Thursday, June 29, 2006

The cogwheels

A new day starts in the morning, and thus comes a new Perelman problem for you.

There are two cogwheels, a large one with 24 teeth, and a small one with 8 teeth. They are geared together such that the large one stays fixed in the center while the small one rotates around. How many revolutions will the small cogwheel perform while going once around the large cogwheel?

5 comments:

Anonymous said...

I'm going for the obvious answer, which probably isn't right. 3 revolutions.

If there is 24 four teeth then there are 24 spaces. For the small cog to do one revolution would be 8 teeth, or eight spaces on the big cog. 8 divided by 24 is 3.

Anonymous said...

Er, 24 divided by 8 is 3. Oops.

Anonymous said...

I'll go for 4 because as the little cog goes around the bigger one it goes in a circle.

It would be 3 if the 24 teeth were laid in a straight track

Anonymous said...

I'm going for 4 as well.

Let's reduce the number of teeth to 12 on the big cog an 4 on the small one.

The 12 teeth (or better the valleys between the teeth) indicate the hours of an analog clock. The 4 teeth on the small cog indicate the hours 3, 6, 9 and 12.

Let's start with 6 of the small cog diving into 12 of the big cog. This means that 3 of the small cog looks in the direction of 3 o'clock.

Now move the small cog from 12 on the big cog to 1 on the big cog. This means that now 3 of the small cog dives into 1 of the big cog, which means that the 3 now looks in the direction of 7 o'clock.

7 - 3 = 4, i.e. moving one tooth ahead means moving 4 hours ahead on the small cog. After 3 moves the 3 of the small cog again looks in the direction 3 o'clock.

As there are 12 teeth and thus 12 valleays on the big clog, the small clog does 12 / 3 = 4 revolutions.

Andres said...

Counterintuitively, the answer is 4 revolutions.

I really like the approach in comment #4. Fantastic!