Brush Loads

4911 /www/nrich/html/content/id/4911/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Well, you have a whole collection of cubes and we're going to paint the outside that's visible once we've put them together. A few simple rules: <ul> <li>the cubes must be together;</li> </ul> <ul> <li>they must not be toppling over.</li> </ul> We'll need to look at how much paint is needed. One Brush Load (a kind of unit that we'll use) will paint one square face. Well here are $5$ cubes: <mdo:image height="93" width="146" alt="n5cubesFlat" src="n5CubesFlat.jpg"></mdo:image> Counting the faces to be painted comes to $15$, so $15$ Brush Loads (remember we're only counting visible faces, so not those that are touching the surface where the cubes are placed). But of course we could have placed the $5$ cubes differently: <mdo:image height="129" width="143" alt="n5CubesOrd" src="N5CubesOrd.jpg"></mdo:image> Counting the faces to be painted we now have $17$, so $17$ Brush Loads. And, how about: <mdo:image height="248" width="84" alt="n5CubesUp" src="N5CubesUp.jpg"></mdo:image> Now we'll need $21$ Brush Loads. O.K. Now let's explore different numbers of cubes, say $6, 7, 8, 9,$ etc. Each time we count the number of square faces showing and so find the number of Brush Loads (BLs). The cubes - as in the case of $5$ - can be arranged in different ways, sometimes giving different BLs. So I'm setting the challenge of finding ways of arranging cubes so that first of all there are the least number of BLs needed and then the biggest number of BLs needed. Try first with $5$ cubes. How about $6, 7, 8$ and $9$ cubes? If you wish to go further, then increase the number of cubes. See if you can get all the numbers between the largest and the smallest e.g. if $15$ is the smallest number for $5$ cubes and $21$ is the largest, can you get arrangements that'll give $16, 17, 18, 19,$ and $20$ BLs? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Callum and Josh submitted some ways of making the largest number of strokes and the smallest number of strokes for 5, 6 and 7 cubes: <mdo:image height="500" width="500" src="Brush%20strokes%20callum%20and%20josh.gif" alt="Callum and Josh"></mdo:image> Lan explored the pattern for greater numbers of cubes, as did Rohaan from New Zealand. Lan describes the method used: From counting cubes, we find: With 5 cubes, the least number of BLs is 15, and the largest number of BLs is 21. With 6 cubes, the least number of BLs is 16, and the largest number of BLs is 25. With 7 cubes, the least number of BLs is 19, and the largest number of BLs is 29. With 8 cubes, the least number of BLs is 20, and the largest number of BLs is 33. With 9 cubes, the least number of BLs is 23, and the largest number of BLs is 37. We remark the pattern: With 5 cubes, the least number is 15, and the largest number is 21. With 6 cubes, the least number is 15+1 = 16, and the largest number is 21+4 = 25. With 7 cubes, the least number is 16+3 = 19, and the largest number is 25+4 = 29. With 8 cubes, the least number is 19+1 = 20, and the largest number is 29+4 = 33. With 9 cubes, the least number is 20+3 = 23, and the largest number is 33+4 = 37. This is interesting, isn't it? If we think about the least number of BLs, we are trying to make as close to a rectangular arrangement of cubes as possible. This helps us to see why we only add one face going from an odd to an even number of cubes, but three faces going from an even to an odd. However, with is a rectangular array always the best for minimising the number of brush loads? If you had 9 cubes, how about making a square arrangement? How many brush loads of paint would you need then? Rohaan then spotted that the largest number of strokes is always (4 x number of cubes) +1, and that the smallest number depends on whether the number of cubes is odd or even: Odd: (n-5) x2 +15 Even: (n-6) x2 + 16 Does this work for a square arrangement too? </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div class="embed"> <h2>Brush Loads</h2> Well, you have a whole collection of cubes and we're going to paint the outside that's visible once we've put them together. A few simple rules: <ul> <li>the cubes must be together;</li> </ul> <ul> <li>they must not be toppling over.</li> </ul> We'll need to look at how much paint is needed. One Brush Load (a kind of unit that we'll use) will paint one square face. Well here are $5$ cubes: <mdo:image alt="n5cubesFlat" height="93" src="n5CubesFlat.jpg" width="146"></mdo:image> Counting the faces to be painted comes to $15$, so $15$ Brush Loads (remember we're only counting visible faces, so not those that are touching the surface where the cubes are placed). But of course we could have placed the $5$ cubes differently: <mdo:image alt="n5CubesOrd" height="129" src="N5CubesOrd.jpg" width="143"></mdo:image> Counting the faces to be painted we now have $17$, so $17$ Brush Loads. And, how about: <mdo:image alt="n5CubesUp" height="248" src="N5CubesUp.jpg" width="84"></mdo:image> Now we'll need $21$ Brush Loads. O.K. Now let's explore different numbers of cubes, say $6, 7, 8, 9,$ etc. Each time we count the number of square faces showing and so find the number of Brush Loads (BLs). The cubes - as in the case of $5$ - can be arranged in different ways, sometimes giving different BLs. So I'm setting the challenge of finding ways of arranging cubes so that first of all there are the least number of BLs needed and then the biggest number of BLs needed. Try first with $5$ cubes. How about $6, 7, 8$ and $9$ cubes? If you wish to go further, then increase the number of cubes. See if you can get all the numbers between the largest and the smallest e.g. if $15$ is the smallest number for $5$ cubes and $21$ is the largest, can you get arrangements that'll give $16, 17, 18, 19,$ and $20$ BLs? </div> <h3>Why do this problem?</h3> <div>This <a href="http://nrich.maths.org/public/viewer.php?obj_id=4911&part=">activity</a> is a very practical as well as visualisation situation. It can help pupils' development of the concepts associated with area - surface area. It is an investigation that can be extended very far indeed and some pupils may succeed beyond your expectation.</div> <h3>Possible approach</h3> <div>It is worthwhile for pupils to work in pairs or threes and to have cubes in front of them so as to enable good discussions to take place. The youngsters usually find that keeping shapes flat produces smaller numbers and making the tallest towers produces the largest, but challenge them to explain why this is the case. Look out for children moving one cube about from one place to another - you can get into good quality conversations about how many square faces are being covered by the new position and how many new square faces are added.</div> <div>Some younger children find it hard to count the faces that are visible, and they recount some or miss some out - don't we all? The children have to invent ways of being more and more accurate. It's good to ask them about their ways of counting and you may very well get these kinds of replies: "I look at each cube in turn and count the faces that are visible" or "Well I go all around the edge then count the ones on top". So for example with these $8$ cubes:</div> <div><mdo:image alt="" height="128" src="8CubesFlat.jpg" width="264"></mdo:image></div> <div>they'd count "$3$ plus $3$ plus $3$ plus $3$ plus $8$". It then becomes very interesting that with $9$ cubes there will also be $4$ threes but adding $9$ at the end. Moving to $10$ cubes it would be $3 + 4 + 3 + 4 + 10$. The discussions can go on and on as the confidence grows.</div> <h3>Key questions</h3> <div>How are you counting the faces?</div> <div>Are you able to make a shape with more/fewer faces?</div> <div>Can you explain why or why not?</div> <h3>Possible extension</h3> <div>The more mature youngsters who have no difficulties in counting will probably move on to tabulating their results. You might want to use standard units (cm$^2$) with older children. If you use multilink, $1$BL might be equivalent to $4$ cm$^2$. Here is a spreadsheet of some findings related to this investigation that you may find useful. But, please don't just hand it out to the children!</div> <div style="text-align: center;"><mdo:image alt="spreadSA" height="553" src="SpreadSA.jpg" width="509"></mdo:image></div> <div style="text-align: center;"> </div> <div style="text-align: left;">You might like to read the article <a href="http://nrich.maths.org/public/viewer.php?obj_id=4910&part=index">Back to the Practical?</a> which uses this problem as an example.</div> <h3>Possible support</h3> <div>Probably the best way to support learners who are struggling is to make sure they have a supply of cubes and encourage them to discuss what they are doing.</div> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> You could use Multilink cubes to try out your ideas - but this means that the cubes can't be "staggered". Remember, once the shape is made there is no picking it up to count "underneath" as these faces would not be painted! </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Looking at smallest numbers of BL's for different numbers of cubes we'll probably have something like: <mdo:image width="146" height="93" src="n5CubesFlat.jpg" alt=""></mdo:image><mdo:image width="166" height="97" src="N6CubesFlat.jpg" alt=""></mdo:image><mdo:image width="215" height="118" src="N7CubesFlat.jpg" alt=""></mdo:image> <mdo:image width="203" height="109" src="N8CubesFlat.jpg" alt=""></mdo:image><mdo:image width="212" height="121" src="N9CubesFlat.jpg" alt=""></mdo:image><mdo:image width="253" height="116" src="n10CubesFlat.jpg" alt=""></mdo:image><mdo:image width="239" height="120" src="n11CubesFlat.jpg" alt=""></mdo:image><mdo:image width="258" height="124" src="n12CubesFlat.jpg" alt=""></mdo:image> So for 5, 6, 7, 8, 9, 10, 11, 12 cubes the smallest numbers are 15, 16, 19, 20, 21, 24, 25, 26. Looking at the largest numbers we use towers like: <mdo:image width="147" height="282" alt="" src="5CubeUpCol.jpg"></mdo:image> Which some of the children see as 4 times the number of cubes plus 1. So for 5, 6, 7, 8, 9, 10, 11, 12 cubes the largest numbers are 21, 25, 29, 33, 37, 41, 45, 49. </mdoxml> 2 3 0 1 0 0 0 Brush Loads How can you arrange the 5 cubes so that you need the smallest number of Brush Loads of paint to cover them? Try with other numbers of cubes as well. Measures and Mensuration Surface and surface area Using, Applying and Reasoning about Mathematics Visualising Using, Applying and Reasoning about Mathematics Practical Activity Using, Applying and Reasoning about Mathematics Investigations Using, Applying and Reasoning about Mathematics Working systematically Mathematics Tools Interlocking cubes Admin Upper primary mapping document