A two-digit number is such that the product of its digits is 20. When the digits are reversed, the number formed is greater than the original number by 9. Find the number.

Let the two-digit number be $10t + u$, where $t$ is the tens digit and $u$ is the units digit.

Condition 1: The product of its digits is 20.
$$
t \times u = 20 \quad (1)
$$
Condition 2: When the digits are reversed, the number formed is greater than the original number by 9.  
The reversed number is $10u + t$.
$$
10u + t = (10t + u) + 9
$$
$$
10u + t = 10t + u + 9
$$
$$
9u - 9t = 9
$$
Divide by 9:
$$
u - t = 1 \quad (2)
$$

From equation (2), express $u$ in terms of $t$:
$$
u = t + 1
$$
Substitute this into equation (1):
$$
t(t + 1) = 20
$$
$$
t^2 + t = 20
$$
$$
t^2 + t - 20 = 0
$$

Factor the quadratic equation:
$$
(t + 5)(t - 4) = 0
$$
This gives two possible values for $t$: $t = -5$ or $t = 4$.  
Since $t$ must be a positive digit, we take $t = 4$.

Using $u = t + 1$:
$$
u = 4 + 1 = 5
$$

The number is $10t + u = 10(4) + 5 = 45$.

Check: Product of digits $4 \times 5 = 20$. Reversed number is $