Last visit was: 21 May 2024, 14:19 It is currently 21 May 2024, 14:19
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
SORT BY:
Date
Manager
Manager
Joined: 05 Oct 2016
Posts: 68
Own Kudos [?]: 126 [35]
Given Kudos: 135
Location: United States (OH)
GPA: 3.58
Send PM
Most Helpful Reply
Math Expert
Joined: 02 Sep 2009
Posts: 93373
Own Kudos [?]: 625613 [11]
Given Kudos: 81918
Send PM
General Discussion
Manager
Manager
Joined: 05 Oct 2016
Posts: 68
Own Kudos [?]: 126 [2]
Given Kudos: 135
Location: United States (OH)
GPA: 3.58
Send PM
Math Expert
Joined: 02 Sep 2009
Posts: 93373
Own Kudos [?]: 625613 [3]
Given Kudos: 81918
Send PM
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher [#permalink]
2
Kudos
1
Bookmarks
Expert Reply
SandhyAvinash wrote:
Bunuel wrote:
SandhyAvinash wrote:
Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer?

A)1
B)2
C)3
D)4
E)6


4k + 2 = 2(2k + 1) = 2*odd. So, we are looking for even factors which are not multiples of 4.

180 = 2^2*3^2*5. Consider the part without 2^2. Now, 3^2*5 has (2 + 1)(1 + 1) = 6 factors: 1, 3, 5, 9, 15, 45. Any of them paired with 2 will be even factor of 180 which is not a multiple of 4, so 2, 6, 10, 18, 30 and 90.

Answer: E.


Bunuel how did you find 6 factors as 1,3,5,9,15,45. please help me in this.


I just manually listed the factors of 90 = 3^2*5. You don't have to find the factors for this question though, it's done here just to illustrate. The fact that 90 has 6 factors (does not matter what they really are) is enough to get that 180 will have twice as many even factors which are not multiples of 4.
Intern
Intern
Joined: 09 Aug 2017
Posts: 20
Own Kudos [?]: 1 [1]
Given Kudos: 0
Send PM
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher [#permalink]
1
Kudos
Bunuel wrote:
SandhyAvinash wrote:
Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer?

A)1
B)2
C)3
D)4
E)6


4k + 2 = 2(2k + 1) = 2*odd. So, we are looking for even factors which are not multiples of 4.

180 = 2^2*3^2*5. Consider the part without 2^2. Now, 3^2*5 has (2 + 1)(1 + 1) = 6 factors: 1, 3, 5, 9, 15, 45. Any of them paired with 2 will be even factor of 180 which is not a multiple of 4, so 2, 6, 10, 18, 30 and 90.

Answer: E.

How did you got (2+1)(1+1) ?? Pls explain

Sent from my Redmi 3S using GMAT Club Forum mobile app
Math Expert
Joined: 02 Sep 2009
Posts: 93373
Own Kudos [?]: 625613 [4]
Given Kudos: 81918
Send PM
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher [#permalink]
2
Kudos
2
Bookmarks
Expert Reply
Spongebob02 wrote:
Bunuel wrote:
SandhyAvinash wrote:
Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer?

A)1
B)2
C)3
D)4
E)6


4k + 2 = 2(2k + 1) = 2*odd. So, we are looking for even factors which are not multiples of 4.

180 = 2^2*3^2*5. Consider the part without 2^2. Now, 3^2*5 has (2 + 1)(1 + 1) = 6 factors: 1, 3, 5, 9, 15, 45. Any of them paired with 2 will be even factor of 180 which is not a multiple of 4, so 2, 6, 10, 18, 30 and 90.

Answer: E.

How did you got (2+1)(1+1) ?? Pls explain

Sent from my Redmi 3S using GMAT Club Forum mobile app


Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.

2. Properties of Integers




5. Divisibility/Multiples/Factors



For other subjects:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread
Intern
Intern
Joined: 04 Jul 2018
Posts: 1
Own Kudos [?]: 0 [0]
Given Kudos: 2
Send PM
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher [#permalink]
There are 6 possibilities, where f(x)=4x+2 is a factor of 180.

f(0)=2
F(1)=6
f(2)=10
f(4)=18
f(7)=30
f(22)=80
Senior Manager
Senior Manager
Joined: 11 Mar 2018
Posts: 264
Own Kudos [?]: 349 [0]
Given Kudos: 271
Location: India
GMAT 1: 710 Q49 V37 (Online)
Send PM
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher [#permalink]
SandhyAvinash wrote:
Find the number of factors of 180 that are in the form (4*k + 2), where k is a non-negative integer?

A)1
B)2
C)3
D)4
E)6


I did this question with simple substitution like below -

we know K >=0 and is an integer.

therefore for
K = 0 ; (4*k + 2) = 2 - A factor of 180
K = 1 ; (4*k + 2) = 6 - A factor of 180
K = 2 ; (4*k + 2) = 10 - A factor of 180
K = 4 ; (4*k + 2) = 18 - A factor of 180

No we know answer can be D or E.

If I am able to find one more my answer will be E.

Hence I factorized 180 = 2x3x3x2x5

Next number after 18 I can form is 30 that will be a factor of 180 and I get that with k = 7.

Hence E.

FYI - I did this question under two minutes. So IMO not a lengthy method.
User avatar
Non-Human User
Joined: 09 Sep 2013
Posts: 33140
Own Kudos [?]: 829 [0]
Given Kudos: 0
Send PM
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher [#permalink]
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.
GMAT Club Bot
Re: Find the number of factors of 180 that are in the form (4*k + 2), wher [#permalink]
Moderator:
Math Expert
93373 posts