It is currently 09 Jul 2020, 23:14

### 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

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

# S be the set of all positive integers n such that n^2

Author Message
TAGS:
Retired Moderator
Joined: 07 Jun 2014
Posts: 4804
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 163

Kudos [?]: 2762 [1] , given: 394

S be the set of all positive integers n such that n^2 [#permalink]  04 Jan 2016, 16:45
1
KUDOS
Expert's post
00:00

Question Stats:

55% (02:17) correct 44% (02:51) wrong based on 59 sessions
Let S be the set of all positive integers n such that $$n^2$$ is a multiple of both 24 and 108. Which of the following integers are divisors of every integer n in S ?

Indicate all such integers.

A. 12
B. 24
C. 36
D. 72

Practice Questions
Question: 14
Page: 159
Difficulty: hard

[Reveal] Spoiler: OA
A,C

_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Retired Moderator
Joined: 07 Jun 2014
Posts: 4804
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 163

Kudos [?]: 2762 [3] , given: 394

Re: S be the set of all positive integers n such that n^2 [#permalink]  04 Jan 2016, 18:48
3
KUDOS
Expert's post
Now here we are looking for a Set S of numbers n such that $$n^2$$ is multiple of 24 and 108.

Lets take one such case Least Common Multiple of 108 and 24 is 216.

$$n^2$$ = 216
n = 6$$\sqrt{6}$$.

This is not an integer however if we multiply 216 by any number it is still divisible by 24 and 108. So we multiply by 6 and repat the above steps to find n. So n =36.

Now n= 36 is one of the number in set S.

Hence all the options that divide n =36 are correct. Hence option A and C.
_________________

Sandy
If you found this post useful, please let me know by pressing the Kudos Button

Try our free Online GRE Test

Intern
Joined: 28 Mar 2016
Posts: 11
Followers: 0

Kudos [?]: 3 [0], given: 0

Re: S be the set of all positive integers n such that n^2 [#permalink]  29 Mar 2016, 09:38
Manager
Joined: 23 Jan 2016
Posts: 133
Followers: 4

Kudos [?]: 145 [1] , given: 15

Re: S be the set of all positive integers n such that n^2 [#permalink]  15 May 2016, 21:09
1
KUDOS
LCM of 108 and 24 is 216 = 2^3*3^3. As n^2 is a square, the least value n^2 can take is 2^4*3^4 = 6*216 = 36 * 36, so n should be at least 36 or multiples of 36. So any factor of 36 should always divide n. Hence A and C are correct.
Manager
Joined: 17 Aug 2019
Posts: 243
Followers: 2

Kudos [?]: 30 [0], given: 59

Re: S be the set of all positive integers n such that n^2 [#permalink]  07 Nov 2019, 00:49
LCM of 108 and 24 is 216 = 2^3*3^3. As n^2 is a square, the least value n^2 can take is 2^4*3^4 = 6*216 = 36 * 36, so n should be at least 36 or multiples of 36. So any factor of 36 should always divide n. Hence A and C are correct.

Though I understand what you wrote but I need clearer explanation and a step by step pls. Now, after we find out LCM 216 and it is not a square of any number what do we do ?
VP
Joined: 20 Apr 2016
Posts: 1279
WE: Engineering (Energy and Utilities)
Followers: 20

Kudos [?]: 1235 [4] , given: 242

Re: S be the set of all positive integers n such that n^2 [#permalink]  07 Nov 2019, 02:12
4
KUDOS
Asmakan wrote:
LCM of 108 and 24 is 216 = 2^3*3^3. As n^2 is a square, the least value n^2 can take is 2^4*3^4 = 6*216 = 36 * 36, so n should be at least 36 or multiples of 36. So any factor of 36 should always divide n. Hence A and C are correct.

Though I understand what you wrote but I need clearer explanation and a step by step pls. Now, after we find out LCM 216 and it is not a square of any number what do we do ?

don't get confused with LCM and GCF, use ur own logic

Back to ques:

$$24 = 2^3 * 3$$

$$108 = 2^2 * 3^3$$

For $$2$$, we have maximum value = $$2^3$$

For $$3$$, we have maximum value =$$3^3$$

We need the value of n in such a way that squaring n will contains both maximum value of $$2$$(i.e $$2^3$$) and$$3 (i.e 3^3)$$

But, squaring a number won't raise the power to $$3$$

so the minimum value should be $$2^2$$ and squaring this will be = $$2^4$$

Similarly for $$3$$, the minimum value =$$3^2$$and squaring will be = $$3^4$$

therefore, the minimum value = $$2^2 * 3^2 = 36$$. This is divisible by $$36$$ and $$12$$
_________________

If you found this post useful, please let me know by pressing the Kudos Button

Rules for Posting

Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months GRE Prep Club tests

Director
Joined: 22 Jun 2019
Posts: 521
Followers: 4

Kudos [?]: 99 [0], given: 161

Re: S be the set of all positive integers n such that n^2 [#permalink]  07 Nov 2019, 03:21
VP
Joined: 20 Apr 2016
Posts: 1279
WE: Engineering (Energy and Utilities)
Followers: 20

Kudos [?]: 1235 [2] , given: 242

Re: S be the set of all positive integers n such that n^2 [#permalink]  07 Nov 2019, 03:56
2
KUDOS
huda wrote:
Official explanation.

PLZ, don't attach screenshot of official explanation. This is not entertained in this forum.

you can write it down
_________________

If you found this post useful, please let me know by pressing the Kudos Button

Rules for Posting

Got 20 Kudos? You can get Free GRE Prep Club Tests

GRE Prep Club Members of the Month:TOP 10 members of the month with highest kudos receive access to 3 months GRE Prep Club tests

Director
Joined: 22 Jun 2019
Posts: 521
Followers: 4

Kudos [?]: 99 [0], given: 161

Re: S be the set of all positive integers n such that n^2 [#permalink]  07 Nov 2019, 04:27
pranab01 wrote:
huda wrote:
Official explanation.

PLZ, don't attach screenshot of official explanation. This is not entertained in this forum.

you can write it down

Well said. Fortunately, There is no Easy way to write something long here. Make it easy after-all.
_________________

New to the GRE, and GRE CLUB Forum?
Posting Rules: QUANTITATIVE | VERBAL

Questions' Banks and Collection:
ETS: ETS Free PowerPrep 1 & 2 All 320 Questions Explanation. | ETS All Official Guides
3rd Party Resource's: All In One Resource's | All Quant Questions Collection | All Verbal Questions Collection | Manhattan 5lb All Questions Collection
Books: All GRE Best Books
Scores: Average GRE Score Required By Universities in the USA
Tests: All Free & Paid Practice Tests | GRE Prep Club Tests
Extra: Permutations, and Combination
Vocab: GRE Vocabulary

Founder
Joined: 18 Apr 2015
Posts: 12020
Followers: 254

Kudos [?]: 2998 [0], given: 11253

Re: S be the set of all positive integers n such that n^2 [#permalink]  11 Nov 2019, 01:22
Expert's post
It is quite easy if you do know how to format a question and the tag usage.

Regards
_________________

Need Practice? 20 Free GRE Quant Tests available for free with 20 Kudos
GRE Prep Club Members of the Month: Each member of the month will get three months free access of GRE Prep Club tests.

GRE Instructor
Joined: 10 Apr 2015
Posts: 3532
Followers: 133

Kudos [?]: 3995 [5] , given: 65

Re: S be the set of all positive integers n such that n^2 [#permalink]  11 Nov 2019, 10:42
5
KUDOS
Expert's post
GIVEN: n² is a multiple of 24
24 = (2)(2)(2)(3)
So, n² must have at least three 2's and one 3 in its prime factorization.
What does this tell us about n?
Since n² = (n)(n), we can conclude that n must have at least two 2's and one 3 in its prime factorization.

GIVEN: n² is a multiple of 108
108 = (2)(2)(3)(3)(3)
So, n² must have at least two 2's and three 3's in its prime factorization.
What does this tell us about n?
Since n² = (n)(n), we can conclude that n must have at least one 2 and two 3's in its prime factorization.

So, when we combine both pieces of information, we can see that n must have at least two 2's and two 3's in its prime factorization.
In other words, n = (2)(2)(3)(3)(k), where k is some positive integer

Now let's check the answer choices...
A. 12 = (2)(2)(3)
Since n = (2)(2)(3)(3)(k), we can see that n IS divisible by 12

B. 24 = (2)(2)(2)(3)
Since n = (2)(2)(3)(3)(k), we can see that n need NOT be divisible by 24

C. 36 = (2)(2)(3)(3)
Since n = (2)(2)(3)(3)(k), we can see that n IS divisible by 36

D. 72 = (2)(2)(2)(2)(3)
Since n = (2)(2)(3)(3)(k), we can see that n need NOT be divisible by 72

Cheers,
Brent
_________________

Brent Hanneson – Creator of greenlighttestprep.com

Intern
Joined: 11 Nov 2019
Posts: 29
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: S be the set of all positive integers n such that n^2 [#permalink]  14 Nov 2019, 09:08
I don't understand this one bit why so difficult.

Posted from my mobile device
Intern
Joined: 11 Nov 2019
Posts: 29
Followers: 0

Kudos [?]: 0 [0], given: 0

Re: S be the set of all positive integers n such that n^2 [#permalink]  14 Nov 2019, 09:09

Posted from my mobile device
Re: S be the set of all positive integers n such that n^2   [#permalink] 14 Nov 2019, 09:09
Display posts from previous: Sort by