6274 /www/nrich/html/content/id/6274/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> Place each of the numbers $1$ to $5$ in the V shape below so that the two arms of the V have the same total.<br></br> <br></br> <div style="text-align: center;"><mdo:image alt="" height="193" src="V.gif" width="300"></mdo:image></div> <br></br> <br></br> How many different possibilities are there?<br></br> What do you notice about all the solutions you find?<br></br> <br></br> Can you explain what you see?<br></br> <br></br> Can you convince someone that you have all the solutions?<br></br> <br></br> What happens if we use the numbers from $2$ to $6$? From $12$ to $16$? From $37$ to $41$? From $103$ to $107$?<br></br> <br></br> What can you discover about a V that has arms of length $4$ using the numbers $1-7$?<br></br> <br></br> <br></br> <a href="http://nrich.maths.org/public/viewer.php?obj_id=6814&amp;part=">Click here for a poster of this problem</a>.<br></br> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p><span class="editorial">I am extremely impressed by the solutions sent in to the Magic Vs problem, so well done to you all. Tom from Crudgington Primary, Krishan from Buckingham College Prep School, Harriet and Laura from Shincliffe C of E Primary, Nicholas and Jared from Clifton Hill Primary, Andy from Garden International School, Jonathan from DGS, Lucas from Grinling Gibbons Primary and Scott from Echline Primary explained that there were three different Magic Vs for the numbers 1-5.</span></p> <p class="editorial">Emma and Charlotte from Greenacre School for Girls said:</p> The solution we found was that you always put an odd number at the bottom when it was 1-5.<br></br> There were three odd numbers and two even numbers. To make it balance you've got to get rid of an odd number so you put an odd number at the bottom, because then you have two odds and two evens.<br></br> You have to put the biggest remaining odd number with the smallest even number on one arm, and the smallest remaining odd number with the biggest even number on the other arm to make it work.<br></br> <br></br> <p><span class="editorial">This is very clearly explained, thank you. Yujin from Dubai International Academy explained this in a slightly different way:</span></p> <p>... if there are two even numbers and three odd numbers; you put the odd number at the bottom: even + even = odd + odd.</p> <p>You can't put the even at the bottom because even + odd does not equal to odd + odd.</p> <p><span class="editorial">That's right. Gemma, Dylan, Faazil and Gabriel from Dubai English Speaking School had a good way of working which helped them explain why odd numbers have to go at the bottom:</span></p> <div>We first started off by putting 1 in the bottom of the V and then adding up the remaining digits (2, 3, 4, 5) which came to 14. So, half of 14 is 7 and 2+5 and 3+4 make 7 so they go either side of the V.</div> <div>We then tried 2 and realised that the remaining digits (1, 3, 4, 5) added up to 13 which meant we couldn't 'even up' either side of the V.</div> <div>We then thought that only odd numbers could go at the bottom of the V.</div> <br></br> <p class="editorial">Bazahir from Ray Lodge had a good way of approaching the problem:</p> <div>I forgot about the base and found two sums that have the same total using the numbers 1-5 then I put the base number.</div> <br></br> <p class="editorial">That's a good idea, Bazahir.</p> <p><span class="editorial">So, the three solutions will be one with 1 at the bottom of the V, one with 3 at the bottom and one with 5 at the bottom:</span></p> <br></br> <br></br> <mdo:image height="94" width="500" alt="" src="sol1.gif"></mdo:image><br></br> <br></br> <p><span class="editorial">However, some of you decided that you can rearrange the numbers on the arms of the V, keeping the same number at the bottom, to make different solutions. Alex, Zara, Frankie, Akeel and Molly from Talbot Primary School, Yujin and Gijs from Dubai International Academy and James from Crudgington Primary argued that there are 24 Magic Vs in total for the numbers 1-5. Gijs took the example with the 3 at the bottom of the V. He wrote:</span></p> 5 add 1 is 6 and 4 add 2 is also 6. There are eight possibilities when there is a number at the bottom. The numbers on the sides can be switched around lots of different ways and they still make 6. The numbers have to stay together, though, you can't do 5 and 2 because that makes 7.<br></br> <br></br> <p><span class="editorial">Here are the eight different Magic Vs with 3 in the bottom:</span></p> <br></br> <br></br> <mdo:image height="218" width="581" src="sol2.gif" alt=""></mdo:image><br></br> <p class="editorial">Can you see how all these have been generated? Perhaps you can do the same with 1 at the bottom of the V to convince yourself. So, if we assume these are all different Magic Vs, there would indeed be 24 in total.</p> <p class="editorial"></p> <p class="editorial">Many of you then looked into Magic Vs which used different numbers and Magic Vs which were different sizes.Gemma, Dylan, Faazil and Gabriel from the Dubai English Speaking School did a very thorough investigation:</p> <br></br> We then tried 2, 3, 4, 5, 6 and put the odd number at the bottom and soon realised that the remaining digits added up to an odd number which meant we couldn't fill in the V evenly. So, we decided that when you start a five-digit sequence with an odd number, the number at the bottom of the V has to be an odd number, if you start a sequence with an even number then the number at the bottom of the V has to be even.<br></br> <br></br> Our teacher then told us to do the same with seven digits (1, 2, 3, 4, 5, 6, 7) and we soon realised that this time if you started off with an odd number at the bottom of the V the remaining digits added up to an odd number so it wouldn't work but, if we started off with an even number at the bottom of the V then we could even up both sides. So the solution for a seven-digit V was opposite to the one for a five-digit V.<br></br> <br></br> <p><span class="editorial">Very interesting! Did you try seven digits which started with an even number, I wonder? What would happen with 2, 3, 4, 5, 6, 7, 8, for example? They continued:</span></p> <div>Our homework was to find out how we could predict whether a V with 35 digits would have to have an odd or an even number at the bottom of the V to work. Gemma came up with most of the solution as she said that if you add up all the digits and the sum was an even number then it would need an even number at the bottom of the V to work. If the sum was odd then the number at the bottom of the V would have to be odd.</div> <br></br> <p class="editorial">Many of you I've already mentioned, and Milne from East Hoathly, agreed with what they have said about the numbers 2-6.</p> <div class="editorial">George sent in a general method for finding Magic Vs:</div> <br></br> <div>Add all numbers of the consecutive series - which need to be an odd number of numbers, whether there are five numbers, seven numbers.</div> <br></br> <div>Let the addition of these numbers equal A (AE for an EVEN number) (AO for an ODD number)</div> <br></br> <div>If the total answer is an even number, subtract one of the even numbers from that set of numbers. Let this remaining number be B.</div> <div>Else, if the total answer is an odd number, subtract an odd number from that set of numbers. This remaining number can also be called B since it to is an EVEN number. [The even or odd number that is initially subtracted will be the number at the bottom of the Magic V.]</div> <br></br> <div>Next divide B, the total left, by TWO. i.e. B/2</div> <div>From the EVEN quantity of numbers left to be considered after removing the initial EVEN or ODD number, take half of the numbers and determine whether it is possible to add them so that they will equal B/2.</div> <div>If half the quantity of number left does equal B/2, then the numbers left after adding these numbers together will also equal B/2. One set of these numbers that equal B/2, can fill the spaces on the left leading down to the bottom of the Magic V. The other numbers that equal B/2, can fill the spaces on the right leading down to the bottom of the Magic V.</div> <br></br> <div>After the first attempt try subtracting a different EVEN or ODD number from the total AE or AO and go through the sequence to see if any other combinations are possible.</div> <br></br> <p class="editorial">You can see a summary of George's results <a href="/content/id/6274/MagicVGeorge.xls">here</a> . Will it always be possible to create Magic Vs in this way, do you think? How do we know that we're going to be able to get each arm to add to the same total?</p> <p class="editorial">Ross and Brandon from Ashford Hill represented the five-digit Magic Vs alebraically which you may find helpful. They said:</p> <div>If n is the first number then the other numbers are n+1 n+2 n+3 n+4, the three solutions for every number are:</div> <br></br> <mdo:image height="116" width="550" alt="" src="sol3.gif"></mdo:image><br></br> <br></br> <p><span class="editorial">What a wealth of responses - thank you to everyone who sent something in and I'm sorry that we can't mention everyone. If you would like to contribute anything more to the solution, do</span> <a class="editorial" href="mailto:%20nrich@damtp.cam.ac.uk">email</a> <span class="editorial">us.</span></p> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><div class="embed"> <h2>Magic Vs</h2> <br></br> Place each of the numbers $1$ to $5$ in the V shape below so that the two arms of the V have the same total.<br></br> <br></br> <div style="text-align: center;"><mdo:image alt="" height="193" src="V.gif" width="300"></mdo:image></div> <br></br> <br></br> How many different possibilities are there?<br></br> What do you notice about all the solutions you find?<br></br> <br></br> Can you explain what you see?<br></br> <br></br> Can you convince someone that you have all the solutions?<br></br> <br></br> What happens if we use the numbers from $2$ to $6$? From $12$ to $16$? From $37$ to $41$? From $103$ to $107$?<br></br> <br></br> What can you discover about a V that has arms of length $4$ using the numbers $1-7$?<br></br> <br></br> <br></br> <a href="http://nrich.maths.org/public/viewer.php?obj_id=6814&amp;part=">Click here for a poster of this problem</a>.<br></br>  </div> <br></br> <h3><span style="font-weight: bold;">Why do this problem?</span></h3> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=6274&amp;part=index">This problem</a> gives opportunities for children to make conjectures, prove these conjectures and make generalisations. They will be practising addition and subtraction, and applying their knowledge of odd/even numbers.</div> <br></br> <h3>Possible approach</h3> <div>You can see Lynne McClure introducing Magic Vs to a small group of teachers on YouTube <a href="http://www.youtube.com/watch?v=-JrZcMbsNdA">here</a> and she takes the ideas a bit further in <a href="http://www.youtube.com/watch?v=3fx7A8WRfk8">this clip</a>.</div> <div> </div> <div>You could start by having two Vs displayed on the board (interactively if possible), one which is "magic" (i.e. whose arms have the same total) and one which is not. Ask the children to talk about what they see. If it doesn&#39;t come up naturally, draw their attention to the total of each arm and introduce the term "magic V".</div> <br></br> <div>Ask learners to suggest some questions they could ask about magic Vs and then direct their attention to finding out how many other magic Vs there are, using the numbers $1-5$. Children could work in pairs, using digit cards to try out their ideas. They could record magic Vs on <a href="/content/id/6274/Magic%20Vs.pdf">this sheet</a> . After a some time, bring them together to share some of the magic Vs they have found so far. You could record them on the board and then invite learners to comment on what they notice. Can they offer explanations?</div> <br></br> <div>Having given them more time to explore and make generalisations, you could allow learners to pursue one of the questions they asked at the start of the lesson. These might include, for example, investigating different ranges of numbers or Vs which have four numbers in each arm.</div> <br></br> <h3>Key questions</h3> <div>What do your magic Vs have in common?</div> <div>Can you explain why?</div> <div>What would happen if we used five different consecutive numbers?</div> <div>Can you explain why?</div> <br></br> <h3>Possible extension</h3> <div>Children can be challenged to investigate Vs of different sizes with different ranges of numbers. Are they always possible to solve and can they predict the number of solutions? Further extensions to this problem include other arrangements of the numbers, for example a magic cross, or the use of negative numbers, or how about a multiplication Magic V?</div> <br></br> <h3>Possible support</h3> <div>Using mini-whiteboards and digit cards may &#39;free up&#39; some children so that they don&#39;t worry about getting a magic V straight away.</div> <br></br> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> You could use <a href="/content/id/6274/MagicV.pdf">this sheet</a> to record your magic Vs.<br></br> Have a go at putting the numbers into the V and then adding up each arm. Perhaps you can adjust the arrangement slightly to make the totals the same?<br></br> What do you notice about the number at the base of the V in all your magic Vs?<br></br> <br></br></mdoxml> 2 4 0 1 0 0 0 Can you put the numbers 1-5 in the V shape so that both 'arms' have the same total? Numbers and the Number System Odd and even numbers Calculations and Numerical Methods Addition & subtraction Using, Applying and Reasoning about Mathematics Working systematically Using, Applying and Reasoning about Mathematics Generalising Using, Applying and Reasoning about Mathematics Making and proving conjectures Mathematics Tools Digit cards