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?

Question

Shikha Rani · Accepted Answer

Now, let's find all the combinations of digits (a, b, c) such that a &times; b &times; c = 36.
Start with the prime factorization: - 36 = 2 &times; 2 &times; 3 &times; 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.