Last visit was: 10 Jul 2025, 10:03 It is currently 10 Jul 2025, 10:03
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
User avatar
idontknowwhy94
Joined: 03 Oct 2016
Last visit: 15 Mar 2017
Posts: 61
Own Kudos:
301
 [88]
Given Kudos: 6
Concentration: Technology, General Management
WE:Information Technology (Computer Software)
Products:
Posts: 61
Kudos: 301
 [88]
7
Kudos
Add Kudos
81
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 13 May 2024
Posts: 6,756
Own Kudos:
34,049
 [46]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,756
Kudos: 34,049
 [46]
23
Kudos
Add Kudos
23
Bookmarks
Bookmark this Post
General Discussion
avatar
Joc456
Joined: 26 Jan 2016
Last visit: 08 May 2017
Posts: 75
Own Kudos:
51
 [1]
Given Kudos: 55
Location: United States
GPA: 3.37
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 13 May 2024
Posts: 6,756
Own Kudos:
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,756
Kudos: 34,049
Kudos
Add Kudos
Bookmarks
Bookmark this Post
joannaecohen
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 )
User avatar
goforgmat
Joined: 09 Feb 2015
Last visit: 02 Nov 2019
Posts: 246
Own Kudos:
Given Kudos: 232
Location: India
Concentration: Social Entrepreneurship, General Management
GMAT 1: 690 Q49 V34
GMAT 2: 720 Q49 V39
GPA: 2.8
Products:
GMAT 2: 720 Q49 V39
Posts: 246
Kudos: 104
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATPrepNow
idontknowwhy94
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?
User avatar
Aim800score
Joined: 30 Sep 2017
Last visit: 09 Dec 2018
Posts: 30
Own Kudos:
50
 [1]
Given Kudos: 27
Location: India
Concentration: Entrepreneurship, General Management
GMAT 1: 700 Q50 V37
GPA: 3.7
WE:Engineering (Energy)
GMAT 1: 700 Q50 V37
Posts: 30
Kudos: 50
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
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
User avatar
hellosanthosh2k2
Joined: 02 Apr 2014
Last visit: 07 Dec 2020
Posts: 362
Own Kudos:
565
 [3]
Given Kudos: 1,227
Location: India
Schools: XLRI"20
GMAT 1: 700 Q50 V34
GPA: 3.5
Schools: XLRI"20
GMAT 1: 700 Q50 V34
Posts: 362
Kudos: 565
 [3]
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
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
User avatar
gurmukh
Joined: 18 Dec 2017
Last visit: 24 Jan 2025
Posts: 260
Own Kudos:
244
 [1]
Given Kudos: 20
Posts: 260
Kudos: 244
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Total number of factors = 4×3×3=36
Factors which are multiple of 21
= 3×3×2 =18
Required Probability =1/2
Option E is the answer

Posted from my mobile device
avatar
kitzi
Joined: 07 Apr 2020
Last visit: 16 Jul 2020
Posts: 7
Given Kudos: 3
Posts: 7
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Determining the sample space/all the possible numbers that could be made from those factors

1. There are 12 ways to choose different numbers from the factors of 3,5,7 (for instance 3x5x7, 3x5x7^2, 3x5^2x7, and so on).
2. There are 6 ways to choose different numbers from the factors of 3,5 (3x5, 3x5^2, and so on).
3. There are 4 ways to choose different numbers from the factors of 5, 7 (5x7, 5x7^2, and so on).
4. There are 6 ways to choose different numbers from the factors of 3,7 (3x7, 3x7^2, 3^2 x 7, 3^2 x 7^2, and so on)
5. There is a total of 7 numbers if we only account for 1 number from the factors (3, 3^2,3^3, 5, 5^2) and so on)
6. And indeed 1 is a factor of any number.

The event that we desire is only from numbers 1 and 4.

Pr(the number is divisible by 21) = 18/36 = 1/2
User avatar
sathriyan
Joined: 22 Feb 2019
Last visit: 29 Jan 2023
Posts: 7
Own Kudos:
15
 [1]
Given Kudos: 293
Status:Stay Hungry, Stay Foolish
Posts: 7
Kudos: 15
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
BrentGMATPrepNow
idontknowwhy94
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



Why are we factoring out that 21 before calculating the number of factors? [Highlighted Part]
User avatar
samc88
Joined: 09 Aug 2021
Last visit: 18 Jun 2024
Posts: 16
Own Kudos:
Given Kudos: 71
Products:
Posts: 16
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I did it a slightly different way:

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

21 = (3)(7)

so if we look at all the factors of (3's and 5's) and (5's and 7's) and remove the overlap then that will give us the number that DON'T have 21 as a factor

(3³)(5²) number of factors = 4x3 = 12
(5²)(7²) number of factors = 3x3 = 9

but 1,5,25 are common to both, so the number of factors that don't have 21 as a factor is (12 + 9 - 3) = 18

(36/18)/36 = 1/2
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,375
Own Kudos:
Posts: 37,375
Kudos: 1,010
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Math Expert
102619 posts
PS Forum Moderator
685 posts