How many positive divisors $X = 2^2 \cdot 3^3 \cdot 4^4 \cdot 5^5 \cdot6^6 \cdot 7^7$ are divisible by $35$?

How many positive divisors $X = 2^2 \cdot 3^3 \cdot 4^4 \cdot 5^5 \cdot 6^6 \cdot 7^7$ are divisible by $35$?

Since the prime factorization of $N$ is $2^{16} \cdot 3^9 \cdot 5^5 \cdot 7^7$, $N$ has $(16+1)(9+1)(5+1)(7+1) = 8160$ divisors only.

I don't know how to get the count of those numbers that are divisible by 35.

Please feel free to share your ideas on how can I solve this problem.

Divide $X$ by $35$. Now all the divisors of this new number, $2^{16} \cdot 3^5 \cdot 5^4 \cdot 7^6$ correspond one-to-one with the divisors of $X$ which are multiples of $35$, as Will Jagy said.