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Bunuel
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How many positive integers with distinct digits have the product of th 14 Apr 2020, 09:00
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Difficulty:

75% (hard)

Question Stats:

47% (01:49) correct 53% (01:45) wrong based on 223 sessions

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How many positive integers with distinct digits have the product of their digits equal to 28?

A. 4
B. 6
C. 7
D. 8
E. 12


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D
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How many positive integers with distinct digits have the product of th Updated on: 14 Apr 2020, 09:17
Originally posted by GMATinsight on 14 Apr 2020, 09:07.
Last edited by GMATinsight on 14 Apr 2020, 09:17, edited 1 time in total.
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Bunuel
How many positive integers have the product of their digits equal to 28?

A. 4
B. 6
C. 7
D. 8
E. 12


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28 = 4*7 or 1*4*7


Numbers made up of digits 4, 7 = 2! = 2. {47 and 74}

Numbers made up of digits 1, 4, 7 = 3! = 6

Total Numbers = 2+6 = 8

Answer: Option D
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How many positive integers with distinct digits have the product of th 14 Apr 2020, 09:10
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GMATinsight
Bunuel
How many positive integers have the product of their digits equal to 28?

A. 4
B. 6
C. 7
D. 8
E. 12


Are You Up For the Challenge: 700 Level Questions

28 = 4*7 or 2*2*7 or 1*4*7

Numbers made up of digits 2, 2, 7 = 3!/2! = 3. {227, 272 and 722}

Numbers made up of digits 4, 7 = 2! = 2. {47 and 74}

Numbers made up of digits 1, 4, 7 = 3! = 6

Total Numbers = 3+2+6 = 11

But there can be many. ore such Numbers with digits 1, 2, 2, 7 = 4!/2! = 12

in fact we can use 1 as many times as we want so there will be countless cases

Bunuel There are infinite cases possible considering digit 1 as many times as we want along with 4 and 7 or along with 2, 2 and 7
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Fixed. Thank you.
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How many positive integers with distinct digits have the product of th 14 Apr 2020, 09:31
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Two digit numbers; The two digits can be 4 and 7: Two possibilities 47 and 74.
Three-digit numbers: The three digits can be 1, 4 and 7: 3! Or 6 possibilities.
We cannot have three digits as (2, 2, 7) as the digits have to be distinct.
We cannot have numbers with 4 digits or more without repeating the digits.
So, there are totally 8 numbers.

correct answer D
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How many positive integers with distinct digits have the product of th 14 Apr 2020, 11:04
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Bunuel
How many positive integers with distinct digits have the product of their digits equal to 28?

A. 4
B. 6
C. 7
D. 8
E. 12


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Asked: How many positive integers with distinct digits have the product of their digits equal to 28?

28 = 1*2*2*7= 1*4*7

Numbers = {47,74,147,174,714,417,471,741}

IMO D
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How many positive integers with distinct digits have the product of th 27 Jan 2026, 09:12
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Failed to understand why 8 is the answer ?
Question says distinct digits are required to be multiplied to get 28.
Prime factors 2^2 * 7^1
(4,7)(7,4)
(14,2)(2,14)

I am not sure if we can include 28,1 and 1,28 because if we involve 1 then we can get more possible outcomes for above 2 cases.
but in either way, I am not getting 8. Can someone help please ?
Bunuel
How many positive integers with distinct digits have the product of their digits equal to 28?

A. 4
B. 6
C. 7
D. 8
E. 12


Are You Up For the Challenge: 700 Level Questions
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Bunuel
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How many positive integers with distinct digits have the product of th 27 Jan 2026, 09:18
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tryorcry
Failed to understand why 8 is the answer ?
Question says distinct digits are required to be multiplied to get 28.
Prime factors 2^2 * 7^1
(4,7)(7,4)
(14,2)(2,14)

I am not sure if we can include 28,1 and 1,28 because if we involve 1 then we can get more possible outcomes for above 2 cases.
but in either way, I am not getting 8. Can someone help please ?

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The numbers are:

47
74

147
174
417
471
714
741
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