Find the number of divisors of the number `N=2^3 .3^5 .5^7 .7^9` which are perfect squares.

Explanation

Since the divisor is a perfect square, each prime factor must occur even number of times.
2 can be taken in 2 ways `(2^(0) " or" 2^(2))`
3 can be taken in 3 ways `(3^(0) " or" 3^(2) " or" 3^(4))`
Similarly, 5 can be taken in 4 ways `(5^(0) " or" 5^(2) " or" 5^(4) " or " 5^(6))` and 7 can be taken in 5 ways `(7^(0) " or " 7^(2) " or" 7^(4) " or " 7^(6) " or " 7^(8))`
Hence, total divisors which are perfect square `=2xx3xx4xx5=120`