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  <path>/www/nrich/html/content/00/07/penta3/</path>
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  <last_published>2011-02-01T00:00:01</last_published>
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&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;div style=&quot;text-align: center;&quot;&gt;&lt;br&gt;&lt;/br&gt;
&lt;mdo:image src=&quot;numbers.png&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;
&lt;p&gt;To find the &lt;strong&gt;factors&lt;/strong&gt; of a number, you have to find &lt;strong&gt;all&lt;/strong&gt; the pairs of numbers that multiply together to give that number.&lt;/p&gt;
&lt;p&gt;The factors of $48$ are:&lt;/p&gt;
&lt;p&gt;$1$ and $48$&lt;/p&gt;
&lt;p&gt;$2$ and $24$&lt;/p&gt;
&lt;p&gt;$3$ and $16$&lt;/p&gt;
&lt;p&gt;$4$ and $12$&lt;/p&gt;
&lt;p&gt;$6$ and $8$&lt;/p&gt;
&lt;p&gt;If we leave out the number we started with, $48$, and add all the other factors, we get $76$:&lt;/p&gt;
&lt;p&gt;$1 + 2 + 3 + 4 + 6 + 8 + 12 + 16 + 24 = 76$&lt;/p&gt;
&lt;p&gt;So .... $48$ is called an &lt;strong&gt;abundant&lt;/strong&gt; number because it is less than the sum of its factors (without itself). ($48$ is less than $76$.)&lt;/p&gt;
&lt;p&gt;See if you can find some more abundant numbers!&lt;/p&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 class=&quot;editorial&quot;&gt;Georgia, from Holy Trinity School, found two
more abundant numbers:&lt;/p&gt;

$36$: total is $55$ ($1+2+3+4+6+9+12+18$)&lt;br&gt;&lt;/br&gt;
$70$: total is $74$ ($1+2+5+7+10+14+35$)&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
&lt;p&gt;&lt;span class=&quot;editorial&quot;&gt;Rachael, Jamie, Heledd, Sian, Dafydd,
Tom, Edward and Isaac from Ysgol Bryncrug clearly worked hard on
this problem.  They told us:&lt;/span&gt;&lt;/p&gt;

We decided to find out which of the numbers from $1$ to $100$ are
abundant numbers.&lt;br&gt;&lt;/br&gt;
We decided that prime numbers are not abundant numbers:&lt;br&gt;&lt;/br&gt;
$2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59,
61, 67, 71, 73, 79, 83, 89, 97$&lt;br&gt;&lt;/br&gt;
because without themselves $1$ is their only factor.  &lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
We crossed out all the prime numbers from our $100$ square:&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
 &lt;mdo:image height=&quot;213&quot; width=&quot;302&quot; alt=&quot;&quot; src=&quot;primes.png&quot;&gt;&lt;/mdo:image&gt;&lt;br&gt;&lt;/br&gt;
Then we looked at all the remaining numbers. Here is an
example:&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
$27 = 1\times27$, $3\times9$&lt;br&gt;&lt;/br&gt;
So the factors of $27$ are:&lt;br&gt;&lt;/br&gt;
$1, 3, 9, 27$&lt;br&gt;&lt;/br&gt;
The sum of its factors without itself is:&lt;br&gt;&lt;/br&gt;
$1+3+9 = 13$&lt;br&gt;&lt;/br&gt;
  &lt;br&gt;&lt;/br&gt;
$13$ is &lt;span style=&quot;font-weight: bold;&quot;&gt;less than&lt;/span&gt; $27$
therefore $27$ &lt;span style=&quot;font-weight: bold;&quot;&gt;is not&lt;/span&gt; an
abundant number&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
And another example:&lt;br&gt;&lt;/br&gt;
$48 = 1\times48$, $2\times24$, $3\times16$, $4\times12$,
$6\times8$&lt;br&gt;&lt;/br&gt;
So the factors of $48$ are:&lt;br&gt;&lt;/br&gt;
$1, 2, 3, 4, 6, 8, 12, 16, 24, 48$.&lt;br&gt;&lt;/br&gt;
The sum of its factors without itself is:&lt;br&gt;&lt;/br&gt;
$1+2+3+4+6+8+12+16+24 = 76$&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
$76$ is &lt;span style=&quot;font-weight: bold;&quot;&gt;greater than&lt;/span&gt; $48$
therefore $48$ &lt;span style=&quot;font-weight: bold;&quot;&gt;is&lt;/span&gt; an
abundant number&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Something different happened with $6$:&lt;br&gt;&lt;/br&gt;
$6 = 1\times6$, $2\times3$&lt;br&gt;&lt;/br&gt;
So the factors of $6$ are:&lt;br&gt;&lt;/br&gt;
$1, 2, 3$ and $6$&lt;br&gt;&lt;/br&gt;
The sum of its factors without itself is: $1+2+3 = 6$&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
$6$ is equal to $6$!!   &lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
This also happened with $28$:&lt;br&gt;&lt;/br&gt;
$1+2+4+7+14 = 28$  &lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
Edward wanted to find out what we call a number where the sum of
its factors (without itself) is equal to the number. He asked his
Dad and he told him that it is called a PERFECT NUMBER.&lt;br&gt;&lt;/br&gt;
He also found out that the next perfect number after $28$ is $496$.
Then $8128$!!  &lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
We continued marking the numbers on the $100$ square: &lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
 &lt;mdo:image height=&quot;341&quot; width=&quot;413&quot; src=&quot;100sq.png&quot; alt=&quot;&quot;&gt;&lt;/mdo:image&gt;&lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
There are twenty two abundant numbers on our $100$
square. &lt;br&gt;&lt;/br&gt;
 &lt;br&gt;&lt;/br&gt;
&lt;p class=&quot;editorial&quot;&gt;Thank you for letting us know how you
approached this problem.  I like the way you discovered
perfect numbers along the way.  Well done!&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;Abundant Numbers&lt;/h2&gt;
&lt;br&gt;&lt;/br&gt;
&lt;p&gt;To find the &lt;strong&gt;factors&lt;/strong&gt; of a number, you have to find &lt;strong&gt;all&lt;/strong&gt; the pairs of numbers that multiply together to give that number.&lt;/p&gt;
&lt;p&gt;The factors of $48$ are:&lt;/p&gt;
&lt;p&gt;$1$ and $48$&lt;/p&gt;
&lt;p&gt;$2$ and $24$&lt;/p&gt;
&lt;p&gt;$3$ and $16$&lt;/p&gt;
&lt;p&gt;$4$ and $12$&lt;/p&gt;
&lt;p&gt;$6$ and $8$&lt;/p&gt;
&lt;p&gt;If we leave out the number we started with, $48$, and add all the other factors, we get $76$:&lt;/p&gt;
&lt;p&gt;$1 + 2 + 3 + 4 + 6 + 8 + 12 + 16 + 24 = 76$&lt;/p&gt;
&lt;p&gt;So .... $48$ is called an &lt;strong&gt;abundant&lt;/strong&gt; number because it is less than the sum of its factors (without itself). ($48$ is less than $76$.)&lt;/p&gt;
&lt;p&gt;See if you can find some more abundant numbers!&lt;/p&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;
This activity helps to reinforce the ideas surrounding factors. It could be used to help pupils learn to pursue calculations for a longer period of time and you could decide to focus on working systematically. It offers a lot of engaging arithmetic work from a bvery briefly described starting point. Systematic recording of results and conclusions is helpful in tackling this problem.&lt;br&gt;&lt;/br&gt;
&lt;h3&gt;Possible approach&lt;/h3&gt;
Introduce the idea of abundant numbers using the problem and then work with the whole class to explore a couple of other examples. You could try $12$ which has the factors $1$ and $12$, $2$ and $6$, $3$ and $4$. If you add together $1$, $2$, $3$, $4$ and $6$ you get $10$ which is less than $12$ so $12$ is not abundant. Then try $18$ which is abundant. There are plenty of other examples you could
use and the children could be encouraged to make suggestions. Once they have the idea, they can explore on their own.&lt;br&gt;&lt;/br&gt;
&lt;h3&gt;Key questions&lt;/h3&gt;
&lt;div&gt;What are the factors of...?&lt;br&gt;&lt;/br&gt;
Can you predict whether they will be abundant?&lt;br&gt;&lt;/br&gt;
How have you decided which numbers to choose?&lt;/div&gt;
&lt;div&gt;I see you seem to have a system for doing this, can you tell me about it?&lt;/div&gt;
&lt;h3&gt;Possible extension&lt;/h3&gt;
Children could be encouraged to find all the abundant numbers below a certain target or to develop strategies for choosing numbers that may be abundant.&lt;br&gt;&lt;/br&gt;
&lt;h3&gt;Possible support&lt;/h3&gt;
A table square to $100$ may help to support some children in identifying multiples. They may need support in finding the pairs of factors by using cubes or counters to help them. They could be encouraged to try to find the factors of numbers to $20$ first.&lt;/mdoxml&gt;</noteXML>
  <clueXML/>
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&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
Many people worked well on this problem. Some of the best efforts
came from, Rebecca , Rebecca G , Chung , Jasmine , Liana and
Stephie from Crofton Junior School, Kent, Jessica , Jaimee and
Christopher from Tattingstone School, UK. These people did well
because they tried to find lots of Abundant Numbers and showed all
the factors they'd used. Rachel from Crofton Junior School, Kent
did a particularly good job of testing the first 80 counting
numbers. &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Here is a list of abundant numbers. I'm sure there are many more!
&lt;br&gt;&lt;/br&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&lt;mdo:image width=&quot;445&quot; height=&quot;587&quot; src=&quot;abundant%201011.jpg&quot; alt=&quot;abundant&quot;&gt;&lt;/mdo:image&gt;&lt;/div&gt;
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Rowena noticed that the example given (48) had a lot of factors,
and so used this idea to choose some numbers that she knew had lots
of factors to test (12 and 24). &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Peter has spotted another three abundant numbers under 100: 66, 88
and 96. Thanks for letting us know, Peter. &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
Daniel from Anglo-Chinese School, Singapore noticed something else
about Abundant numbers. Does this always work? Why or why not?
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
&lt;mdo:image width=&quot;720&quot; height=&quot;520&quot; alt=&quot;1&quot; src=&quot;pentasol-1.gif&quot;&gt;&lt;/mdo:image&gt;&lt;br&gt;&lt;/br&gt;
Chung from Crofton Junior School, Kent found enough factors of one
million to be sure is was an abundant number! &lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</canonXML>
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  <title>Abundant Numbers</title>
  <description>48 is called an abundant number because it is less than the sum of
its factors (without itself). Can you find some more abundant
numbers?</description>
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  <spec_group>Calculations and Numerical Methods
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