UKMT Cayley Olympiad Answers

Don't trust these answers! These are the answers I got when I had finished. In all honesty, they may be all wrong!

1) 784
3) 2
4) 142857
5) 5
6) Two ninths.

I would be really interested in hearing what you guys got!

Remember, I'm definitely not sure about Questions 5 and 6.

Reply 1

Shailinder

How did you get question's 4 and 6, if you don't mind explaining? Did you do all 6 questions? I assume you did all of them. I could do 1 to which the answer is 784. 2 is an easy proof question. As for 3, I got two sequences as well. As four 4, I'm confused how you got the answer but it's brilliant because the answer is correct! How did you get there? As for 5, I answered 5 as well, but then again, I'm not sure at all for question 4. Then, I had no idea how to do question 6. How did you do it?

Reply 2

tigerater

4) Let n be the number of digits that m has.

(m-10^(n))x10=3m

Let us trial n=1
After some simplification, you should end up with 7m=99
If we trial n=2, we end up with 7m=999
Eventually, I found proof that it continues 9999, then 99999, but I can't remember it now =(.
The smallest number is 7m=999999
This means that the smallest such m is 142857.

Reply 3

tigerater

6) Mirror each alternate semicircle around the edge so that there are 6 identical circles surrounding the center circle.

If we draw a line from center to center of the surrounding circles, we end up with a hexagon. If we draw a line from each vertex to the center of the middle circle, we can prove that the center circle is the same as the surrounding circles.

From then on, we can just find that the area of the outer circle is 9 x Pi x Radius^2

The area of one of the smaller circles is Pi x Radius^2, so all seven would be 7 x Pi x Radius^2, thus the area of the shaded area would be 2 x Pi x Radius^2.

2 x Pi x Radius^2 over 9 x Pi x Radius^2 simplifies to 2 over 9.

A really clever boy told me afterwards that he had gotten 11 over 14. If you got this answer, then you're probably right.

(edited 11 years ago)

Reply 4

tigerater

3) All you had to do was replace the first 2 numbers with X and Y. I am pretty sure that the 8th term was something like 8X+13Y.

From there on it was just trial and error to find solutions that worked.

I think that there were only 2 sequences which worked, but I'm not positive, as my friends got different answers.