Hint: Try breaking each cog size into prime factors.
Answer: 30 revolutions.
If we break each wheel into its prime factors, we get:
63 = 3 x 3 x 7
42 = 2 x 3 x 7
35 = 5 x 7
27 = 3 x 3 x 3
We now think of rotating the large wheel just once, and this is 3 x 3 x 7 teeth moved (3 x 21), and we can see that 42 tooth wheel also has a 3 x 7 (21 teeth) in it, with an extra 2. If we therefore rotate the 63 toothed wheel twice, the 42 will have rotated three times.
The answer involves cancelling any common factors from the large wheel. We can cancel 3, 3, 7 from any of the smaller ones to leave 2 (from the 42), 5 (from the 35) and 3 (from the 27). 2 x 5 x 3 = 30.
Recently Michael had a party and because it was his birthday he wanted a mathematical cake.
He ordered a cake in the shape of a cube, and had it completely iced (but of course he didn't have the underneath iced because that just isn't very sensible).
Michael cut every side of the cake into three equal pieces, giving him 27 slices of cake.
Michael wanted to eat all of the pieces of cake that had exactly 2 sides with icing.
So, how many pieces did that leave for his guests?
Hint: How many pieces on the top layer had two sides with icing?
Answer: 15 pieces.
On each layer there are 4 pieces with exactly two sides of icing.
Therefore Michael ate 12 pieces, leaving 15 for his guests.