Three children are going to buy some plants for their birthdays. They will plant them within circular paths. How could they do this?
Use the number weights to find different ways of balancing the equaliser.
If you have ten counters numbered 1 to 10, how many can you put
into pairs that add to 10? Which ones do you have to leave out?
We had some good solutions sent in with some diagrams to support them. From Isseya St Andrews School Sukhumvit inThailand we had this sent in;
We had correct solutions from Ethan at the Good Shepherd Lutheran School in Australia, Rebecca, Ryan and Georgia from Falcon Junior School in England.
Emerald and Brooke from Myland School in England sent in the following interesting account of their work;
First I tried putting any numbers into the circles. After a few goes I made this triangle where the sides add up to $11$, $12$ and $13$. Because $11$, $12$, $13$ was so close, I swapped the $3$ for the $2$ to make $13$ one smaller and $11$ one higher. Then all the sides total $12$.
I tried to add up to nine lots of different ways only using numbers $1$ to $6$. There were only $3$ different ways $6 + 1 + 2 = 9$, $1 +5 + 3 = 9$ and $3 + 4 + 2 = 9$. I found that I couldn?t have $5$ and $2$ on the same side because you would need another $2$ to make it add up to $9$. I put the three sets of numbers into the triangle to see if they match up to make nine. I found that I needed to
hide the numbers $4, 5$ and $6$ on a middle, not a corner because they were only used once in the ways of making $9$. For the $10$ triangle the ways to make $10$ are $5 + 3 + 2 = 10$, $5 + 1 + 4 = 10$, $6 + 1 + 3 = 10$. You can't have a $4, 2$ or $6$ on a corner because they are only in one way of making $10$.
To make the $11$ triangle I started using Brooke's method. I made $11$ by adding $6 + 4 + 1$ and $5 + 2 + 4$. I put those sides on a triangle with the $4$ at the top because it was in both sentences. The only number missing was $3$, so that had to go in the empty circle. That made the bottom side total $14$. I swapped the $5$ and the $2$ to make the bottom side less. All the sides
I guessed that we could make other triangles with the numbers $1$ to $6$. I tried to make triangle totals of $3, 5, 8, 13$ and $15$, but none of them could be done. For the $8$ triangle there were not two numbers to go with $6$ that would make $8$. For the $13$ and $15$ triangles there were not two numbers to go with $1$ to make $13, 15$ or higher numbers.
From Valley Road Primary School the following came from Lauren;
Well in school I was quite scared when I saw it but then we had to work with our talk partners and then I really enjoyed it I didn't want to stop. My solution is: I had counters $1$ to $6$ then I put them in any old order but then I then knew that I had to have them all equal to $11$ or $10$ so then I got them all.
Thank you for being so honest Lauren and thank you for all the emails that were sent in, on this
occasion no one sent any ideas into the blog .