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2011-02-01T00:00:01
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<span style="font-style: italic;">This problem is in two parts. The first part provides some building blocks which will help you to solve the final challenge. These can be attempted in any order. Of course, you are welcome to go straight to the Final Challenge!</span><br></br>
<br></br>
Click a question from below to get started.<br></br>
<br></br>
<a href="#" id="hideShow1">Question A</a><br></br>
<div id="hidden1" style="display: none;">Choose any two numbers from the $7$ times table. Add them together. Repeat with some other examples. Notice anything interesting?<br></br>
Now do the same with a different times table. What do you notice this time? Convince yourself it always happens.</div>
<br></br>
<a href="#" id="hideShow2">Question B</a>
<div id="hidden2" style="display: none;">Choose two digits and arrange them to make two double-digit numbers.<br></br>
For example, if you choose $5$ and $2$, you can make $52$ and $25$.<br></br>
Now add your two-digit numbers.<br></br>
Repeat with some other examples.<br></br>
Notice anything interesting? Convince yourself it always happens.</div>
<br></br>
<br></br>
<a href="#" id="hideShow3">Question C</a><br></br>
<div id="hidden3" style="display: none;">Look at this sequence of numbers: $11, 101, 1001, 10001, 100001, ...$<br></br>
Divide numbers in this sequence by $11$, WITHOUT using a calculator.<br></br>
Notice anything interesting? Convince yourself it always happens.</div>
<br></br>
<a href="#" id="hideShow4">FINAL CHALLENGE</a><br></br>
<br></br>
<div id="hidden4" style="display: none;">Take any four-digit number, move the first digit to the 'back of the queue' and move the rest along. For example $5238$ would become $2385$.<br></br>
Now add your two numbers.<br></br>
Is the answer always a multiple of $11$? Can you convince yourself?<br></br>
What happens when you do this with three-digit numbers? Five-digit numbers? Six-digit numbers? 38-digit numbers ... ?<br></br>
Prove your findings!</div>
<br></br>
<a href="http://nrich.maths.org/6821">Click here for a poster of this problem</a>.<br></br>
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<p class="editorial" xmlns="http://www.w3.org/1999/xhtml">This
problem offers a fun way of practising column addition and short
division. Furthermore, it is a great way to test and develop your
understanding of place value. The problem is divided into two
parts: the first part consists of three "building-block" questions,
which lead up to the final challenge in the second part. Some
students submitted solutions to both parts, whilst others plunged
straight in to the final challenge. The solutions submitted were
all great, with very clear explanations, so thank you very much to
everyone for this!</p>
<p xmlns="http://www.w3.org/1999/xhtml"><span class="editorial">Rebecca from Battle Primary School, and Akintunde
from Wilson's Grammar School both submitted good solutions to the
first part of the problem. Here is Akintunde's solution (note that
the editor has added comments in between questions):</span></p>
Question A:<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
I noticed two things from answering question A.<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
$1$) Whatever numbers from a particular times table you decide to
add together, the sum would be a multiple of the times table number
(i.e. a member of the times table). An example is if you add
multiples of $7$ together, e.g. $7$ and $14$, the sum will be $21$,
also a multiple of $7$.<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
$2$) The other thing I noticed is that, when you add multiples of
odd numbers together, the sum will either be odd or even, but when
you add multiples of even numbers together, the sum will always be
even.<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
An example of the odd multiples addition is $7$ + $14$, which
equals $21$ (odd) and $105$ + $7$, which equals $112$ (even). This
is an example of the multiples of $7$.<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
An example of the even multiples addition is $6$ + $12$, which
equals $18$ and $60$ + $72$, which equals $132$. Both of these sums
are even numbers.<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
<p class="editorial" xmlns="http://www.w3.org/1999/xhtml">This
could also be shown more generally. For example, let us consider
the $n$ times table. One number from this table is $xn$, and
another is $yn$, where $x$ and $y$ are numbers that $n$ is
multiplied by to give a multiple of $n$. Adding the two members of
the table together gives: $xn + yn=n(x+y)$. From this, we can see
that the sum will always be a multiple of $n$ i.e. it will always
be a member of the $n$ times table.</p>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
Question B<br xmlns="http://www.w3.org/1999/xhtml"></br>
I noticed from question B that when you add two numbers with the
same two digits (e.g. $26$ and $62$) you always get a multiple of
$11$. An example is $53$ + $35$, which equals $88$. $88$ is a
multiple of $11$.<br xmlns="http://www.w3.org/1999/xhtml"></br>
<p class="editorial" xmlns="http://www.w3.org/1999/xhtml">This can
also be shown more generally. For example, consider the number AB.
Remember that A is in the tens position and B is in the units
position. Thus, the number AB can be written as $10$A+B. Now, let
us reverse the digits as described in the question. This gives BA,
which can again be separated into tens and units: $10$B + A.
Finally, we add the two numbers together: AB+BA= ($10$A+B) +
($10$B+A)= $11$B+$11$A. This can be written as $11$(A+B), and so
again we can see that the sum of these two-digit numbers that we
created will always be a multiple of $11$.</p>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
Question C<br xmlns="http://www.w3.org/1999/xhtml"></br>
I noticed that if an even number of zeros is in between the two
$1$s, the answer will be an integer when I divide by $11$.<br xmlns="http://www.w3.org/1999/xhtml"></br>
If an odd number of zeros is in between the two $1$s, I will not
end up with a whole number when I divide by $11$.<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
Furthermore, when you divide a number with an odd number of zeros
by $11$, you'll get a decimal number, and then when you divide a
number with one extra number of zeros, you'll get a number that is
the last number multiplied by $10$, but without the rest of the
decimal numbers.<br xmlns="http://www.w3.org/1999/xhtml"></br>
Here is an example: when you divide $101$ by $11$, you get
$9.1818...$, and when you divide $1001$ by $11$, you get $91$.<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
<p class="editorial" xmlns="http://www.w3.org/1999/xhtml">The final
challenge built on these three questions. Several students
submitted good solutions to this. These included: Thomas from
A.Y.Jackson, Kang from Garden International School (in Malaysia!
Hello Malaysia!) , Richard from Comberton Village College, Andrew
from the Whitby Maths Club, Akintunde and Janahan from Wilson's
Grammar School, Nick from St Stephen's Carramar, and Susannah from
Wimbledon High School. Thomas explained:</p>
With any four-digit number, it can be rewritten as
$1000$A+$100$B+$10$C+D.<br xmlns="http://www.w3.org/1999/xhtml"></br>
Moving the A to the back gives $1000$B+$100$C+$10$D+A...<br xmlns="http://www.w3.org/1999/xhtml"></br>
...and the sum of the two is $1001$A+$1100$B+$110$C+$11$D.<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
It became evident that the numerical coefficient for every term had
a common factor of $11$ ($1001=11\times7\times13$). Therefore,
every such four digit number ABCD added to BCDA will be divisible
by eleven (as suggested by the title).<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
It works because as you move a digit to the back, all the other
digits (except for the first) will be added to its tenfold (since
they are moved one place to the left) thus giving $11$, $110$,
$1100$...<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
The first digit will increase by its unit-place value as in
($3000+3$) for $3456$, and generally (K000+K for KBCD). Further,
with any two digit number ($10$A+B+$10$B+A) = $11$A+$11$B is also
divisible by eleven, while any three digit number
($100$A+$10$B+C+$100$B+$10$C+A) =$101$A+$110$B+$11$C isn't
divisible by eleven as $11$ isn't a factor of $101$ ($101$ is
prime).<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
A pattern begins to emerge: all values other than the first: B, C,
D... will always become multiples of eleven as they are added to
their tenfold: $1$=> $11$ $20$=> $220$, $500$=> $5500$.
Therefore only the value of the <span style="font-weight: bold; font-style: italic;" xmlns="http://www.w3.org/1999/xhtml">first</span> digit in the sum
will determine the divisibility by $11$.<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
Two-digit: ($10+1$)A=$11$A, Three-digit: ($100+1$)=$101$A, Four
digit: ($1000+1$)A=$1001$A<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
It becomes clear that any number with an EVEN number of digits
added to its counterpart will be divisible by $11$ while any number
with an ODD number of digits will not. So it will work for a $38$
digit number as well as a $100$...(Billion) digit number!<br xmlns="http://www.w3.org/1999/xhtml"></br>
<br xmlns="http://www.w3.org/1999/xhtml"></br>
<p xmlns="http://www.w3.org/1999/xhtml"><span class="editorial">Thank you once again for these great solutions. I
look forward to reading next month's! In the meantime, if you
enjoyed this problem, have a go at these extension problems:</span>
<a href="http://nrich.maths.org/116&part=Double%20Digit" class="editorial">Double Digit</a><span class="editorial">,</span>
<a href="http://nrich.maths.org/2129&part=Special%20Number" class="editorial">Special Number</a><span class="editorial">,
and/or</span> <a href="http://nrich.maths.org/524&part=Repeaters" class="editorial">Repeaters</a><span class="editorial">. If you
would like additional practice, try</span> <a href="http://nrich.maths.org/2791&part=Diagonal%20Sums" class="editorial">Diagonal Sums</a><span class="editorial">.</span></p>
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<h3>Why do this problem?</h3>
<div>This problem involves a significant 'final challenge' which
can either be tackled on its own or after working on a set of
related 'building blocks' designed to lead students to helpful
insights. At one level it provides self-checking practice of the
algorithms for column addition and short division. At its heart it
is about challenging and developing students' understanding of
place value.</div>
<br></br>
<div>Initially working on the building blocks then gives students
the opportunity to work on harder mathematical challenges than they
might otherwise attempt.</div>
<br></br>
<div>The problem is structured in a way that makes it ideal for
students to work on in small groups.</div>
<h3>Possible approach</h3>
<div>Hand out a set of building block cards (<a href="/content/98/04/six1/Legs%20Eleven.doc">Word</a>, <a href="/content/98/04/six1/Legs%20Eleven.pdf">PDF</a>) to each group
of three or four students. (The final challenge will need to be
removed to be handed out later.) Within groups, there are several
ways of structuring the task, depending on how experienced the
students are at working together.</div>
<br></br>
<div>Each student, or pair of students, could be given their own
building block to work on. After they have had an opportunity to
make progress on their question, encourage them to share their
findings with each other and work together on each other's
tasks.</div>
<br></br>
<div>Alternatively, the whole group could work together on all the
building blocks, ensuring that the group doesn't move on until
everyone understands.</div>
<br></br>
<div>When everyone in the group is satisfied that they have
explored in detail the challenges in the building blocks, hand out
the final challenge.</div>
<br></br>
<div>The teacher's role is to challenge groups to explain and
justify their mathematical thinking, so that all members of the
group are in a position to contribute to the solution of the
challenge.</div>
<br></br>
<div>It is important to set aside some time at the end for students
to share and compare their findings and explanations, whether
through discussion or by providing a written record of what they
did.</div>
<br></br>
<h3>Key questions</h3>
<div>What important mathematical insights does my building block
give me?</div>
<div>How can these insights help the group tackle the final
challenge?</div>
<h3>Possible extension</h3>
<div>Of course, students could be offered the Final Challenge
without seeing any of the building blocks.</div>
<div>After establishing a general proof, students might like to try
some similar problems:</div>
<div><a href="http://nrich.maths.org/116&part=">Double
Digit</a></div>
<div><a href="http://nrich.maths.org/2129&part=">Special
Number</a></div>
<div><a href="http://nrich.maths.org/524&part=">Repeaters</a></div>
<h3>Possible support</h3>
<div>Encourage groups not to move on until everyone in the group
understands. The building blocks could be distributed within groups
in a way that plays to the strengths of particular students.</div>
<br></br>
<div>For students who have a weak grasp of place value, <a href="http://nrich.maths.org/2791&part=">Diagonal Sums</a>
gives addition practice in a context where the place value
structure is quite explicit.</div>
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The number $5374$ is $5000+300+70+4$<br></br>
What does this become once the $5$ has moved to the back of the
queue?<br></br>
How many 5s, 3s, 7s and 4s are there when you combine the two
numbers?<br></br></mdoxml>
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<tbody>
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</table>
<div><span class="editorial">Stella from GIS noticed
that</span></div>
<div class="editorial"></div>
<div>"all of the numbers can be divided by 11."</div>
<br></br>
<div class="editorial">Ram explained why this is the case:</div>
<br></br>
<div>1) Let's take a number 'abcd'</div>
<div>Value of this number in decimal notation = a*1000 + b*100 +
c*10 + d</div>
<br></br>
<div>2) After moving the first digit to the last position, the
number becomes 'bcda'.</div>
<div>Value of this number in decimal notation = b*1000 + c*100 +
d*10 + a</div>
<br></br>
<div>Adding up the values of two numbers in decimal notation</div>
<br></br>
<div>(a*1000 + b*100 + c*10 + d) + (b*1000 + c*100 + d*10 +
a)</div>
<div>= a(1000+1) + b(100+1000) + c(10+100) + d(1+10)</div>
<div>= 1001a + 1100b + 110c + 11d</div>
<div>= 11(91a + 100b + 10c + d)</div>
<br></br>
<div>As can be seen, the total has a factor of 11 and is divisible
by 11.</div>
<br></br>
<div><span class="editorial">Mikey from Tadcaster Grammar School
offered a very similar explanation, added that this only works for
numbers with an even number of digits and gave a few examples of
numbers with an odd number of digits where it does not work.
Click</span> <a class="editorial" href="/content/98/04/six1/Solution%20to%20Eleven%20Legs.doc">here</a>
<span class="editorial">to see his solution.</span></div>
<br></br>
<div class="editorial">Andy from Garden International School came
to the same conclusion:</div>
<br></br>
<div>They are all multiples of eleven. It works for all numbers
with even digits because it will be adding up multiples of 11
together.</div>
<br></br>
<div>Take 8937 for example:</div>
<div>8937+9378 = 8008+9900+330+77 = 18315</div>
<div>and 8008, 9900, 330 and 77 are all multiples of eleven.</div>
<br></br>
<div>It won't work for odd digits because numbers that have odd
numbers of 0 in the middle would not be a multiple of 11:</div>
<div>Take 135 for example:</div>
<div>135+351 = 101+330+55 = 486</div>
<div>which is not a multiple of 11 because 101 is not a multiple of
11 (and the others are).</div>
<br></br>
<div class="editorial">Kang from Garden International School came
to the same conclusion and showed why it works for two digit
numbers:</div>
<br></br>
<div>First digit = a</div>
<div>Second digit = b</div>
<div>If a + b = c</div>
<div>a b + b a = c c</div>
<br></br>
<div class="editorial">Thomas from A.Y. Jackson School sent us a
very full and clear explanation:</div>
<br></br>
<div>Any four-digit number,can be rewritten as
1000A+100B+10C+D.</div>
<div>Moving the A to the back gives 1000B+100C+10D+A, and the sum
of the two is 1001A+1100B+110C+11D.</div>
<br></br>
<div>It became evident that the numerical coefficient for every
term had a common factor of 11 (1001=11*7*13).</div>
<div>Therefore, every such four digit number ABCD added to BCDA
will be divisible by eleven (as suggested by the title).</div>
<div>It works because as you move a digit to the back, all the
other digits (except for the first) will be added to its tenfold
(since they are moved one place to the left) thus giving 11, 110,
1100...</div>
<div>The first digit will increase by its unit-place value as in
(3000+3) for 3456, and generally (K000+K for KBCD).</div>
<br></br>
<div>Further, with any two digit number</div>
<div>(10A+B+10B+A) = 11A+11B is also divisible by eleven,</div>
<div>while any three digit number</div>
<div>(100A+10B+C+100B+10C+A) =101A+110B+11C isn't divisible by
eleven as 11 isn't a factor of 101.</div>
<br></br>
<div>A pattern begins to emerge: all values other than the first:
B, C, D... will always become multiples of eleven as they are added
to their tenfold:</div>
<div>1=> 11 20=> 220, 500=> 5500.</div>
<div>Therefore only the value of the first digit in the sum will
determine the divisibility of 11.</div>
<div>Two-digit: (10+1)A=11A</div>
<div>Three-digit: (100+1)=101A</div>
<div>Four digit: (1000+1)A=1001A</div>
<br></br>
<div>It becomes clear that any number with an EVEN number of digits
added to its counterpart will be divisible by 11 while any number
with an ODD number of digits will not.</div>
<div>So it will work for a 38 digit number as well as a
100...(Billion) digit number!</div>
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Legs Eleven
Take any four digit number. Move the first digit to the 'back of
the queue' and move the rest along. Now add your two numbers. What
properties do your answers always have?
Algebra
Creating expressions/formulae
Numbers and the Number System
Place value
Numbers and the Number System
Divisibility
Using, Applying and Reasoning about Mathematics
Disproof by counterexample
Calculations and Numerical Methods
Number operations - generally
Admin
Workshop
Information and Communications Technology
smartphone
Admin
Stage 3&4 Investigation
Secondary Mapping Document
Number operations and calculation methods