Question

If p and q are distinct primes. Find the number of positive divisors of
p^2 * q^2

Answer 1

Answer:

9

Step-by-step explanation:

We know that for $p^mq^n$  where p and q are different prime numbers the number of positive divisors are (m+1) (n+!)

We have given $p^2q^2$

So here m=2 and n=2

So number of positive divisors $=(2+1)\times (2+1)$ =9

So the number of positive divisors of  $p^2q^2$ is 9

Answer 2

Answer:

The number of positive divisors are 7.

Step-by-step explanation:

Since it is given that p and q are prime numbers thus they will not have any divisors other that itself and 1

The number $p^{2}\times q^{2}$ can be represented as

$p^{2}\times q^{2}=p\times (p\times q^{2})\\p^{2}\times q^{2}=(p\times q)\times (p\times q)\\p^{2}\times q^{2}=(p^{2}\times q)\times q$

Thus the number $(pq)^{2}$ can be divided by:

1) $1$

2) $p$

3) $q$

4) $pq$

5)$p\times q^{2}$

6)$p^{2}\times q$

7)$(pq)^{2}$

Thus the number of positive divisors are 7.