It seems to be a good strategy for Player 1 to choose a prime number. Here is one way to justify that.
Claim: If \(p\) is prime and \(p\) divides \(ab\), then \(p\) divides \(a\) or \(p\) divides \(b\).
Note that ‘or’ in maths is inclusive, so you should read the conclusion above as “\(p\) divides \(a\) or \(p\) divides \(b\) or both”.
Assume that \(p\) is prime and that \(p\) divides \(ab\).
If \(p\) divides \(a\) then we are done, so assume that \(p\) does not divide \(a\). We want to show that \(p\) must divide \(b\).
Since \(p\) is prime, the highest common factor of \(a\) and \(p\) is \(1\).
There are integers \(m\) and \(n\) such that \(am + pn = 1\).
You might have found this while exploring the open-ended investigation Buckets and ponds.
So \(abm + pbn = b\).
The left-hand side is a multiple of \(p\).
So \(p\) divides \(b\), as required.
One good thing about this argument is that it doesn’t assume any knowledge about prime factorisation, so we can use this result to prove facts about prime factorisation. You might like to try the resource The Fundamental Theorem of Arithmetic to see that in action.
We might wonder whether Player 1 can ever have a winning strategy if the number is not prime. That would be a good question to consider…