# Able Mathematicians: Cambridge University "nrich" challenge

Able mathematicians from Years 3 to 6 meet weekly with Mrs Roopun to challenge themselves with investigations from Cambridge University’s ‘nrich’ programme.

This term we are working on mathematical reasoning, and we have found that explaining and communicating our solutions is often harder than solving the challenges in the first place!

We submitted our solutions to the nrich website and received an acknowledgement on the solutions page: "*Thank you to Haider, Isabel, Matthew, Kanika and Arjan all at St Matthew's C of E Primary School, Redhill, Surrey. Thank you for those excellent thoughts and things that you noticed. "*

This term we are hoping to get a full solution published.

Here is an example of one of our challenges and several of year 5/6’s solutions, (and coming soon… some investigations from the3 / 4 able maths group):

**Make 37**

**Stage: 2 and 3 Challenge Level: **

Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags so that their total is 37 .

__Make 37: solution by Kanika__

This challenge is impossible.

I didn’t know this at first, I thought trying to make 37 with the 1s, 3s, 5s and 7s would be easy, especially when I found out that we could use 10 numbers.

I decided to use 7s first, because they were the biggest number. However, if I used all 7s the answer would be too big. So I thought I would have to use 4 of the 7s, which equals 28. This meant I had 6 numbers left to use and I had to make 9 with them. Then I realised that’s not possible, because the only other numbers left were 1s, 3s and 5s and you can’t make 9 using 6 of these.

This challenge is impossible because 37 is odd , all of the numbers you can add are odd as well, and odd plus odd is always even.

__Make 37 solution by Isabel__

The Make 37 challenge is impossible because you are given only odd numbers in the four bags, and you are only allowed to use 10 numbers. It is impossible because a pair of odd numbers always equals an even number, eg 7+5=12, 1+3=4 and 10 is an even number. Making 37 with 10 odd numbers would be 5 pairs of odd numbers so the answer will always be an even number.

You can make 36 and 38 in any different ways, but you will never get 37. That is why it is impossible.

__The solution for ‘Make 37’__ by Haider

Here are a couple of my first tries, which, later on, help to understand the reason why it is impossible to make 37 using 10 odd numbers:

7+1+5+3+7+5+3+3+1+1=36

7+5+7+1+1+1+7+5+3+1=38

3+1+7+5+1+1+5+3+5=38

If you look up there you’ll realise that whatever you do the answer will always be an even number. This happens because a pair of odd numbers will equal an even number. The same happens with 5 pairs of odd numbers.

But, we can make 37 by using an odd number of odd numbers eg 7+5+7+1+1+1+7+5+3=37

This will work because an odd number of odd numbers will always equal an odd number (eg 3+3+3=9)

__The solution for Make 37 by Matthew__

In the beginning I started to try some solutions eg

5+5+5+5+5+7+1+1+1+1=36 and 1+1+1+1+1+1+1+7=7=22.

A little while later I realised that I never reached 37.

After a while I realised I could never get 37 because all the numbers in the bag are odd, and odd plus odd equals even.

As 37 is odd, and 10 is 5 pairs of numbers, the total will always be even and the whole question is impossible.