Hi All,

We're told that the product of the digits of the four-digit number H is 36 and no two digits of H are identical. We're asked for the number of different possible values of H. While this question might seem complex, most of the work is based on basic Arithmetic (and some Permutation math) , so you just need to be careful with your notes and calculations.

To start, we need to find 4 DIFFERENT one-digit numbers that give us a product of 36. You might use prime-factorization or simply 'play around' with the numbers until you find the exact digits...

36 = (2)(2)(3)(3) but we are NOT allowed to have duplicate numbers, so we can 'combine' a 2 and a 3 to give us (2)(3) = a 6 and include the number 1. This gives us...
36 = (1)(2)(3)(6)

How many 4-digit numbers can we create by using each of the digits 1, 2, 3 and 6 just one time each? That's ultimately a Permutation....

There are 4 options for the 1st digit. Once we choose one...
there are 3 options for the 2nd digit. Once we choose one....
there are 2 options for the 3rd digit. Once we choose one...
there is just 1 option for the 4th digit.
(4)(3)(2)(1) = 24 different 4-digit numbers.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

swerve wrote:The product of the digits of the four-digit number h is 36. No two digits of h are identical. How many different numbers are possible values of h?

A. 6
B. 12
C. 24
D. 36
E. 48

The OA is C

Source: Manhattan Prep

In order for 4 distinct digits to multiply to 36 we have:

1, 2, 3, 6

Since the 4 digits can be arranged in 4! = 24 ways, there 24 possible values for h.

Answer: C