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# M31-23

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Math Expert
Joined: 02 Sep 2009
Posts: 52294

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14 Jun 2015, 12:20
00:00

Difficulty:

85% (hard)

Question Stats:

36% (01:57) correct 64% (02:23) wrong based on 36 sessions

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How many positive integers less than 10,000 are such that the product of their digits is 30?

A. 12
B. 24
C. 36
D. 38
E. 50

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Math Expert
Joined: 02 Sep 2009
Posts: 52294

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14 Jun 2015, 12:20
Official Solution:

How many positive integers less than 10,000 are such that the product of their digits is 30?

A. 12
B. 24
C. 36
D. 38
E. 50

$$30 = 2*3*5 = 6*5$$ (only $$2*3$$ gives single digit number 6).

So, we should count the number of positive integers less than 10,000 with the digits {2, 3, 5} and {5, 6} and any number of 1's with each set.

2-digit numbers:

{5, 6} - the number of combinations = 2: 56 or 65.

3-digit numbers:

{1, 5, 6} - the number of combinations = 3! = 6: 156, 165, 516, 561, 615, or 651.

{2, 3, 5} - the number of combinations = 3! = 6.

4-digit numbers:

{1, 1, 5, 6} - the number of combinations = 4!/2! = 12.

{1, 2, 3, 5} - the number of combinations = 4! = 24.

Total = 2 + 6 + 6 + 12 + 24 = 50.

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Senior Manager
Joined: 31 Mar 2016
Posts: 384
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

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29 Jul 2016, 10:47
I think this is a high-quality question and I agree with explanation.
Current Student
Joined: 10 Jan 2016
Posts: 17
Location: United States (WV)
Concentration: Social Entrepreneurship, Sustainability
GMAT 1: 690 Q47 V38
GPA: 3.72
WE: Project Management (Non-Profit and Government)

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03 Aug 2016, 13:27
Bunuel wrote:
Official Solution:

How many positive integers less than 10,000 are such that the product of their digits is 30?

A. 12
B. 24
C. 36
D. 38
E. 50

$$30 = 2*3*5 = 6*5$$ (only $$2*3$$ gives single digit number 6).

So, we should count the number of positive integers less than 10,000 with the digits {2, 3, 5} and {5, 6} and any number of 1's with each set.

2-digit numbers:

{5, 6} - the number of combinations = 2: 56 or 65.

3-digit numbers:

{1, 5, 6} - the number of combinations = 3! = 6: 156, 165, 516, 561, 615, or 651.

{2, 3, 5} - the number of combinations = 3! = 6.

4-digit numbers:

{1, 1, 5, 6} - the number of combinations = 4!/2! = 12.

{1, 2, 3, 5} - the number of combinations = 4! = 24.

Total = 2 + 6 + 6 + 12 + 24 = 50.

Just to clarify:

For {1, 1, 5, 6} - the number of combinations = 4!/2! = 12. you choose 2! because two of the four places would produce the same number?
Math Expert
Joined: 02 Sep 2009
Posts: 52294

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03 Aug 2016, 21:23
SemperLiberi wrote:
Bunuel wrote:
Official Solution:

How many positive integers less than 10,000 are such that the product of their digits is 30?

A. 12
B. 24
C. 36
D. 38
E. 50

$$30 = 2*3*5 = 6*5$$ (only $$2*3$$ gives single digit number 6).

So, we should count the number of positive integers less than 10,000 with the digits {2, 3, 5} and {5, 6} and any number of 1's with each set.

2-digit numbers:

{5, 6} - the number of combinations = 2: 56 or 65.

3-digit numbers:

{1, 5, 6} - the number of combinations = 3! = 6: 156, 165, 516, 561, 615, or 651.

{2, 3, 5} - the number of combinations = 3! = 6.

4-digit numbers:

{1, 1, 5, 6} - the number of combinations = 4!/2! = 12.

{1, 2, 3, 5} - the number of combinations = 4! = 24.

Total = 2 + 6 + 6 + 12 + 24 = 50.

Just to clarify:

For {1, 1, 5, 6} - the number of combinations = 4!/2! = 12. you choose 2! because two of the four places would produce the same number?

Yes.

THEORY:

Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:

$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$.

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.
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Intern
Joined: 28 Mar 2017
Posts: 4

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27 Feb 2018, 09:15
I think this is a high-quality question and I agree with explanation. This is a very tricky question and a true rep of GMAT type Qs. Thanks, Bunuel.
Intern
Joined: 07 Nov 2017
Posts: 2

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27 Aug 2018, 11:02
in (1, 1 , 5, 6) why it is a permutation while the order does not matter ???
Math Expert
Joined: 02 Sep 2009
Posts: 52294

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28 Aug 2018, 04:19
saranasser wrote:
in (1, 1 , 5, 6) why it is a permutation while the order does not matter ???

We are interested in numbers less than 10,000 such that the product of their digits is 30.

One combination which gives the product of 30 is (1, 1 , 5, 6). But with this combination gives different numbers, isn't it? 1156 is different from 6511 and each of them has the product of their digits equal to 30. Thus we need all numbers which we can get from each group.

Hope it helps.
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Re: M31-23 &nbs [#permalink] 28 Aug 2018, 04:19
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# M31-23

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