ways that sticks of six, eight and nine cubes can be broken into two pieces. Then they should record their findings. Supply squared paper for those who wish to record using it.</div> <div>When learners feel ready to generalise they can go on to working out the number of ways with $20$, $50$ and $100$ cubes and then ANY number of cubes.</div> <div>At the end of the lesson all can come together to discuss their findings. They can be asked how they knew they had found all the ways of breaking a stick into two pieces. Those who know a good way of expressing "any number" can explain their reasoning.</div> <h3>Key questions</h3> <div>How are you going to record what you have done?</div> <div>How do you know you have found all the ways of breaking it into two pieces?</div> <div>If you break it into $0 + 7$, does this give you two pieces?</div> <div>How many ways do you think you can break $20/50/100$ cubes?</div> <div>Can you see a connection between the total number of cubes and the number of ways you can break the stick into two pieces?</div> <div>How could you express that "for any number"?</div> <h3>Possible extension</h3> <div>Learners who can generalise the first part of the problem could then go on to exploring the second part - if all the cubes are the same colour, looking at the difference between odd and even numbers.</div> <div>Then they could try a different problem in which generalisation is required such as <a bitly="BITLY_PROCESSED" href="http://nrich.maths.org/88&part=">Sticky Triangles</a>.</div> <h3>Possible support</h3> Suggest working with the interlocking cubes and recording on squared paper. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">How are you going to record what you have done? How do you know you have found all the ways of breaking it into two pieces? Can you see a connection between the total number of cubes and the number of ways you can break the stick into two pieces? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> First part: Number of ways of breaking the stick = n - 1 (where n is the number of cubes) Second part: Number of ways of breaking the stick = n/2 if n is even Number of ways of breaking the stick = (n-1)/2 if n is odd old notes 4 + 2 and 2 + 4 will look the same if all the cubes are the same colour. The last part of the questions asks you to investigate this. </mdoxml> 2 3 1 1 0 0 0 Break it up! In how many different ways can you break up a stick of 7 interlocking cubes? Now try with a stick of 8 cubes and a stick of 6 cubes. 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