Is there an efficient way to work out how many factors a large number has?
Choose any 3 digits and make a 6 digit number by repeating the 3
digits in the same order (e.g. 594594). Explain why whatever digits
you choose the number will always be divisible by 7, 11 and 13.
Find the number which has 8 divisors, such that the product of the
divisors is 331776.
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):
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.
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 $11$.
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:
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.