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Rhiannon from St Mary Redcliffe Primary School sent a very well-explained solution to this problem. She wrote:

On the 15th beat, you will be doing a click. I know this because 15 is in the three times table and there is a click every three beats.
On the 20th beat, you will be doing a clap, because 20 is not in the three times table.
On the 99th beat you will be doing a click, because 99 is in the three times table. I know this because 9+9 = 18 and 1+8=9 and 9 is in the three times table.
On the 100th beat you will be doing a clap because you did a click on the 99th beat.

If two people do the following rhythms:
person 1: clap clap click
person 2: click clap clap
then you won't click at the same time.
On person 1, you click if the number is dividable by three (we can say "divisible by 3") .
On person 2, you click if the number has remainder 1 when divided by three. Since a number can never ever be dividable by three and remainder 1, you won't click at the same time.
To have the same result, person 1 could go "clap click clap", person 2 could go "clap clap click".
So that you do click at the same time you could:
person 1: clap clap click
person 2: clap clap clap click
You would both click at the same time every 12 beats.

Fantastic, Rhiannon! Rohan from Longbay Primary School suggested an alternative:

Change your friend's rhythm to: click clap click etc. and keep your rhythm the same. 9.
A click on your rhythm occurs every 3rd beat and a click on your friend's rhythm will occur every 1st and 3rd beat so you will both click at the same time on the 3rd beat.

Well done to everyone who sent in solutions.