1866
/www/nrich/html/content/03/10/six3/
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2011-02-01T00:00:01
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<p><span style="font-style: italic;">This problem builds on</span> <a href="/7405" style="font-style: italic;">What Numbers Can We Make?</a><br></br>
</p>
<p>Take a look at the video below. Will Charlie always find three numbers that add up to a multiple of three?</p>
<video controls="controls" height="315" src="1866%20Take%20Three%20From%20Five.mp4" width="420"></video>
<div> </div>
<p align="left"><em>If you can't see the video, click below to read a description.</em></p>
<p align="left"> </p>
<div align="left" class="toggle">Charlie invites James and Caroline to give him sets of five whole numbers. Each time he chooses three of their numbers that add together to make a multiple of three:
<p align="left"> </p>
<table style="width: 500px; border-spacing: 1px;" border="1">
<tbody>
<tr>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;">TOTAL</td>
</tr>
<tr>
<td style="padding: 1px; text-align: center;">3</td>
<td style="padding: 1px; text-align: center;"><strong> 6</strong></td>
<td style="padding: 1px; text-align: center;"><strong>5</strong></td>
<td style="padding: 1px; text-align: center;"><strong>7</strong></td>
<td style="padding: 1px; text-align: center;">2</td>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;">18</td>
</tr>
<tr>
<td style="padding: 1px; text-align: center;"><strong>7</strong></td>
<td style="padding: 1px; text-align: center;"><strong>17</strong></td>
<td style="padding: 1px; text-align: center;"><strong>15</strong></td>
<td style="padding: 1px; text-align: center;">8</td>
<td style="padding: 1px; text-align: center;">10</td>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;">39</td>
</tr>
<tr>
<td style="padding: 1px; text-align: center;">20</td>
<td style="padding: 1px; text-align: center;"><strong>15</strong></td>
<td style="padding: 1px; text-align: center;"><strong>6</strong></td>
<td style="padding: 1px; text-align: center;">11</td>
<td style="padding: 1px; text-align: center;"><strong>12</strong></td>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;">33</td>
</tr>
<tr>
<td style="padding: 1px; text-align: center;"><strong>23</strong></td>
<td style="padding: 1px; text-align: center;"><strong>16</strong></td>
<td style="padding: 1px; text-align: center;"><strong>9</strong></td>
<td style="padding: 1px; text-align: center;">21</td>
<td style="padding: 1px; text-align: center;">36</td>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;">48</td>
</tr>
<tr>
<td style="padding: 1px; text-align: center;"><strong>99</strong></td>
<td style="padding: 1px; text-align: center;"><strong>57</strong></td>
<td style="padding: 1px; text-align: center;">5</td>
<td style="padding: 1px; text-align: center;"><strong>72</strong></td>
<td style="padding: 1px; text-align: center;">23</td>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;">228</td>
</tr>
<tr>
<td style="padding: 1px; text-align: center;"><strong>312</strong></td>
<td style="padding: 1px; text-align: center;"><strong>97</strong></td>
<td style="padding: 1px; text-align: center;">445</td>
<td style="padding: 1px; text-align: center;"><strong>452</strong></td>
<td style="padding: 1px; text-align: center;">29</td>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;">861</td>
</tr>
<tr>
<td style="padding: 1px; text-align: center;">-1</td>
<td style="padding: 1px; text-align: center;"><strong>-1</strong></td>
<td style="padding: 1px; text-align: center;"><strong>0</strong></td>
<td style="padding: 1px; text-align: center;"><strong>1</strong></td>
<td style="padding: 1px; text-align: center;">1</td>
<td style="padding: 1px; text-align: center;"> </td>
<td style="padding: 1px; text-align: center;">0</td>
</tr>
</tbody>
</table>
<p align="left"><br></br>
Charlie challenges Caroline and James to find a set of five whole numbers that doesn't include three that add up to a multiple of three.<br></br>
</p>
</div>
<p align="left"> </p>
<p align="left"><strong>Can you come up with a set of five whole numbers that don't include three that add up to a multiple of three?</strong></p>
<p align="left">You can use the interactivity below to input sets of five numbers and test whether there are three numbers that add up to a multiple of three.</p>
<p align="left"><br></br>
<a href="/content/03/10/six3/ThreeFromFive.swf">Full Screen Version</a></p>
<p><mdo:flash height="400" id="/content/03/10/six3/ThreeFromFive.swf" width="550"><param name="allowfullscreen" value="true"></param><param name="allowfullscreen" value="true"></param>
<param name="movie" value="/content/03/10/six3/ThreeFromFive.swf"></param>
<param name="flashplayerversion" value="9"></param>
</mdo:flash></p>
<p> </p>
<p><strong>If you can't find a set of five whole numbers that doesn't include three that add up to a multiple of three, can you convince us that no such set exists?</strong></p>
<p> </p>
<p><a href="http://nrich.maths.org/7215&part=">Click here for a poster of this problem</a>.</p>
<p> </p></mdoxml>
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<mdoxml version="1.0"><span class="editorial">The group at Henry Cavendish Primary School thought they found 5 numbers that worked - 6, 17, 5, 41, 96 - and asked for some feedback! This was a really good try, but unfortunately there is a multiple of three in there: 17 + 5 + 41 is equal to 3 times 21. Never mind!<br></br>
<br></br>
The answer is that it's not possible. Lots of people got this, but David, from Sha Tin College, gave a particularly well written answer:</span><br></br>
<br></br>
Firstly, each one of the five integers can be expressed in the form 3x + 0, 3x + 1 or 3x + 2, where x is an integer.<br></br>
<br></br>
When we take three integers (say 3x + a, 3y + b, 3z + c) from a group of 5 and add them together, we will always get an integer in the following form:<br></br>
<br></br>
3x + 3y + 3z + k, which equals 3(x + y +z) + k (where k is the sum of a, b and c)<br></br>
<br></br>
It is obvious that the term 3(x + y + z) must be a multiple of 3. Therefore, we only need to consider k in order to see whether 3(x + y + z) + k is a multiple of 3. For 3(x + y + z) + k to be a multiple of 3, k must therefore also be a multiple of 3.<br></br>
<br></br>
As a, b and c are each equal to 0, 1 or 2, there are 4 different ways in which k can be a multiple of 3:<br></br>
<br></br>
1. 0 + 0 + 0 (i.e. a=0, b=0, c=0)<br></br>
2. 1 + 1 + 1<br></br>
3. 2 + 2 + 2<br></br>
4. 0 + 1 + 2<br></br>
<br></br>
Out of a group of 5 integers, it is always possible to find an a, b and c that add up to a multiple of 3. This is because in a group of 5 zeros, ones or twos, there is either at least one of each number (0,1,2), or at least 3 of just one of the numbers (0,0,0 or 1,1,1 or 2,2,2).<br></br>
<br></br>
Therefore it is always possible to choose three numbers that will add up to a multiple of 3 from any group of 5 numbers.<br></br>
<br></br>
<span class="editorial">The Year 5 group at St. John's C of E Primary remarked that it is possible to make a set of 5 numbers like this if you don't insist that they have to be <strong>whole</strong> numbers: they came up with 0.2, 0.2, 0, 0, 1. That's definitely thinking outside the box!</span><br></br>
<br></br>
<span class="editorial">Thanks to everyone - lots of clever answers here!</span></mdoxml>
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<h3>Why do this problem?</h3>
<p><a href="http://nrich.maths.org/public/viewer.php?obj_id=1866">This problem</a> looks like a number task, possibly revision about multiples, but it becomes a question about establishing why something can never happen, and creating a real proof of this. When it comes, the proof often feels powerful, satisfying and complete, and students leave feeling they have achieved something.</p>
<p> </p>
<h3>Possible approach</h3>
<div> </div>
<div>It would be very useful for students to work on <a href="/7405">What Numbers Can We Make?</a> before attempting this problem.</div>
<div> </div>
<div>Introduce the problem the way Charlie does in the video, by inviting students to suggest sets of five whole numbers, circle three of them that add up to a multiple of three, and write down the total.</div>
<div> </div>
<div>Don't say anything - let students work out what is special about the sum of the numbers you select. Suggest that if they know what is going on they may like to choose $5$ numbers that stop you achieving your aim. At some stage check that they all know what is going on.</div>
<div> </div>
<div>Challenge them to offer five numbers that don't include three that add up to a multiple of $3$. Allow them time to work on the problem in pairs or small groups, and suggest that they write any sets they find up on the board. Students may enjoy spotting errors among the suggestions on the board.</div>
<p><br></br>
Allow negative numbers, as long as they will allow you negative multiples of $3$ (and zero).<br></br>
<br></br>
At some stage there may be mutterings that it's impossible. A possible response might be:<br></br>
"Well if you think it's impossible, there must be a reason. If you can find a reason then we'll be sure."<br></br>
<br></br>
Once they have had sufficient thinking time, bring the class together to share ideas.</p>
<p>If it hasn't emerged, share with students Charlie's representation from <a href="/7405">What Numbers Can We Make?</a></p>
<p>All numbers fall into one of these 3 categories:<br></br>
<br></br>
Type A (multiple of $3$)</p>
<p><mdo:image src="Charlie3n.png"></mdo:image><br></br>
<br></br>
Type B (of the form $3n+1$)<br></br>
<mdo:image src="Charlie3n%2B1.png"></mdo:image><br></br>
<br></br>
Type C (of the form $3n+2$)</p>
<p><mdo:image src="Charlie3n%2B2.png"></mdo:image></p>
<p><br></br>
<span style="font-style: italic;">We have found that trying to use algebraic expressions as above, is tricky, students often end up with n having two or more values at once. Students are unlikely to know the notation of modular arithmetic, but the crosses notation above is sufficient for the context, and it suggests a geometrical image that students can use in explaining their ideas.</span></p>
<p>"Which combinations of A, B and C give a multiple of three?"<br></br>
"Can you find examples in our list on the board where you gave me one of those combinations?"</p>
<p>A few minutes later...</p>
<p>"Great, then all you have to do is find a combination of As, Bs and Cs that doesn't include AAA, BBB, CCC or ABC!"</p>
<p>Later still...</p>
<p><em>"It's impossible! All the combinations will include AAA, BBB, CCC or ABC!"</em></p>
<p>"OK, but can you prove it? Can you convince me that it's impossible?"<br></br>
</p>
<h3>Possible extension</h3>
<div> </div>
<div><a href="/8280">What Numbers Can We Make Now?</a> is a suitable follow-up task.</div>
<div> </div>
<div>A challenging extension:</div>
<div>You can guarantee being able to get a multiple of $2$ when you select $2$ from $3$.</div>
<div>You can guarantee being able to get a multiple of $3$ when you select $3$ from $5$.</div>
<div>Can you guarantee being able to get a multiple of $4$ when you select $4$ from $7$?</div>
<div> </div>
<h3>Possible support</h3>
<div> </div>
<div>Select sets of $3$ numbers. Your sets will always include two numbers that add up to an even number. Why?</div>
<p><br></br>
<br></br>
<br></br>
<br></br>
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<mdoxml version="1.0"><p>Start with the problem <a href="/7405">What Numbers Can We Make?</a></p>
<p>Think of a simpler problem:<br></br>
If you choose $2$ from $3$ numbers you can always select $2$ numbers that add up to an even number. Why?<br></br>
<br></br>
</p></mdoxml>
<?xml version="1.0" encoding="UTF-8"?>
<mdoxml version="1.0"><br></br>
<p class="editorial">This problem has been solved by Felix from the German Swiss International School, Hong Kong, Curt from Reigate College, Tom who does not tell us his school and David from Sha Tin College, also in Hong Kong.<br></br>
<br></br>
Here is David's solution:</p>
<div>Firstly, each one of the five integers can be expressed in the form $3x + c$, where $x$ is an integer and $c$ is either 0, 1 or 2. When we take three integers (say $3x +c_1$, $3y + c_2$, $3z + c_3$) from a set of 5 and add them together, we will always get an integer in the following form: $$3x + 3y + 3z + k,$$ which equals $$3(x + y +z) + k$$ (where k is the sum of $c_1$, $c_2$ and
$c_3$)</div>
<br></br>
<div>It is obvious that the term $3(x + y + z)$ must be a multiple of 3. Therefore, we only need to consider $k$ in order to see whether $3(x + y + z) + k$ is a multiple of 3.</div>
<br></br>
<div>For $3(x + y + z) + k$ to be a multiple of 3, $k$ must therefore also be a multiple of 3.</div>
<br></br>
<div>As $c = 0, 1$ or 2 therefore there are 4 different ways in which $k$ can be a multiple of 3:</div>
<br></br>
<div>1. 0 + 0 + 0</div>
<div>2. 1 + 1 + 1</div>
<div>3. 2 + 2 + 2</div>
<div>4. 0 + 1 + 2</div>
<br></br>
<div>Out of a group of 5 integers, it is always possible for three values of $c$ to add up to a multiple of 3. This is because in a group of 5 zeros, ones or twos, there is either at least one of each number (0,1,2), or at least 3 of just one of the numbers (0,0,0 or 1,1,1 or 2,2,2).</div>
<br></br>
<div>Therefore it is always possible to choose three numbers that will add up to a multiple of 3 from any group of 5 numbers.</div>
<br></br></mdoxml>
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Take Three From Five
Caroline and James pick sets of five numbers. Charlie chooses three of them that add together to make a multiple of three. Can they stop him?
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