Helen made the conjecture that "every multiple of six has more
factors than the two numbers either side of it". Is this conjecture
I start with a red, a green and a blue marble. I can trade any of my marbles for two others, one of each colour. Can I end up with five more blue marbles than red after a number of such trades?
I start with a red, a blue, a green and a yellow marble. I can
trade any of my marbles for three others, one of each colour. Can I
end up with exactly two marbles of each colour?
Adam and Dylan had some great ideas, Sam
explains his thoughts clearly:
Using $3z$ and $4z$ the highest number that can't be made is 5.
This is because you can make 6, 7 and 8.
Mrs Dillon's year 7 class had a different way
of showing 7 is the largest amount that can't be made;
Tze Liang Chee looked into other possible
combinations for other countries: