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  <resource>
  <id>6779</id>
  <path>/www/nrich/html/content/id/6779/</path>
  <resourceTypeID>1</resourceTypeID>
  <last_published>2011-02-01T00:00:01</last_published>
  <indexXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
You may like to have a go at the problem &lt;a href=&quot;http://nrich.maths.org/6778&quot;&gt;So It's $28$&lt;/a&gt; before trying
this challenge.&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image height=&quot;74&quot; width=&quot;75&quot; src=&quot;the%2028.jpg&quot; alt=&quot;the 28&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;

&lt;div style=&quot;text-align: center;&quot;&gt;We are about to explore the number
$28$.&lt;/div&gt;

&lt;div&gt;In $2010$ the month of February will have four weeks of $7$
days making $28$ days altogether.&lt;/div&gt;

&lt;div style=&quot;text-align: center;&quot;&gt;----------&lt;/div&gt;

&lt;div&gt;Half a pack of cards with the jokers makes $28$ cards
altogether:&lt;/div&gt;

&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image height=&quot;236&quot; width=&quot;334&quot; src=&quot;pic%2028%20cards.jpg&quot; alt=&quot;cards&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;

&lt;div style=&quot;text-align: center;&quot;&gt;----------&lt;/div&gt;

&lt;div&gt;The seventh triangular number is also $28$:&lt;/div&gt;

&lt;br&gt;&lt;/br&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image height=&quot;231&quot; width=&quot;229&quot; alt=&quot;tri sq&quot; src=&quot;tri%20sq.jpg&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;

&lt;div&gt;But all these examples of $28$ are just flat - or
you might call them $2$D.&lt;/div&gt;

&lt;br&gt;&lt;/br&gt;
 

&lt;div&gt;I'm suggesting we try making something $3$D.&lt;/div&gt;

&lt;div&gt;How about  if  we put some small cubes
together and count the faces which are visible. Can we arrange the
cubes so that there are exactly $28$ faces?&lt;/div&gt;

&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image height=&quot;201&quot; width=&quot;379&quot; alt=&quot;ex cubes&quot; src=&quot;2%20egs%20of%20cubes%20small.jpg&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;

&lt;div style=&quot;text-align: left;&quot;&gt;In the pictures above, we're not
counting faces that are 'on the table'. I've coloured the cubes
(according to how many faces are showing) to make the counting a
little easier.&lt;/div&gt;

&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image height=&quot;69&quot; width=&quot;486&quot; alt=&quot;key&quot; src=&quot;key%202%20colours.jpg&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;

&lt;div style=&quot;text-align: left;&quot;&gt;It would probably be good to make
one of these shapes and do your own counting.&lt;/div&gt;

&lt;div style=&quot;text-align: left;&quot;&gt;Can you arrange some cubes so that
$28$ of their faces are visible?&lt;/div&gt;

&lt;div style=&quot;text-align: center;&quot;&gt;----------&lt;/div&gt;

&lt;div style=&quot;text-align: left;&quot;&gt;You might want to try another way
with a new shape resting on a glass table (you could just pretend!)
so you can count the faces underneath too. Here are two
shapes:&lt;/div&gt;

&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image height=&quot;133&quot; width=&quot;254&quot; alt=&quot;2 on glass&quot; src=&quot;2%20egs%20on%20Glass%20small.jpg&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;

&lt;div style=&quot;text-align: left;&quot;&gt;Cubes that click together might make
it easier.&lt;/div&gt;

&lt;div style=&quot;text-align: center;&quot;&gt;----------&lt;/div&gt;

&lt;div&gt;So this is your challenge: make a shape that has
$28$ little square faces visible.&lt;/div&gt;

&lt;div&gt;Now  try  on a pretend glass
table  and  see how many you can make then.
Any surprises?&lt;/div&gt;

&lt;div&gt;You may like to take photos of them.&lt;/div&gt;

&lt;div&gt;Please send in any thoughts, ideas or pictures relating to
your explorations.&lt;/div&gt;

&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</indexXML>
  <solutionXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
&lt;p&gt;&lt;span class=&quot;editorial&quot;&gt;We had some lovely ideas sent in. It
would have been good to be there when they were created and to know
how you got your solutions. Here are just a few. I was interested
in both Alexander's and George's good desciptions in words to say
what they created.&lt;/span&gt;&lt;/p&gt;

&lt;p&gt;&lt;span class=&quot;editorial&quot;&gt;Alexander wrote:&lt;/span&gt;&lt;/p&gt;

&lt;br&gt;&lt;/br&gt;
I used $9$ bricks. I put $8$ of them together in a $3$ by $3$
square with the middle one missing. I then put the $9$th brick on
the middle brick of one of the rows. So I had one brick with $5$
faces showing, $7$ bricks with $3$ faces showing and $1$ brick with
$2$ faces showing.&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;p&gt;&lt;span class=&quot;editorial&quot;&gt;and George wrote;&lt;/span&gt;&lt;/p&gt;

I used $11$ bricks. I had $1$ row of $3$ bricks joined to a row of
$4$ bricks and the $4$ brick row was joined to another row of $3$
bricks. On one of the rows of $3$ bricks I put the $11$th brick at
the end on top.&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;p&gt;&lt;span class=&quot;editorial&quot;&gt;Ewan, Fraser, Jenny Lee and Ayan had a
good idea of only counting the faces you could see, so they sent in
this picture&lt;/span&gt;;&lt;/p&gt;

&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image width=&quot;280&quot; height=&quot;150&quot; src=&quot;ewan%2CFraser%20etc%20.jpg&quot; alt=&quot;ewan etc.&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;

&lt;br&gt;&lt;/br&gt;
&lt;p&gt;&lt;span class=&quot;editorial&quot;&gt;Kristy and Amy used the computer and
sent in&lt;/span&gt;;&lt;/p&gt;

&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image width=&quot;338&quot; height=&quot;173&quot; alt=&quot;Amy&quot; src=&quot;Amy%20%26%20Kristy%20WBTrtu.jpg&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;

&lt;br&gt;&lt;/br&gt;
&lt;p&gt;&lt;span class=&quot;editorial&quot;&gt;Zain sent in two ideas, one of which
is;&lt;/span&gt;&lt;/p&gt;

&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image width=&quot;208&quot; height=&quot;322&quot; src=&quot;Zain%2028.jpg&quot; alt=&quot;Zain&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;

&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;p&gt;&lt;span class=&quot;editorial&quot;&gt;Finally Lily and Alex sent
in:&lt;/span&gt;&lt;/p&gt;

&lt;br&gt;&lt;/br&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image width=&quot;466&quot; height=&quot;216&quot; alt=&quot;Lily&quot; src=&quot;Lily%20%26%20Alex.jpg&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;

&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;p class=&quot;editorial&quot;&gt;Well done everyone! I'm sure that many more of
you who did not send in solutions came up with some interesting
ideas and if your class/group did not have a go try it now!&lt;/p&gt;

&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</solutionXML>
  <noteXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;div class=&quot;embed&quot;&gt;
&lt;h2&gt;28 and It&amp;#39;s Upward and Onward&lt;/h2&gt;
&lt;br&gt;&lt;/br&gt;
You may like to have a go at the problem &lt;a href=&quot;http://nrich.maths.org/6778&quot;&gt;So It&amp;#39;s $28$&lt;/a&gt; before trying this challenge.&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image alt=&quot;the 28&quot; height=&quot;74&quot; src=&quot;the%2028.jpg&quot; width=&quot;75&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;We are about to explore the number $28$.&lt;/div&gt;
&lt;div&gt;In $2010$ the month of February will have four weeks of $7$ days making $28$ days altogether.&lt;/div&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;----------&lt;/div&gt;
&lt;div&gt;Half a pack of cards with the jokers makes $28$ cards altogether:&lt;/div&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image alt=&quot;cards&quot; height=&quot;236&quot; src=&quot;pic%2028%20cards.jpg&quot; width=&quot;334&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;----------&lt;/div&gt;
&lt;div&gt;The seventh triangular number is also $28$:&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image alt=&quot;tri sq&quot; height=&quot;231&quot; src=&quot;tri%20sq.jpg&quot; width=&quot;229&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;
&lt;div&gt;But all these examples of $28$ are just flat - or you might call them $2$D.&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div&gt;I&amp;#39;m suggesting we try making something $3$D.&lt;/div&gt;
&lt;div&gt;How about if we put some small cubes together and count the faces which are visible. Can we arrange the cubes so that there are exactly $28$ faces?&lt;/div&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image alt=&quot;ex cubes&quot; height=&quot;201&quot; src=&quot;2%20egs%20of%20cubes%20small.jpg&quot; width=&quot;379&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;In the pictures above, we&amp;#39;re not counting faces that are &amp;#39;on the table&amp;#39;. I&amp;#39;ve coloured the cubes (according to how many faces are showing) to make the counting a little easier.&lt;/div&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image alt=&quot;key&quot; height=&quot;69&quot; src=&quot;key%202%20colours.jpg&quot; width=&quot;486&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;It would probably be good to make one of these shapes and do your own counting.&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;Can you arrange some cubes so that $28$ of their faces are visible?&lt;/div&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;----------&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;You might want to try another way with a new shape resting on a glass table (you could just pretend!) so you can count the faces underneath too. Here are two shapes:&lt;/div&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image alt=&quot;2 on glass&quot; height=&quot;133&quot; src=&quot;2%20egs%20on%20Glass%20small.jpg&quot; width=&quot;254&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;Cubes that click together might make it easier.&lt;/div&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;----------&lt;/div&gt;
&lt;div&gt;So this is your challenge: make a shape that has $28$ little square faces visible.&lt;/div&gt;
&lt;div&gt;Now try on a pretend glass table and see how many you can make then. Any surprises?&lt;/div&gt;
&lt;div&gt;You may like to take photos of them.&lt;/div&gt;
&lt;div&gt;Please send in any thoughts, ideas or pictures relating to your explorations.&lt;/div&gt;
&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;h3&gt;&lt;span style=&quot;font-weight: bold;&quot;&gt;Why do this problem?&lt;/span&gt;&lt;/h3&gt;
&lt;div&gt;You may have tried &lt;a href=&quot;http://nrich.maths.org/public/viewer.php?obj_id=60&amp;amp;part=&quot;&gt;Month Mania&lt;/a&gt; in the past and &lt;a href=&quot;http://nrich.maths.org/public/viewer.php?obj_id=6779&amp;amp;part=&quot;&gt;this activity&lt;/a&gt; takes the action further. This challenge is a great open-ended investigation which has many opportunties for children to explore their own ideas at length. It requires visualisation
and a systematic approach.&lt;/div&gt;
&lt;div&gt; &lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;h3&gt;Possible approach&lt;/h3&gt;
&lt;div&gt;You could introduce this activity by sharing the examples of $28$ in the problem and asking if any of them know something special about the number $28$.&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div&gt;Show the group an arrangement of cubes resting on a surface and explain that $28$ small faces are visible. Check that everyone is happy that we&amp;#39;re not counting the faces that are resting on the surface. Encourage learners to work in pairs to find other arrangements which have $28$ faces. Depending on your supplies of cubes, you can suggest children keep each arrangement once it is made, or
that they record it in some way. (Isometric paper may be useful.)&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div&gt;Once everyone has found some examples, draw the group together for a few minutes and ask them how they are checking each of their arrangements is different. How can they try to find all the ways? Encourage them to share some different systems, for example sticking with the same number of cubes and looking for ways of changing their arrangement. Other children may find ways to add successive
cubes but keeping the number of visible faces the same.&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;div&gt;You may like to keep this challenge as an ongoing &amp;#39;simmering&amp;#39; activity which children add to over a period of a number of days/weeks.&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;h3&gt;Key questions&lt;/h3&gt;
&lt;div&gt;Tell me about this.&lt;/div&gt;
&lt;div&gt;Have you anything to say about this shape that you have made?&lt;/div&gt;
&lt;div&gt;How could you change your shape and still have $28$ faces showing?&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;h3&gt;Possible extension&lt;/h3&gt;
&lt;div&gt;Children could explore other notable things about $28$, for example it&amp;#39;s a &lt;a href=&quot;/content/id/6779/hex%20nos.jpg&quot;&gt;hexagonal number&lt;/a&gt;. $28$ is the sum of the first five consecutive primes: $2, 3, 5, 7, 11$. You might like to introduce them to &lt;a href=&quot;/content/id/6779/perfect.jpg&quot;&gt;perfect numbers&lt;/a&gt;.&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;h3&gt;Possible support&lt;/h3&gt;
&lt;div&gt;Plenty of cubes will be needed for this activity. If children are struggling to record their arrangements, digital photographs may be the answer.&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</noteXML>
  <clueXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
How many cubes will you start with?&lt;br&gt;&lt;/br&gt;
Try making a shape and then counting the number of faces. Are there
more than $28$ or fewer than $28$? Can you see how to change your
shape to get exactly $28$ faces?&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</clueXML>
  <canonXML>&lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt;
&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
&lt;a href=&quot;/content/id/6779/28%203D.doc&quot;&gt;This doc&lt;/a&gt;  gives
some ideas as to a solution.&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</canonXML>
  <end_user_role>2</end_user_role>
  <difficulty>3</difficulty>
  <keystage1>0</keystage1>
  <keystage2>1</keystage2>
  <keystage3>0</keystage3>
  <keystage4>0</keystage4>
  <keystage4plus>0</keystage4plus>
  <title>28 and it's upward and onward</title>
  <description>Can you find ways of joining cubes together so that 28 faces are
visible?</description>
  <spec_group>Using, Applying and Reasoning about Mathematics
    <specifier>Representing</specifier>
  </spec_group>
  <spec_group>Using, Applying and Reasoning about Mathematics
    <specifier>Investigations</specifier>
  </spec_group>
  <spec_group>Using, Applying and Reasoning about Mathematics
    <specifier>Visualising</specifier>
  </spec_group>
  <spec_group>Numbers and the Number System
    <specifier>Properties of numbers</specifier>
  </spec_group>
</resource>