If you pick any two integers at random, what is that probability that they will be relatively prime? ("relatively prime" means that the two numbers share no divisors except 1)

Tell how you end up with the answer.

Each number has a 1/2 chance of being divisible by 2, so there is a 1/4 chance of them both being divisible by 2.

Similarly there's a 1/9 chance of both being divisible by 3, a 1/16 chance of both being divisible by 4, etc etc etc.

So my first though was along the lines of...

1 - (1/4 + 1/9 + 1/16 +...+ 1/n^2 +.....)

But this has duplication. For example if both numbers are divisible by 2 and divisible by 3, then they must both be divisible by 6 - which perhaps means that the 1/36 element shouldn't be included. This would give a solution of...

1 - (1/4 + 1/9 + 1/25 + 1/49 +...+ 1/n^2 +...) for all prime n

I have absolutely no idea though as to the value of this sum!

Posted by fwaff on 2003-07-10 08:00:37