positive integers

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GmatKiss: Legendary Member; Posts: 2789; Joined: Tue Jul 26, 2011 12:19 amFollowed by: 43 members; Location: Chennai, India; Thanked: 206 times; GMAT Score:640

by GmatKiss » Mon May 21, 2012 5:27 am

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

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Source: Beat The GMAT — Problem Solving | Mon May 21, 2012 5:27 am

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon May 21, 2012 6:57 am

GmatKiss wrote: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

Edited:

210 = 2x3x5x7

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 2x3=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

Last edited by Brent@GMATPrepNow on Tue May 22, 2012 6:55 am, edited 2 times in total.

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by Brent@GMATPrepNow » Mon May 21, 2012 7:11 am

Edited:

I should point out that, in the above solution, I used the nice rule that says "n unique objects can be arranged in n! ways"

Alternatively, if we don't know the above rule, we can use the Fundamental Counting Principle (FCP). For example, in case 1, we need to arrange the digits 2, 3, 5, 7.
To apply the FRC, we'll take the task of arranging the 4 digits and break it into 4 stages:

Stage 1: Select a digit for the units position.
Stage 2: Select a digit for the tens position.
Stage 3: Select a digit for the hundreds position.
Stage 4: Select a digit for the thousands position.

Stage 1: there are 4 digits, so this stage can be accomplished in 4 ways
Stage 2: once we have completed stage 1, there are 3 digits remaining, so stage 2 can be accomplished in 3 ways
Stage 3: once we have completed stage 2, there are 2 digits remaining, so stage 3 can be accomplished in 2 ways
Stage 4: once we have completed stage 3, there is 1 digit remaining, so stage 4 can be accomplished in 1 ways

From here, the FCP tells us that all 4 stages can be completed in 4x3x2x1 ways (24 ways)

If you're interested in learning more, here's a free video on the FCP: https://www.gmatprepnow.com/module/gmat-counting?id=775

Cheers,
Brent

Last edited by Brent@GMATPrepNow on Tue May 22, 2012 6:57 am, edited 1 time in total.

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hey_thr67: Master | Next Rank: 500 Posts; Posts: 299; Joined: Tue Feb 15, 2011 10:27 amFollowed by: 2 members; Thanked: 9 times

by hey_thr67 » Mon May 21, 2012 8:04 am

IMO: [spoiler]B (30)[/spoiler]

4 digits that will give their product as 210 are 2x3x5x7
So number of arrangements are : 4! = 24

3 digits that will give their product as 210 are 6x5x7
So number of arrangements are : 3! = 6

So, total number of arrangements are 24+6 = 30

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon May 21, 2012 8:42 am

GmatKiss wrote: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

210 = 2*3*5*7.

Case 1: 4-digit integers composed of the digits 2, 3, 5 and 7.
Number of ways to arrange these 4 digits = 4! = 24.

Case 2: 4-digit integers composed of the digits 1, 5, 6, and 7.
Number of ways to arrange these 4 digits = 4! = 24.

Case 3: 3-digit integers composed of the digits 5, 6 and 7.
Number of ways to arrange these 3 digits = 3! = 6.

Total ways = 24+24+6 = 54.

The correct answer is D.

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Mon May 21, 2012 9:12 am

GMATGuruNY wrote:
GmatKiss wrote: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
210 = 2*3*5*7.

Case 1: 4-digit integers composed of the digits 2, 3, 5 and 7.
Number of ways to arrange these 4 digits = 4! = 24.

Case 2: 4-digit integers composed of the digits 1, 5, 6, and 7.
Number of ways to arrange these 4 digits = 4! = 24.

Case 3: 3-digit integers composed of the digits 5, 6 and 7.
Number of ways to arrange these 3 digits = 3! = 6.

Total ways = 24+24+6 = 54.

The correct answer is D.

Good catch - I completely missed the fact that 2x3 creates a new digit. My bad

Cheers,
Brent

Brent Hanneson - Creator of GMATPrepNow.com

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GmatKiss: Legendary Member; Posts: 2789; Joined: Tue Jul 26, 2011 12:19 amFollowed by: 43 members; Location: Chennai, India; Thanked: 206 times; GMAT Score:640

by GmatKiss » Mon May 21, 2012 10:22 am

Brent@GMATPrepNow wrote:
GMATGuruNY wrote:
GmatKiss wrote: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
210 = 2*3*5*7.

Case 1: 4-digit integers composed of the digits 2, 3, 5 and 7.
Number of ways to arrange these 4 digits = 4! = 24.

Case 2: 4-digit integers composed of the digits 1, 5, 6, and 7.
Number of ways to arrange these 4 digits = 4! = 24.

Case 3: 3-digit integers composed of the digits 5, 6 and 7.
Number of ways to arrange these 3 digits = 3! = 6.

Total ways = 24+24+6 = 54.

The correct answer is D.
Good catch - I completely missed the fact that 2x3 creates a new digit. My bad

Cheers,
Brent

A video explanation is provided in this link

https://gre.magoosh.com/answers/1798768?prompt_id=831

TIA,
GK

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