4 Dom

179 /www/nrich/html/content/00/09/letme2/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Use these four dominoes to make a square that has the same number of dots on each side. <mdo:image src="lmt2-1.gif" alt=""></mdo:image> <mdo:image src="lmt2-2.gif" alt=""></mdo:image> </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Well done to all of you who found a solution to this challenge. Chloe and Mia from Scotts Primary School both managed to solve this challenge whilst in competition with Mrs Briggs, the teaching assistant. They both finished really quickly apparently and were really pleased with themselves (quite rightly!). Their teacher, Miss Lilley, and the rest of the class were very proud of them! Zareah from St Joseph's Convent gave us some good general advice: You just need to even it out. Don't put all the larger numbers in one corner and all the smaller numbers in another corner. Put a small number with a large number and soon you'll get it out! Milo and Darren from SVS used a similar method to Zareah. Angus from West Linton Primary used this strategy for solving the problem: First I added up the dots ($5+1+6+4+2+6+4+3=31$) then divided it by $4$ to find the average dots on a domino. There are one and a half dominoes in a side. $31\div4 =8$ (roughly) $8\times 1\frac{1}{2}= 12$ Then I fitted $12$ domino spots into each side. What a good idea, Angus. This estimation really helped you! Here is Angus' picture of the solution: <mdo:image height="252" width="276" src="AngusSol.gif" alt=""></mdo:image> Lisa, who didn't give her school, described a very convincing way of working out the solution: <div style="float: left;"><mdo:image height="135" width="191" alt="" src="Lisa1.gif"></mdo:image></div> This is the lowest side total possible using the domino with the highest total: $6+4+1=11$. The sides can therefore not add up to less then $11$. However, for the $5+1$ domino to be part of a side totalling $11$, there would need to be another $5$. Because there is not another $5$ the lowest possible total of the sides is $12$. <div style="float: left;"><mdo:image height="197" width="130" alt="" src="Lisa2.gif"></mdo:image></div> This is the highest total possible using the domino with the lowest total: $5+1+6=12$. Combining this with the information we already have, we know the sides must add up to $12$. It is now simply the case of working out the number needed to make $12$. <div style="float: left;"><mdo:image height="191" width="192" alt="" src="Lisa3.gif"></mdo:image></div> $6+4+2=12$ There is only one $2$, so this domino must go here. $6+2+4=12$ The remaining domino must therefore be placed this way round. $3+4+5=12$ If the $5+1$ domino was the other way round it would need to be rotated to make the final side add up. Very well explained, Lisa. We can see that this is the same solution as Angus', just rotated. A pupil from Wootton Upper School had a slightly different method: Part 1: How many dots on each side? The arrangement shows that each side of the square is composed of the sum of both sides of a single domino plus one of the sides of another domino. The sum of the dots on each single domino is as follows: ($5$,$1$) = $6$, ($6$,$4$) = $10$, ($4$,$3$) = $7$ and ($6$,$2$) = $8$. Therefore, the smallest sum possible is $8$ i.e. $6$ (from $5$,$1$) + $2$ (from $6$,$2$) whereas the largest sum possible is $16$ i.e. $10$ (from $6$,$4$) + $6$ (from $5$,$1$). She then conjectured that, since the four sides of the square must have equal numbers of dots, a compromise of the extreme values is needed so each side could total $12$ dots. Part 2: Arrangement The ($5$,$1$) domino needs a $6$ from one side of another domino since $6 + 6 = 12$ Similarly, ($6$,$4$) needs a $2$ as $10 + 2 = 12$ ($4$,$3$) needs a $5$ as $7 + 5 = 12$ ($6$,$2$) needs as $4$ as $8 + 4 = 12$ For $6 + 6$, the $6$ can be obtained from ($6$,$4$) or ($6$,$2$) For $10 + 2$, the $2$ can only be obtained from ($6$,$2$) For $7 + 5$, the $5$ can only be obtained from ($5$,$1$) For $8 + 4$, the $4$ can be obtained from ($6$,$4$) or ($4$,$3$) Looking back at the square, it should be acknowledged that the dominoes are connected from head to tail like a chain. This is what we need to do with our numbers - arrange them so they form a nice chain. For the above reason, we use ($6$,$4$) to obtain the $6$, and ($4$,$3$) to obtain the $4$ so that each domino is linked up consistently and they all get their required number of dots to total $12$. The result obtained should be: ($6$,$4$) - ($2$,$6$) - ($4$,$3$) - ($5$,$1$) - ($6$,$4$) etc. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div class="embed"> <h2>4 Dom</h2> Use these four dominoes to make a square that has the same number of dots on each side. <mdo:image alt="" src="lmt2-1.gif"></mdo:image> <mdo:image alt="" src="lmt2-2.gif"></mdo:image> </div> <h3>Why do this problem?</h3> Dominoes are such a versatile resource. The discussion of the thinking involved in solving <a href="http://nrich.maths.org/public/viewer.php?obj_id=179&part=index">this tricky problem</a>, and the use of appropriate vocabulary, is a good reason to do it. And there are tons of similar problems to be made up too. <h3>Possible approach</h3> <div>If you are using an IWB you may wish to use the <a href="http://nrich.maths.org/public/viewer.php?obj_id=6361&part=">Dominoes Environment</a>.</div> The question needs little introduction, but you may wish to check that everyone understands it by displaying the four dominoes and placing them into a square (as in the problem) but in incorrect positions. Ask, "How many dots are there altogether? How many dots are there on each side?". Confirm that each side adds up to a different number and then set the challenge. You may wish to suggest that if they solve the problem they should keep quiet so that they don't spoil it for everyone else - can they work on making up a similar problem using different dominoes from the set? Draw the children back together and spend some time discussing the way that they started and what strategies they used. Most children will use trial and improvement and may strike lucky quite quickly. Listen for statements which give an indication that some logical thinking is going on - perhaps in terms of the biggest number a side could add to, or the smallest, or some acknowledgement of the number of odds and evens which would affect the totals. <h3>Key questions</h3> How many dots are there altogether? How many dots could there be on each side? <h3>Possible extension</h3> <div>As suggested above, children could use other dominoes from the set to make up a similar puzzle. What other rules could they try? What if each side had to total to an even number? What sort of dominoes would they need? What if each side had to add to an odd number?</div> <div>You could collect the children's own puzzles and collate them into a book for the rest of the class to use.</div> <h3>Possible support</h3> Children who are struggling can be asked to place the dominoes in any position and record the sums for each side. How many different totals can they find? Which is the biggest? Why? Which is the smallest? Why? Handouts for teachers are available here (<a href="/content/00/09/letme2/4%20Dom%20Notes.doc">word document</a>, <a href="/content/00/09/letme2/4%20Dom%20Notes.pdf">pdf document</a>), with the problem on one side and the notes on the other. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">What is the biggest number a side could total to? And the smallest? </mdoxml> 2 5 1 0 0 0 0 4 Dom Use these four dominoes to make a square that has the same number of dots on each side. Calculations and Numerical Methods Addition & subtraction Using, Applying and Reasoning about Mathematics Trial and improvement Mathematics Tools Dominoes Admin Lower primary mapping document