Term 3 - July 15th (F-2)

Foundation to Year 2: School Fair Necklaces

Rob and Jennie were making necklaces to sell at the school fair.

They decided to make them very mathematical.

Each necklace was to have eight beads, four of one colour and four of another.

And each had to be symmetrical, like this.

Kimie put her blue sticks end to end in a long line. Sebastian put his red sticks end to end in a line underneath Kimie’s.

How many different necklaces could they make?

Can you find them all?

How do you know there aren’t any others?

What if they had 9 beads, five of one colour and four of another?

What if they had 10 beads, five of each?

Kimie needs three sticks, Sebastian needs two, the line is 6 altogether.

Other solutions are:

I started with nine beads and worked out that because it has to be symmetrical the bead at the bottom must be of one colour because the clue says four beads of one colour and five beads of the other colour. So I thought yellow could be at the bottom and my pattern was:
Y,G,Y,G,(Y),G,Y,G,Y.

Then I worked out 13 beads. I thought because it is another odd number the bead at the bottom must be the colour of bead with the most in it. So I worked out the pattern which was
Y,G,Y,G,Y,G,(Y),G,Y,G,Y,G,Y.

Then I tried to work out what 10 was. Then I tried it and thought that it wouldn’t work because it was even but I was wrong. So I tried it again and got the pattern right which was:
Y,G,Y,G,(Y)(Y),G,Y,G,Y.

I then moved onto 12. Because 10 worked 12 MUST work so I worked out the pattern which was: Y,G,Y,G,Y,(G)(G),Y,G,Y,G,Y.

Finally, I did 11 beads. And worked out the pattern which was:
G,Y,G,Y,G,(Y),G,Y,G,Y,G.


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