The number of 5-digit natural numbers, such that the product of their digits is 36, is ______.

We need the number of 5-digit numbers whose product of digits is

$$36$$

Prime factorization:

$$36=2^2\cdot3^2$$

Possible non-zero digit combinations are:

$$36=1\cdot1\cdot1\cdot1\cdot36\quad (\text{not possible})$$

Now list valid digit multisets using digits $$0-9.$$

Case 1:

$$36=1\cdot1\cdot1\cdot4\cdot9$$

Number of arrangements:

$$\frac{5!}{3!}=20$$

Case 2:

$$36=1\cdot1\cdot1\cdot6\cdot6$$

Number of arrangements:

$$\frac{5!}{3!2!}=10$$

Case 3:

$$36=1\cdot1\cdot2\cdot2\cdot9$$

Number of arrangements:

$$\frac{5!}{2!2!}=30$$

Case 4:

$$36=1\cdot1\cdot2\cdot3\cdot6$$

Number of arrangements:

$$\frac{5!}{2!}=60$$

Case 5:

$$36=1\cdot1\cdot3\cdot3\cdot4$$

Number of arrangements:

$$\frac{5!}{2!2!}=30$$

Case 6:

$$36=1\cdot2\cdot2\cdot3\cdot3$$

Number of arrangements:

$$\frac{5!}{2!2!}=30$$

Total numbers:

$$20+10+30+60+30+30=180$$

Hence, the required number of 5-digit numbers is

$$\boxed{180}$$.