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We had lots of solutions sent in for this activity, so thank you to everybody who shared their ideas with us.
Anika from the National Academy for Learning in Bangalore, India sent in a table of possibilities for the numbers in each envelope, with this explanation:
I made five columns, one each for the envelopes. For each number, I wrote down the possible pairs that could be in the envelope. I found that the envelope 8 can contain the numbers (0,8),(5,3),(1,7),(6,2).
Well done for finding all these possibilities, Anika. In Anika's table we can see all the pairs of numbers that make the total on each envelope.
Leo from Kings' School in Dubai, UAE used the same idea as Anika to draw a table of all the different possibilities. He then worked out which of these possibilities would work with each other, as the same number can't be in more than one envelope:
I started off by creating a table of all of the possible number bonds to each of the targets on the envelopes.
I then worked systematically so that I tried every potential pathway across the table. Some of them proved to be impossible but I was able to find all 3 solutions. I have also included an example of a pathway that was impossible; it was where I started the process.
I know that there are only 3 possible solutions as I worked through every possibility in a systematic fashion.
Well done for working systematically through all of the different possibilities, Leo. Leo's solution shows that the "8" envelope can contain 0 and 8, 5 and 3, or 1 and 7, but it looks like the cards 6 and 2 can't be in the "8" envelope. I wonder why?
James and Oais from the UK found the same solutions as Leo, and they explained why 6 and 2 can't be in the "8" envelope:
We made a chart with all the possibilities for making the numbers on each envelope, then we tested the different ways it could go.
If you start by looking at how to make 14, there are two: 5/9 or 6/8. Then, if you try to make 13, you can see that there is only one way that works for each. When you choose 5/9 for 14, you have to choose 6/7 for 13. When you choose 6/8 to make 14, you have to choose 4/9 to make 13 because you've used up the other numbers.
The numbers you have to make 8 are 0/8, 1/7, 2/6 or 3/5 but you'll never be able to use 2/6 because you always have to use the 6 to make 13 or 14. So the answer is 0,8,1,7,3 or 5!
Good explanation! I wonder if it's more helpful to start with the smallest numbered envelopes like Leo did, or to start with the largest numbered envelopes like James and Oais did?
Thank you as well to these children who sent in excellent solutions: James from Hamstel Junior School in England; Mia and Hari from Kings' School in Dubai; Anna from Sahuaro Elementary in the USA; Lauren from Australia; K from Crossflatts in the UK; Saanvi from Newcastle-under-Lyme in the UK; Isla from Walton and Holymoorside Primary in the UK; and Ci Hui Minh Ngoc from Kong Hwa School in Singapore.
An investigation involving adding and subtracting sets of consecutive numbers. Lots to find out, lots to explore.
Roll two red dice and a green dice. Add the two numbers on the red dice and take away the number on the green. What are all the different possible answers?
Use the 'double-3 down' dominoes to make a square so that each side has eight dots.