7. If $p$ and $q$ are both prime numbers, the number of divisors $d(a)$ of the natural number $a=p^{\alpha} q^{\beta}$ is given by the formula
$$
d(a)=(\alpha+1)(\beta+1)
$$

For example, $12=2^{2} \times 3^{1}$, so the number of divisors of 12 is
$$
d(12)=(2+1)(1+1)=6,
$$

with the divisors being $1, 2, 3, 4, 6$, and 12.
According to the given formula, please answer: Among the divisors of $20^{30}$ that are less than $20^{15}$, how many are not divisors of $20^{15}$? $\qquad$
7.450 .

Notice that, $20^{30}=2^{60} \times 5^{30}$.
Then $d\left(20^{30}\right)=(60+1)(30+1)=1891$.
Also, $20^{30}=20^{15} \times 20^{15}$ is a perfect square, and except for $20^{15}$, for any divisor greater than $20^{15}$, there is a corresponding divisor less than $20^{15}$, whose product is exactly $20^{30}$.

Thus, among the divisors of $20^{30}$, there are $\frac{1891-1}{2}=945$ divisors less than $20^{15}$.
Furthermore, the divisors of $20^{15}$ are all divisors of $20^{30}$, and $20^{15}=2^{30} \times 5^{15}$,
so $d\left(20^{15}\right)=(30+1)(15+1)=496$.
Therefore, among the divisors of $20^{30}$ that are less than $20^{15}$, those that are not divisors of $20^{15}$ total $945-496+1=450$ in number.