##### *All hints are included below so please scroll down slowly to avoid revealing them all at once*

****Solutions are given at the very bottom of the page****

#### Puzzle 1 – Pulleys

“You’ll need to stand on the platforms and work out how far the water weight must fall to raise you.” Terry squawks from above. Can you work out the distances in each case?

#### Hint 1

**“Try to use the fixed pulleys as reference points and consider how far the bucket moves if a fixed length of rope passes around them.”**

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##### *Scroll down for hint 2*

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#### Hint 2

**“If 2 metres of rope passes over pulley A, how much does pulley B drop and hence the bucket?”**

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##### *Scroll down for hint 3*

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#### Hint 3

**“For the second setup, is it possible for the distance between the bucket and the smallest pulley to change?”**

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##### *Scroll down for hints to the chameleon puzzle*

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#### Puzzle 2 – chameleons

The chameleons are all green, blue or red and when two chameleons of different colours meet, they both turn into the remaining colour. Right now I count 14, 15 and 16 of each colour respectively. For years I’ve been waiting to see if they could all at some point be the same colour. Can you tell me how they must meet to do so, or if it is even possible?” You gather that in any encounter there are only ever two chameleons present. Can you help Zora out?

#### Hint 1

**“In each encounter, how do the numbers of each colour change? How do they change relative to other colours?”**

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##### *Scroll down for hint 2*

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#### Hint 2

**“It may help to consider the difference between the numbers of green and blue chameleons for example. What values must this difference take if all chameleons become one colour?”**

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##### *Scroll down for hint 3*

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#### Hint 3

**“Start by considering the difference between red and green chameleons, it starts at 2. In each possible encounter, how does this number change? Can this number ever become what it needs to be for all chameleons to be one colour?”**

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##### *Scroll down for solution to the pulley puzzle*

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#### Puzzle 1 (pulleys) – Solution

The first puzzle is a very common setup for lifting heavy loads, though in this case it is reversed. Notice that if the water bucket moves down X metres, then the rope coming up on either side of pulley B must have lengthened by X metres. Thus in total requiring 2X metres of rope to be passed over pulley A. Thus, for the platform to rise 6 metres, the bucket must fall 3. This would require the bucket to weigh twice as much as the platform, not very practical for everyday use.

The second setup is in some sense a trick question. Focus first on the loop around the two largest pulleys, ignoring the smaller pulley for now:

This is just a closed loop of rope and so there is no way to shorten or lengthen it. Instead, it is simply free to rotate around the pulleys like a chain around two gears. Now introduce the pulley with the water weight and the platform as separate objects attached to the loop. Now we are effectively left with a scale, whichever side is heaviest will pull the loop around so that the heaviest weight lies at the bottom of the system. Thus in this case, if the platform moves up 10 metres, the bucket moves down 10 metres, with no rope sliding over the smallest pulley.

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##### *Scroll down for solution to the chameleon puzzle*

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#### Puzzle 2 (chameleons) – Solution

Solving this problem may seem initially difficult but, as with many maths problems, considering the right quantities can hugely simplify the question. Let **D** be the number of red chameleons subtract the number of green chameleons (so **D** starts at 2). How does **D** change with each encounter?

If all the chameleons were the same colour, then they would all be either red, green or blue. This corresponds to **D **= 45, -45 or 0 respectively. Now notice that **D**=2 is not divisible by 3 to start with. So adding or subtracting 3 won’t change this. Thus, no matter what encounters we have, **D **will never be divisible by 3. But 45, -45 and 0 are all multiples of 3. This means **D **will never equal any of these values, so the problem is impossible.

In other words, solving this problem came down to finding a quantity that was constant **modulo 3, **that is **D **always having remainder 2 when divided by 3. This is an example of using modular arithmetic which can be very powerful with the right type of problem. You can find an introduction to it here.