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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  [#permalink]

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New post 19 Nov 2014, 07:24
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Tough and Tricky questions: Combinations.



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


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Re: The product of the digits of the four-digit number h is 36. No two dig  [#permalink]

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New post 19 Nov 2014, 07:43
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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.
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Re: The product of the digits of the four-digit number h is 36. No two dig  [#permalink]

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New post 19 Nov 2014, 13:05
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1x2x3x6 is the only 4 number combo I can think of and 4!=24 so I would say Answer C!
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Re: The product of the digits of the four-digit number h is 36. No two dig  [#permalink]

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New post 19 Nov 2014, 20:01
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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
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Re: The product of the digits of the four-digit number h is 36. No two dig  [#permalink]

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New post 19 Nov 2014, 20:25
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
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Re: The product of the digits of the four-digit number h is 36. No two dig  [#permalink]

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New post 19 Nov 2014, 21:59
Bunuel wrote:

Tough and Tricky questions: Combinations.



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


Kudos for a correct solution.


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
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Re: The product of the digits of the four-digit number h is 36. No two dig  [#permalink]

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New post 20 Nov 2014, 03:10
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
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Re: The product of the digits of the four-digit number h is 36. No two dig  [#permalink]

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New post 20 Nov 2014, 07:50
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.
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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: The product of the digits of the four-digit number h is 36. No two dig  [#permalink]

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New post 22 Jul 2015, 08:17
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do we really need to know each digit of the four?
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Re: The product of the digits of the four-digit number h is 36. No two dig  [#permalink]

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Re: The product of the digits of the four-digit number h is 36. No two dig   [#permalink] 16 Oct 2018, 20:17
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