The product of the digits of the four-digit number h is 36. No two dig

Question

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

PareshGmat · Accepted Answer

Answer = C = 24

Let number = "abcd"

Given that a*b*c*d = 36

a,b,c,d can be 1,2,3,6

Number of ways = 4*3*2 = 24

Bunuel · Answer

Official Solution:

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
 
We start with the prime factorization of  h , which is  2^23^2 . In other words,  h  is the product of two 2's and two 3's. All the factors of 36, except for 1, can be constructed as products of some or all of these 2's and 3's: 2, 3, 4, 6, 9, 12, 18, and 36. 
 
At this stage, we can see that the possible values of any digit of h only include 1, 2, 3, 4, 6, and 9. Any other digit would contain the wrong primes (5, 7) or too many 2's (8), or it would turn the product to zero (0).
 
We now test subgroups of four digits, imposing the conditions that the product of all four digits is 36 and that no two digits are the same. Start with one digit at a time; do not worry about the order of the digits yet. 
 
9: If one of the digits is 9, then the product of the other three digits must be 4. The only possible sets of positive digits that have 3 members and multiply together to 4 are {2, 2, 1} and {4, 1, 1}, but these sets fail the condition that no two digits can be identical. Thus, 9 cannot be one of the digits of  h .
 
6: If one of the digits is 6, then the product of the other three digits must be 6. The only possible set of positive digits that have 3 non-identical members and multiply together to 6 is {3, 2, 1}, again ignoring order. Thus, one possible set of the digits of  h  is {6, 3, 2, 1}.
 
4: If one of the digits is 4, then the product of the other three digits must be 9. The only possible sets of positive digits that have 3 members and multiply together to 9 are {3, 3, 1} and {9, 1, 1}, but these sets fail the condition that no two digits can be identical. Thus, 4 cannot be one of the digits of  h .
 
If we examine the remaining digits 3, 2, and 1, we see that they can only be part of the unordered set {6, 3, 2, 1}, if we are to satisfy all the given conditions.
 
Thus, the possible values of  h  result from the rearrangement of these four digits. Since all the digits are distinct, the number of different rearrangements is simply  4! , or  (4)(3)(2)(1) , which equals 24.
 
Answer: C.

gmatter24 · Answer

It should be C.

Since the product of the four non-repeating digits is 36, the only possibility is if 1, 2, 3 and 6 are used. These digits can be arranged in 4! ways. Therefore, answer is 24.

DangerPenguin · Answer

1x2x3x6 is the only 4 number combo I can think of and 4!=24 so I would say Answer C!

devmillenium2k · Answer

Only way to get the product of the four non identical digits as 36, when numbers are 1,2,3,6.
1*2*3*6 = 36 
We can arrange these numbers in 4! ways. So the different possible values for h are 24.

Answer C

PerfectScores · Answer

In how many ways can we have product of 4 integers as 36?
1 x 2 x 3 x 6 - which can arrange in 4! ways - 24 ways

akumar5 · Answer

Given that,h is a four digit no.

so x*y*z*w=36=1*6*2*3 ; since zero digit repetition.

Hence the possible nos are = 4*3*2*1=24

Answer is C

evdo · Answer

do we really need to know each digit of the four?

EMPOWERgmatRichC · Answer

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

luisdicampo · Answer

Deconstructing the Question

We need a 4-digit number whose digits are all distinct and whose product is 36.

Since the product is 36, all digits must be from 1 to 9.

Step-by-step

Prime factorization:

36 = 2^2 \cdot 3^2

Find 4 distinct digits whose product is 36.

The only valid combination is:

1, 2, 3, 6

since

1 \cdot 2 \cdot 3 \cdot 6 = 36

Now count permutations of these 4 distinct digits:

4! = 24

Answer C

The product of the digits of the four-digit number h is 36. No two dig

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The product of the digits of the four-digit number h is 36. No two dig

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