149 is a 3-digit positive integer, product of whose digits is 1 × 4 × 9 = 36. How many 3-digit positive integers exist, product of whose digits is 36?

Now, let's find all the combinations of digits (a, b, c) such that a × b × c = 36.

Start with the prime factorization:
 - 36 = 2 × 2 × 3 × 3

Consider all possible groupings of these factors into three groups, where each group forms a digit from 1 to 9.

Here are the unique groupings:
 (1, 4, 9)
 (1, 6, 6)
 (2, 3, 6)
 (3, 3, 4)

Now, let's count the permutations of each grouping to find the number of distinct 3-digit numbers for each combination:

(1, 4, 9):
Permutations: 3! = 6

(1, 6, 6):
Permutations: 3! / 2! = 3

(2, 3, 6):
Permutations: 3! = 6

(3, 3, 4):
Permutations: 3! / 2! = 3

Now, sum all the permutations to find the total number of 3-digit positive integers where the product of the digits is 36:
6 + 3 + 6 + 3 = 18

Thus, there are 18 such 3-digit positive integers.