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?
The triangle ABC is equilateral. The arc AB has centre C, the arc
BC has centre A and the arc CA has centre B. Explain how and why
this shape can roll along between two parallel tracks.
You have 27 small cubes, 3 each of nine colours. Use the small cubes to make a 3 by 3 by 3 cube so that each face of the bigger cube contains one of every colour.
Ling Xiang Ning from Singapore sent the following solution to the last
two parts. Nice work Xiang Ning.
Suppose I start the marble trading with 4 marbles one of each
colour red, blue, green and yellow. We can write the numbers of
each colour in brackets, for example (1,2,3,4) stands for 1 red, 2
blue, 3 green and 4 yellow. For one of each colour we write
(1,1,1,1) and after trading a yellow for one of each of the other
three colours we have (2,2,2,0).
The question is can I trade marbles so that I will have exactly
two marbles of each colour?
The total number of marbles at each stage is given by the
sequence 4, 6, 8, 10, 12, ... whichever colours we trade. There is
no way of getting 2 of each colour because I will have 8 marbles
after 2 trades and whatever colours I have traded I always have an
odd number of each colour at that stage.
The next part of the question asks can I have equal even numbers
of each colour after a number of trades? The answer to this is also
no, this is impossible, as we now show.
It is not possible to have equal even numbers of marbles of each
colour after any number of trades. If we first trade a yellow for
one of each of the other three colours we have (2,2,2,0). If you do
a few trades, say for example, (1,1,1,1) -> (2,2,2,0) ->
(1,3,3,1) -> (2,2,4,2) etc. you can see that the number of
marbles of each colour turns from odd to even at each trade so you
can only have an even number of each colour after an odd number of
Each trade adds 2 to the total number of marble so after an odd
number of trades (remember I have to have an odd number of trades
to get an even number of each colour) I have 6, 10, 14, 18, 22,
.... marbles altogether. After 2n+1 trades I have the 4 marbles I
started with plus 2 times (2n+1) more which is 4 + 2(2n+1) = 4n+6
We can now prove that it is impossible to get an equal even
number of each colour by using an argument by contradiction.
Suppose I were to have an even number, say 2k, of each colour at
some stage, that is a total of 8k marbles which is divisible by 4.
This is impossible because 4n + 6 is not divisible by 4.
Xiang Ning has answered the following questions:
What about ending up with 5 marbles of each colour? Is this
possible and if so how many trades does it take? Is it possible to
end up with the same odd number of each colour for any odd
I noticed that:
Take n as a odd number
n, n, n, n
--> n+1, n+1, n+1, n-1( after first trade)
--> n, n+2, n+2, n (after second trade)
--> n+1, n+1, n+3, n+1(after third trade)
--> n+2, n+2, n+2, n+2 (after fourth (2/2*4) trade, all
numbers are odd)
--> n+4, n+4, n+4, n+4 (after 8(4/2*4) trades)
--> n+6, n+6, n+6, n+6 (after 12(6/2*4) trades)
we take 1 as n
so, 5 is n+4
It is possible to end up with 5 marbles of each colour, and
it takes 8 trades.
If we do a few trades, like (1,1,1,1) --> (2,2,2,0) -->
(3,1,3,1) --> (2,2,4,2) --> (3,3,3,3)..., we can see that the
number of marbles of each number turns from odd to even at each
trade so you can only have an odd number of each colour after an
even number of trades.
Each trade adds 2 to the total number of marbles so after an
even number of trades (Remember I can only get an odd number of
each colour after an even number of trades) I have 8, 12, 16, 20...
marbles altogether. After 2n trades I have the 4 marbles I started
with plus 2 times (2n) more which is 4 + 2(2n) = 4n + 4
As 4n+4 is divisible by 4, it is possible to end up with the
same odd number of each colour for any odd number.
So, it is possible to end up with the same odd number of each
colour for any odd number.
Name: Ling Xiang Ning, Allan
School: Tao Nan School