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!
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):
I noticed two things from answering question A.
$1$) 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 $7$.
$2$) 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
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 $7$.
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.
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.
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 $11$.
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 $11$.
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 $11$.
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 $11$.
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
Here is an example: when you divide $101$ by $11$, you get
$9.1818...$, and when you divide $1001$ by $11$, you get $91$.
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:
With any four-digit number, it can be rewritten as
Moving the A to the back gives $1000$B+$100$C+$10$D+A...
...and the sum of the two is $1001$A+$1100$B+$110$C+$11$D.
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).
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 $11$, $110$,
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
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 first
digit in the sum
will determine the divisibility by $11$.
Two-digit: ($10+1$)A=$11$A, Three-digit: ($100+1$)=$101$A, Four
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!
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:
and/or Repeaters. If you
would like additional practice, try Diagonal Sums.