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Joe from Bishop Ramsey School looked at the seven mat problem. He said:

In this problem it doesn't matter where you start on the diagram. My solution is written in a number of stages:1) pick any 3 hexagons (they will turn
blue)

2) pick 2 more hexagons and then pick one
that you chose in the first go. You should now have 4 hexagons
blue.

3) finally pick the three remaining red
hexagons.

It would only have taken you 3
goes.

Many of you answered this part well, including Alistair of Cottenham Primary School.

Kahlia and Amy from Ardingly College Junior School then looked a bit further and tried with other numbers of mats:

If the number of tiles is a multiple of 3 it will divide equally into the number of tileseg: for 60 tiles, remove three each go:
60 divided by 3 = 20 turns

Amelia and Kathryn, also from Ardingly College Junior School, investigated many different numbers of mats in a very systematic way:

6= 2 moves

7= 3 moves

8= 4 moves

9= 3 moves

10= 4 moves

11= 5 moves

12= 4 moves

13= 5 moves

14= 6 moves

15= 5 moves

16= 6 moves

17= 7 moves

18= 6 moves

19= 7 moves

20= 8 moves

Kahlia and Amy identified a pattern:

If there is a number of tiles 1 more than
a multiple of 3 you add 1 to the answer of the multiple below it
eg: 18 tiles = 6 turns; 21 tiles = 7 turns

17 tiles = 7 turns; 20 tiles = 8
turns

Jeff and Raphael from Zion Heights Junior High School relate this back to the strategy for flipping the mats:

For numbers with one remainder after
dividing by three, you follow the strategy for 7 mats stated
above.

For numbers with 2 remainders, you flip
over two mats instead of one on the second move, thus taking one
more move.

So, thinking about this like Kahlia and Amy did, we could say that if the number of tiles is 2 more than a multiple of 3, you add 2 to the answer of the multiple below it.

Well done to everyone who tackled this problem - it wasn't easy at all.