Question

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

Answer 1

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 squares $=2\times 3\times 4\times 5=120$.