GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 15 Jul 2018, 12:31

Close

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel

If a number is selected from the factors of 3^{3} 5^{2} 7^{2}

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

2 KUDOS received
Manager
Manager
avatar
B
Joined: 03 Oct 2016
Posts: 83
Concentration: Technology, General Management
WE: Information Technology (Computer Software)
Reviews Badge
If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink]

Show Tags

New post 06 Oct 2016, 10:24
2
19
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

35% (01:28) correct 65% (01:34) wrong based on 295 sessions

HideShow timer Statistics

If a number is selected from the factors of \(3^{3} 5^{2} 7^{2}\) , what is the probability that the selected number is divisible by 21?

A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2

Keep the Kudos coming in and let the questions go out !!!! :-D

_________________

KINDLY KUDOS IF YOU LIKE THE POST

Most Helpful Expert Reply
Expert Post
Top Contributor
13 KUDOS received
CEO
CEO
User avatar
P
Joined: 12 Sep 2015
Posts: 2630
Location: Canada
If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink]

Show Tags

New post Updated on: 06 Oct 2016, 12:28
13
Top Contributor
10
idontknowwhy94 wrote:
If a number is selected from the factors of \(3^{3} 5^{2} 7^{2}\) , what is the probability that the selected number is divisible by 21?

A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2



Great question!

We'll use the following formula:

If N = (p^a)(q^b)(r^c)..., where p, q, r,...(etc.) are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)...

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) = (5)(4)(2) = 40
--------------------------------------

Okay, now let's answer the question!

P(selected number is divisible by 21) = (# of divisors that are divisible by 21)/(TOTAL # of divisors)


As always, we'll begin with the....

DENOMINATOR
We're given the number: (3³)(5²)(7²)
So, the number of divisors = (3+1)(2+1)(2+1) = (4)(3)(3) = 36

NUMERATOR - i.e., # of divisors that are divisible by 21
Let's factor 21 out of the given number.
(3³)(5²)(7²) = (3¹)(7¹)[(3²)(5²)(7¹)]
= 21[(3²)(5²)(7¹)]
At this point, we can see that, if we multiply 21 by any factor of [(3²)(5²)(7¹)], the product will definitely be divisible by 21
So, the question becomes "how many divisors does [(3²)(5²)(7¹)] have?"
Once again, we'll apply our nice formula.
We have (3²)(5²)(7¹)
So, the number of divisors = (2+1)(2+1)(1+1) = (3)(3)(2) = 18
-------------------

So, P(selected number is divisible by 21) = 18/36
= 1/2

Answer:

RELATED VIDEOS



_________________

Brent Hanneson – Founder of gmatprepnow.com

Image


Originally posted by GMATPrepNow on 06 Oct 2016, 11:44.
Last edited by GMATPrepNow on 06 Oct 2016, 12:28, edited 1 time in total.
General Discussion
1 KUDOS received
Current Student
avatar
B
Joined: 26 Jan 2016
Posts: 110
Location: United States
GPA: 3.37
Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink]

Show Tags

New post 06 Oct 2016, 12:24
1
GMATPrepNow

why where did the other 5 go when you factored out the 21? Shouldn't it be 21[3²*5²5^1)
Expert Post
Top Contributor
CEO
CEO
User avatar
P
Joined: 12 Sep 2015
Posts: 2630
Location: Canada
Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink]

Show Tags

New post 06 Oct 2016, 12:29
Top Contributor
joannaecohen wrote:
GMATPrepNow

why where did the other 5 go when you factored out the 21? Shouldn't it be 21[3²*5²5^1)


Good catch! I've edited my response accordingly (the answer is still E though :-D )
_________________

Brent Hanneson – Founder of gmatprepnow.com

Image

Senior Manager
Senior Manager
avatar
G
Joined: 09 Feb 2015
Posts: 378
Location: India
Concentration: Social Entrepreneurship, General Management
GMAT 1: 690 Q49 V34
GMAT 2: 720 Q49 V39
GPA: 2.8
Premium Member Reviews Badge CAT Tests
Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink]

Show Tags

New post 26 Nov 2017, 07:25
GMATPrepNow wrote:
idontknowwhy94 wrote:
If a number is selected from the factors of \(3^{3} 5^{2} 7^{2}\) , what is the probability that the selected number is divisible by 21?

A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2



Great question!

We'll use the following formula:

If N = (p^a)(q^b)(r^c)..., where p, q, r,...(etc.) are prime numbers, then the total number of positive divisors of N is equal to (a+1)(b+1)(c+1)...

Example: 14000 = (2^4)(5^3)(7^1)
So, the number of positive divisors of 14000 = (4+1)(3+1)(1+1) = (5)(4)(2) = 40
--------------------------------------

Okay, now let's answer the question!

P(selected number is divisible by 21) = (# of divisors that are divisible by 21)/(TOTAL # of divisors)


As always, we'll begin with the....

DENOMINATOR
We're given the number: (3³)(5²)(7²)
So, the number of divisors = (3+1)(2+1)(2+1) = (4)(3)(3) = 36

NUMERATOR - i.e., # of divisors that are divisible by 21
Let's factor 21 out of the given number.
(3³)(5²)(7²) = (3¹)(7¹)[(3²)(5²)(7¹)]
= 21[(3²)(5²)(7¹)]
At this point, we can see that, if we multiply 21 by any factor of [(3²)(5²)(7¹)], the product will definitely be divisible by 21
So, the question becomes "how many divisors does [(3²)(5²)(7¹)] have?"
Once again, we'll apply our nice formula.
We have (3²)(5²)(7¹)
So, the number of divisors = (2+1)(2+1)(1+1) = (3)(3)(2) = 18
-------------------

So, P(selected number is divisible by 21) = 18/36
= 1/2

Answer:

RELATED VIDEOS




wont the factors include 1 as well? shouldn't that be excluded for this question since the q stem asks for numbers divisble by 21?
Intern
Intern
avatar
S
Joined: 30 Sep 2017
Posts: 38
Location: India
Concentration: Entrepreneurship, General Management
Schools: IIM Udaipur '17
GMAT 1: 700 Q50 V37
GPA: 3.7
WE: Engineering (Energy and Utilities)
GMAT ToolKit User CAT Tests
Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink]

Show Tags

New post 27 Nov 2017, 08:10
Total number of factors = (3+1)*(2+1)*(2+1) = 4*3*3
Dividing the number by 21 , n = 3^2*5^2*7^1, Favorable number of factors =(2+1)*(2+1)*(1+1) = 3*3*2

Required probability = 3*3*2/(4*3*3) = 1/2
_________________

If you like my post, motivate me by giving kudos...

1 KUDOS received
Senior Manager
Senior Manager
avatar
G
Joined: 02 Apr 2014
Posts: 486
GMAT 1: 700 Q50 V34
Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2} [#permalink]

Show Tags

New post 30 Jan 2018, 02:00
1
Total number of factors of \(3^3*5^2*7^2\) = \((3+1) * (2+1) * (2+1)\) = 36

to find number of multiples of 21, we find number of factors = \(3^3*5^2*7^2\) / \(21\) = \(3^2 * 5^2 * 7\) = (2+1) * (2+1) * (1+1) = 18

probability of getting number divisible by 21 = 18/36 = 1/2
Re: If a number is selected from the factors of 3^{3} 5^{2} 7^{2}   [#permalink] 30 Jan 2018, 02:00
Display posts from previous: Sort by

If a number is selected from the factors of 3^{3} 5^{2} 7^{2}

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.