Chocolate


The sharing of the three tables with chocolate bars on.
Well this really proved to be quite difficult!
I have three pieces of work that have resulted from trying this problem before it went on the Internet. Robert's solution
  Number A,B or C How much
1 A 3
2 B 2
3 A 1 1/2
4 C 1
5 B 1
6 A 1
7 A 3/4
8 B 2/3
9 A 6/10
10 C 1/2
11 B 1/2
12 A 1/2
13 A 15/35
14 B 4/10
15 A 15/40
16 C 1/3
17 A 1/3
18 B 1/3
19 A 3/10
20 B 2/7
21 A 3/11
22 C 1/4
23 B 1/4
24 A 1/4
25 A 3/13
26 B 2/9
27 A 3/14
28 C 1/5
29 B 1/5
30 A 1/5
You get A B A everytime in the fractions between 1, 1/2, 1/3, 1/4, 1/5: Fraction A Fraction B Fraction A 1 1 1 Fraction A Fraction B Fraction A 1/2 1/2 1/2 Fraction A Fraction B Fraction A etc.
 
This was rather nice and he noticed a good pattern that had emerged in the choice of table visited and the fraction received. Table A, B, C have 3, 2 & 1 piece of chocolate. So he noticed that the people come in and go to table A then B then A getting a different amount each and then the next three people get the same amount; this pattern, as he shows repeats itself. It's also rather nice that the three equal fractions that occur after the A B A gradually go down in order :- 1, 1/2, 1/3, 1/4, 1/5.


This helps you to see very easily how much children would get when there were so many at any table. However, as the note says, she needed to get pieces of paper and do some folding and tearing in order to compare things like 2/3 and 3/4 to find out which was bigger. The last example was done jointly between three 11 year old boys. Looking down the column of 3 bars, 2 bars and 1 bar you see the number of the person entering the room and the amount that they would have received if the sharing had taken place. Doing it this way shows the interesting pattern that comes from the number of those people sitting around a table. So for example, on the 2-bar table at the end were people who had come in as number, 2, 5, 8, 1, etc. all with a rise of 3 each time. It was good to spot that and to notice that the three patterns could be arranged to give a total of 6.


 
These three patterns all add up to 6: 2 + 1 + 3 = 6 3 + 3 = 6 6 = 6 It leads me to want to find out whether if there had been 4 tables and so 10 blocks of chocolate altogether, the patterns would all have added up to ten!These three patterns all add up to 6: 2 + 1 + 3 = 6 3 + 3 = 6 6 = 6 It leads me to want to find out whether if there had been 4 tables and so 10 blocks of chocolate altogether, the patterns would all have added up to ten!