Copyright © University of Cambridge. All rights reserved.

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 Girls said:

The solution we found was that you always put an odd number at the bottom when it was 1-5.There were three odd numbers and two even numbers. To make it balance you've got to get rid of an odd number so you put an odd number at the bottom, because then you have two odds and two evens.

You have to put the biggest remaining odd number with the smallest even number on one arm, and the smallest remaining odd number with the biggest even number on the other arm to make it work.

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 bottom:

We first started off by putting 1 in the bottom of the V and
then adding up the remaining digits (2, 3, 4, 5) which came to 14.
So, half of 14 is 7 and 2+5 and 3+4 make 7 so they go either side
of the V.

We then tried 2 and realised that the remaining digits (1, 3,
4, 5) added up to 13 which meant we couldn't 'even up' either side
of the V.

We then thought that only odd numbers could go at the bottom
of the V.

Bazahir from Ray Lodge had a good way of approaching the problem:

I forgot about the base and found two sums that have the same
total using the numbers 1-5 then I put the base number.

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 wrote:

5 add 1 is 6 and 4 add 2 is also 6. There are eight possibilities when there is a number at the bottom. The numbers on the sides can be switched around lots of different ways and they still make 6. The numbers have to stay together, though, you can't do 5 and 2 because that makes 7.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:

We then tried 2, 3, 4, 5, 6 and put the odd number at the bottom and soon realised that the remaining digits added up to an odd number which meant we couldn't fill in the V evenly. So, we decided that when you start a five-digit sequence with an odd number, the number at the bottom of the V has to be an odd number, if you start a sequence with an even number then the number at the bottom of the V has to be even.

Our teacher then told us to do the same with seven digits (1, 2, 3, 4, 5, 6, 7) and we soon realised that this time if you started off with an odd number at the bottom of the V the remaining digits added up to an odd number so it wouldn't work but, if we started off with an even number at the bottom of the V then we could even up both sides. So the solution for a seven-digit V was opposite to the one for a five-digit V.

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 continued:

Our homework was to find out how we could predict whether a V
with 35 digits would have to have an odd or an even number at the
bottom of the V to work. Gemma came up with most of the solution as
she said that if you add up all the digits and the sum was an even
number then it would need an even number at the bottom of the V to
work. If the sum was odd then the number at the bottom of the V
would have to be odd.

Many of you I've already mentioned, and Milne from East Hoathly, agreed with what they have said about the numbers 2-6.

George sent in a general method for finding
Magic Vs:

Add all numbers of the consecutive series - which need to be
an odd number of numbers, whether there are five numbers, seven
numbers.

Let the addition of these numbers equal A (AE for an EVEN
number) (AO for an ODD number)

If the total answer is an even number, subtract one of the
even numbers from that set of numbers. Let this remaining number be
B.

Else, if the total answer is an odd number, subtract an odd
number from that set of numbers. This remaining number can also be
called B since it to is an EVEN number. [The even or odd number
that is initially subtracted will be the number at the bottom of
the Magic V.]

Next divide B, the total left, by TWO. i.e. B/2

From the EVEN quantity of numbers left to be considered after
removing the initial EVEN or ODD number, take half of the numbers
and determine whether it is possible to add them so that they will
equal B/2.

If half the quantity of number left does equal B/2, then the
numbers left after adding these numbers together will also equal
B/2. One set of these numbers that equal B/2, can fill the spaces
on the left leading down to the bottom of the Magic V. The other
numbers that equal B/2, can fill the spaces on the right leading
down to the bottom of the Magic V.

After the first attempt try subtracting a different EVEN or
ODD number from the total AE or AO and go through the sequence to
see if any other combinations are possible.

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. They said:

If n is the first number then the other numbers are n+1 n+2
n+3 n+4, the three solutions for every number are:

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.