Follow the clues to find the mystery number.
You have 4 red and 5 blue counters. How many ways can they be
placed on a 3 by 3 grid so that all the rows columns and diagonals
have an even number of red counters?
You can trace over all of the diagonals of a pentagon without
lifting your pencil and without going over any more than once. Can
the same thing be done with a hexagon or with a heptagon?
I am extremely impressed by the
solutions sent in to the Magic Vs problem, so well done to you all.
Tom from Crudgington Primary, Krishan from Buckingham College Prep
School, Harriet and Laura from Shincliffe C of E Primary, Nicholas
and Jared from Clifton Hill Primary, Andy from Garden International
School, Jonathan from DGS, Lucas from Grinling Gibbons Primary and
Scott from Echline Primary explained that there were three
different Magic Vs for the numbers 1-5.
Emma and Charlotte from Greenacre School for
This is very clearly explained, thank
you. Yujin from Dubai International Academy explained this in a
slightly different way:
... if there are two even numbers and three odd numbers; you put
the odd number at the bottom: even + even = odd + odd.
You can't put the even at the bottom because even + odd does not
equal to odd + odd.
That's right. Gemma, Dylan, Faazil and
Gabriel from Dubai English Speaking School had a good way of
working which helped them explain why odd numbers have to go at the
Bazahir from Ray Lodge had a good way of
approaching the problem:
That's a good idea, Bazahir.
So, the three solutions will be one with
1 at the bottom of the V, one with 3 at the bottom and one with 5
at the bottom:
However, some of you decided that you
can rearrange the numbers on the arms of the V, keeping the same
number at the bottom, to make different solutions. Alex, Zara,
Frankie, Akeel and Molly from Talbot Primary School, Yujin and Gijs
from Dubai International Academy and James from Crudgington Primary
argued that there are 24 Magic Vs in total for the numbers 1-5.
Gijs took the example with the 3 at the bottom of the V. He
Here are the eight different Magic Vs
with 3 in the bottom:
Can you see how all these have been generated?
Perhaps you can do the same with 1 at the bottom of the V to
convince yourself. So, if we assume these are all different Magic
Vs, there would indeed be 24 in total.
Many of you then looked into Magic Vs which
used different numbers and Magic Vs which were different
sizes.Gemma, Dylan, Faazil and Gabriel from the Dubai English
Speaking School did a very thorough investigation:
Very interesting! Did you try seven
digits which started with an even number, I wonder? What would
happen with 2, 3, 4, 5, 6, 7, 8, for example? They
Many of you I've already mentioned, and Milne
from East Hoathly, agreed with what they have said about the
You can see a summary of George's results
here . Will it
always be possible to create Magic Vs in this way, do you think?
How do we know that we're going to be able to get each arm to add
to the same total?
Ross and Brandon from Ashford Hill represented
the five-digit Magic Vs alebraically which you may find helpful.
What a wealth of responses - thank you
to everyone who sent something in and I'm sorry that we can't
mention everyone. If you would like to contribute anything more to
the solution, do email us.