# Magic Vs

Can you put the numbers 1-5 in the V shape so that both 'arms' have the same total?

## Problem

*The problem is explained below, but you may wish to scroll to the bottom of the page to watch a video of the NRICH team presenting this challenge.*

Place each of the numbers 1 to 5 in the V shape below so that the two arms of the V have the same total.

When the two arms of the V have the same total, we'll call it a 'magic V'.

We'll call the total of the three numbers in an arm of a magic V the 'magic total'.

What is the magic total of each 'arm' in your magic V?

How many other magic Vs can you find with that same magic total?

How do you know you've got them all?

We will assume that if two or more magic Vs have the same magic total, they are the same magic V.

Can you arrange the numbers 1-5 to make some magic Vs with different magic totals?

What do you notice about all the solutions you find?

Can you convince someone that you have all the solutions?

What happens if we use the numbers from 2 to 6?

Can you explain what you notice this time?

Here is a video of the NRICH team presenting the challenge. You might like to watch just the beginning of the video to see a description of the problem, or you might want to watch more of the video, pausing it and working on the task at various points.

Click here for a poster of this problem.

If you have enjoyed this problem, you might like to explore Magic Letters.

## Getting Started

You could use this sheet to record your magic Vs.

Have a go at putting the numbers into the V and then adding up each arm. Perhaps you can adjust the arrangement slightly to make the totals the same?

What do you notice about the number at the base of the V in all your magic Vs?

## Student Solutions

Well done to everybody who had a go at this problem.

Somin, Kim and Georgia from the American International School Chennai in India sent in the following explanation:

To do this problem, we started with the lowest circle on the V, because this would affect both arms. Then, we took the highest of the remaining numbers, and the lowest of the remaining numbers and added them together. We then compared the sum of those to the sum of the two middle/in-between remaining numbers. If the sums were the same, we entered the addends used to make those two sums into the 2 spots on each arm of the V.

We know that we found every solution because when trying to mix up the pairs of numbers to make other sums, none of the sums were equal. The only way you could make another V is if you mirrored one we already made, or switched the places of the top and middle circles of each arm. A pattern we noticed during this project is that every even number in the sequence (the 2nd and 4th numbers, not 2 and 4, for example) did not work as a center number. The other numbers could not be paired to create equal sums. We are including a small picture we made representing our work.

Ethan from Hong Kong International School agreed that 2 and 4 cannot be in the centre of the V:

I think that there are 24 solutions because only odd numbers can be on the bottom circle, and there are three odd numbers. Additionally, the two numbers on both sides can be switched. Also, the two on the right can be switched. Finally, the two on the left can be switched, which makes an equation: 3x2x2x2.

Hwan from Hong Kong International School had some similar ideas:

There are 24 solutions. I tried using equations to solve it. I first found one solution, then found that the left middle and the left top can switch while leaving the right side alone, then switching the right side and leaving the left alone, then switching both sides. Then I switched the entire left side to the entire right side and did the process again. Since I found that there are 4 solutions before switching the entire left to the right, I do 4 times 2 which is 8. Now I have to multiply that by 3 because only the odd numbers can be at the bottom, and there are 3 odd numbers through 1-5. The equation is 2 times 4 times 3 which equals 24.

Here is Maisy and Jessica's submission:

We first found our first three Vs then realised we could find 8 solutions, by just changing one V. This also happened with the other two Vs. We multiplied 8 by 3 because our three vs had 8 solutions. Our answer was 24 solutions.

E.g. To start with, we had 23541 and we swapped the sides to get 41532.

Then we changed round the sides to get 14523. Afterwards, we just swapped one side then another till we couldn't swap it anymore without getting a solution we already had.

Here is Mollie and Amelia's results:

To solve all of the solutions,first we decided to put 1 as the bottom number and worked systematically to get all 8 solutions. Then, we picked then number 3 for the bottom number as we knew it couldn't be even. After finding 8 answers, we got 5 for the number at the bottom. We found all the answers for that number. We stopped here since we couldn't go above 5.There are 24 solutions.

Matty from The British School Manila in the Philippines sent in the following exhaustive solution:

If we have the numbers 1, 2, 3, 4 and 5, and our goal is to arrange all 5 numbers in a way such that the 2 'arms' have the same total. If we get the total of all the give numbers, we get 15. In the middle, we must have an odd number, because if we have an even, say 2, 15-2=13 and 13 is not divisible among the other 2 circles of arms. If we pick 1 as the middle, then each side is required to have a total of 7 on the other 2 circles. For 3, we need a total of 6 on the 2 circles, and for 5, we need a total of 5 on the other 2 circles. Thanks for the question, Nrich.

5,2,1,3,4 5,2,1,4,3 2,5,1,3,4 2,5,1,4,3 3,4,1,5,2 3,4,1,2,5

4,3,1,5,2 4,3,1,2,5 1,5,3,2,4 1,5,3,4,2 5,1,3,2,4 5,1,3,4,2

2,4,3,1,5 2,4,3,5,1 4,2,3,1,5 4,2,3,5,1 1,4,5,2,3 1,4,5,3,2

4,1,5,2,3 4,1,5,3,2 2,3,5,1,4 2,3,5,4,1 3,2,5,1,4 3,2,5,4,1

We have 24 different permutations. Here is the formula:

Slot 1: 4 possibilities

Slot 2: 1 possibility

Slot 3: 2 possibilities

Slot 4: 1 possibility

Middle number: 3 possibilities (3 odd numbers between 1 to 5, inclusive)

So 3*(4*1*2*1) = 24.

We also had some very interesting solutions from North London Collegiate School that are certainly worth viewing here.

magic Vs Y4.pdf magic Vs 3S group.pdf magic Vs Ronis group.pdf

Thank you all so much for these solutions.

## Teachers' Resources

**Why do this problem?**

### Possible approach

In pairs, invite learners to arrange the numbers 1-5 in a V shape. They could use digit cards and/or record using this sheet. Ask them to add the three numbers in each arm. If the arms have the same total, explain that it is a ‘magic V’.

Allow a few minutes for pairs of learners to find some magic Vs. Ask one pair for their magic V and record it on the board for the whole class to see. Explain that you are going to call the sum of the three digits in an arm of a magic V the 'magic total'. What is the magic total of this V?

Challenge the class to find other magic Vs **with this same magic total as the one on the board**. How many can they find? How do they know they have found them all? As they work, look out for those pairs who have developed a system, or way of working, so that they know they don't miss out any possibilities.

After a suitable length of time, encourage a pair to explain how they found all the possibilities, and help the class to articulate how they know they have found them all. Explain to the group that, for this task, we'll consider the **eight possibilities with the same magic total as being the same**.

Invite pairs to use the numbers 1-5 to create magic Vs with **other** magic totals. Bring everyone together again and write up the new magic Vs. Ask children to think about what is the same and what is different about these magic Vs. They might mention that they all use 1-5, but in a different order. They might notice that all the magic Vs have an odd number at the bottom of the V. Challenge learners to explain why an odd number has to go at the bottom of a magic V. If they need a nudge, you could suggest that they find the total of the numbers 1-5. How might this help?

Can they predict what will happen with numbers 2-6? Can they explain their reasoning? Ask them to try it out - were they correct?

### Key questions

What do your magic Vs have in common?

Can you explain why?

What would happen if we used five different consecutive numbers?

Can you explain why?

### Possible extension

For a Christmas version of Magic Vs, see this PowerPoint File, which was created by Rachael Vaughan, a teacher at Bute House Prep School.

### Possible support

Using mini-whiteboards and digit cards may 'free up' some children so that they don't worry about getting a magic V straight away.