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

Explanation

`becauseN=2^(3)*3^(5)*5^(7)*9^(11)`
`=2^(3)*3^(5)*5^(7)*7^(9)*3^(22)`
`=2^(3)*3^(27)*5^(7)*7^(9)`
For perfect square of N, each prime factor must occur even number of times.
2 can be taken in 2 ways (i.e., `2^(0)` or `2^(2)`)
3 can be taken in 14 ways (i.e., `3^(0)` or `3^(2)` or `3^(4)` or `3^(6)` orr `3^(8)` or `3^(10)` orr `3^(12)` or `3^(14)` or `3^(16)` or `3^(18)` or `3^(20)` or `3^(22)` orr `3^(24)` or `3^(26)`)
5 can taken in 4 ways (i.e., `5^(0)` or `5^(2)` or `5^(4)` or `5^(6)`)
and 7 can taken in 5 ways
(i.e., `7^(0)` or `7^(4)` or `7^(4)` or `7^(6)` or `7^(8)`)
Hence, total divisors which are perfect square
`=2xx14xx4xx5=560`