We have a natural square-free number $n$ such that $2^5\cdot 3^6\cdot 5^4\equiv 0 \pmod {\tau(n)}$.
Which is the maximum number of different primes that can divide $n$ ?
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We have that $\tau(n)$ is the number of positive divisors of $n$. So we have that $\tau (n) = \alpha\cdot 2^5\cdot 3^6\cdot 5^4$.
But how does this gives us the number of prime divisors?