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How many positive integers less than 10,000 are such that the product of their digits is 210?

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by BTGmoderatorDC » Tue Nov 23, 2021 6:54 pm

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How many positive integers less than 10,000 are such that the product of their digits is 210?

A. 24
B. 30
C. 48
D. 54
E. 72


OA D

Source: Magoosh
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Re: How many positive integers less than 10,000 are such that the product of their digits is 210?

by swerve » Wed Nov 24, 2021 2:40 pm
BTGmoderatorDC wrote: ↑
Tue Nov 23, 2021 6:54 pm
 How many positive integers less than 10,000 are such that the product of their digits is 210?

A. 24
B. 30
C. 48
D. 54
E. 72


OA D

Source: Magoosh
\(210 = 2\times5\times3\times7 = 5\times6\times7\times1 = 5\times6\times7\)

Those are the only sets of digits we can use to for the numbers (any other combination of factors will have two digit factors).

Numbers using \(2,5,3,7 = 4!\)
Numbers using \(5,6,7,1 = 4!\)
Numbers using \(5,6,7\) (3-digit numbers) \(= 3!\)

Answer \(= 24+24+6 = 54\)
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Re: How many positive integers less than 10,000 are such that the product of their digits is 210?

by Brent@GMATPrepNow » Thu Nov 25, 2021 6:43 am
BTGmoderatorDC wrote: ↑
Tue Nov 23, 2021 6:54 pm
 How many positive integers less than 10,000 are such that the product of their digits is 210?

A. 24
B. 30
C. 48
D. 54
E. 72


OA D

Source: Magoosh
210 = (2)(3)(5)(7)

We need to consider 3 cases:

Case 1: 4-digit numbers using 2, 3, 5, 7
There are 4 digits, so this can be accomplished in 4! (24) ways

Aside: Notice that (2)(3) = 6

Case 2: 4-digit numbers using 1, 6, 5, 7
There are 4 digits, so this can be accomplished in 4! (24) ways

Case 3: 3-digit numbers using 6, 5, 7
There are 3 digits, so this can be accomplished in 3! (6) ways

Add up all 3 cases to get 24 + 24 + 6 = 54

So, the answer is D

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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