It is currently 19 Oct 2017, 07:56

STARTING SOON:

Live Chat with Cornell Adcoms in Main Chat Room  |  R1 Interview Invites: MIT Sloan Chat  |  UCLA Anderson Chat  |  Duke Fuqua Chat (EA Decisions)


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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

If n >2 and n^4-5n^2+4 = 20h, Is h an integer ?

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

Hide Tags

Intern
Intern
avatar
B
Joined: 28 Dec 2010
Posts: 23

Kudos [?]: 44 [0], given: 337

If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

Show Tags

New post 21 Aug 2017, 07:57
2
This post was
BOOKMARKED
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

20% (02:26) correct 80% (02:01) wrong based on 74 sessions

HideShow timer Statistics

If n >2 and n^4-5n^2+4 = 20h, Is h an integer?

1) n/4 is an integer

2) n/5 is an integer
[Reveal] Spoiler: OA

Last edited by chetan2u on 22 Aug 2017, 07:42, edited 1 time in total.
added the OA

Kudos [?]: 44 [0], given: 337

Expert Post
3 KUDOS received
Math Forum Moderator
avatar
P
Joined: 02 Aug 2009
Posts: 4969

Kudos [?]: 5472 [3], given: 112

Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

Show Tags

New post 22 Aug 2017, 07:41
3
This post received
KUDOS
Expert's post
gary391 wrote:
If n >2 and n^4-5n^2+4 = 20h, Is h an integer?

1) n/4 is an integer

2) n/5 is an integer


HI...

please follow the rules for posting a Q..

otherwise for the solution..
\(n^4-5n^2+4 = 20h.............. h=\frac{n^4-5n^2+4}{20}\)
for h to be an integer \(n^4-5n^2+4\) has to be a multiple of 20..

lets see the statements

I..
\(\frac{n}{4}\) is an integer..
let n/4 = s where s is the integer.. n = 4s
so \(h=\frac{n^4-5n^2+4}{20}.......\frac{(4s)^4-5*(4s)^2+4}{20}.....\)
so if s is 1, ans is 180/20=9 yes h is an integer..
but when s is multiple of 5, clearly numerator will be multiple of 5+4, which will not be multiple of 5 and thus 20.. so h is NOT an integer

insufficient

II
\(\frac{n}{5}\) is an integer..
let n/5 = s where s is the integer.. n = 5s
so \(h=\frac{n^4-5n^2+4}{20}.......\frac{(5s)^4-5*(5s)^2+4}{20}.....\)
so whatever be the value of s, the NUMERATOR will be MULTIPLE of 5+4, which will not be a multiple of 5 and thus multiple of 20
ans is always.. NO

sufficient

B
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5472 [3], given: 112

1 KUDOS received
Senior Manager
Senior Manager
avatar
G
Joined: 28 Mar 2017
Posts: 481

Kudos [?]: 121 [1], given: 126

Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

Show Tags

New post 22 Aug 2017, 08:25
1
This post received
KUDOS
Hello chetan2u,

Please clarify my below doubt:

After solving the equation i got it in normalised form as:
(n+1)(n-1)(n+2)(n-2)=20h

Now for statement 1, n=4k. Thus if we put in the values of n=4 and 20, we get:
4 ---> 5*3*6*2; This will be a multiple of 20 so h has to be integer
20 --> 21*19*22*18; This will not be a multiple of 20 since it doesn't have a 5 so h can either be multiple of 5 or a fraction containing 5/multiple of 5 in numerator with a number factor of any of the remaining primes on LHS. Thus insufficient.

Now for statement 1, n=5k. Thus if we put in any values of 5; LHS won't be a multiple of 5 thus h can be either be multiple of 5 or a fraction containing 5/multiple of 5 in numerator with a number factor of any of the remaining primes on LHS. Thus insufficient.

What am I missing here? I got to the crux of the question and almost solved it but I am unable to understand where am I going wrong?

Regards
_________________

Kudos if my post helps!

Helpful links:
1. e-GMAT's ALL SC Compilation

Kudos [?]: 121 [1], given: 126

Expert Post
1 KUDOS received
Math Forum Moderator
avatar
P
Joined: 02 Aug 2009
Posts: 4969

Kudos [?]: 5472 [1], given: 112

Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

Show Tags

New post 22 Aug 2017, 08:35
1
This post received
KUDOS
Expert's post
gmatexam439 wrote:
Hello chetan2u,

Please clarify my below doubt:

After solving the equation i got it in normalised form as:
(n+1)(n-1)(n+2)(n-2)=20h

Now for statement 1, n=4k. Thus if we put in the values of n=4 and 20, we get:
4 ---> 5*3*6*2; This will be a multiple of 20 so h has to be integer
20 --> 21*19*22*18; This will not be a multiple of 20 since it doesn't have a 5 so h can either be multiple of 5 or a fraction containing 5/multiple of 5 in numerator with a number factor of any of the remaining primes on LHS. Thus insufficient.

Now for statement 1, n=5k. Thus if we put in any values of 5; LHS won't be a multiple of 5 thus h can be either be multiple of 5 or a fraction containing 5/multiple of 5 in numerator with a number factor of any of the remaining primes on LHS. Thus insufficient.

What am I missing here? I got to the crux of the question and almost solved it but I am unable to understand where am I going wrong?

Regards


hi..

you have solved it correctly but going wrong on final interpretation in red coloured portion..
LHS is NOT a multiple of 5 say 5s+4 and RHS is 20h..
so if h is a multiple of 5... 5s+4=20*5... NOT possible..
h has to remove the 5 in 20 and that h can do by being a fraction with 5 in denominator..

_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5472 [1], given: 112

Senior Manager
Senior Manager
avatar
G
Joined: 28 Mar 2017
Posts: 481

Kudos [?]: 121 [0], given: 126

Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

Show Tags

New post 22 Aug 2017, 08:55
chetan2u wrote:
hi..

you have solved it correctly but going wrong on final interpretation in red coloured portion..
LHS is NOT a multiple of 5 say 5s+4 and RHS is 20h..
so if h is a multiple of 5... 5s+4=20*5... NOT possible..
h has to remove the 5 in 20 and that h can do by being a fraction with 5 in denominator..


Yes, sorry.
I reviewed my equation and as you pointed out 5 should be in the denominator always. Thus it will never be an integer. Damn !!

Thank you so much for the quick reply.

Regards
_________________

Kudos if my post helps!

Helpful links:
1. e-GMAT's ALL SC Compilation

Kudos [?]: 121 [0], given: 126

Manager
Manager
avatar
S
Joined: 26 Jun 2017
Posts: 104

Kudos [?]: 27 [0], given: 176

Location: Russian Federation
Concentration: General Management, Strategy
WE: Information Technology (Other)
Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

Show Tags

New post 23 Aug 2017, 13:02
chetan2u wrote:
gary391 wrote:
If n >2 and n^4-5n^2+4 = 20h, Is h an integer?
\(\frac{n}{4}\) is an integer..
let n/4 = s where s is the integer.. n = 4s
so \(h=\frac{n^4-5n^2+4}{20}.......\frac{(4s)^4-5*(4s)^2+4}{20}.....\)
so if s is 1, ans is 180/20=9 yes h is an integer..
but when s is multiple of 5, clearly numerator will be multiple of 5+4, which will not be multiple of 5 and thus 20.. so h is NOT an integer

insufficient


Chetan2u,

can you, please explain this moment? It seems, I miss some properties of numbers.
but when s is multiple of 5, clearly numerator will be multiple of 5+4 - from where does it come from?

I solved in next manner:
(n-2)(n-1)(n+1)(n+2) = 20h

|. n=4 -> divisable. n=20 -> not divisable ---> insufficient The disadvantage of this method - I can miss option n = 20. And in your solution it seems impossible.
||. n is divisable by 5 -> n-2, n-1, n+1, n+2 - none of them is divisable by 5 ---> so the answer is: never divisable ---> not integer. Sufficient.

Kudos [?]: 27 [0], given: 176

Expert Post
1 KUDOS received
Math Forum Moderator
avatar
P
Joined: 02 Aug 2009
Posts: 4969

Kudos [?]: 5472 [1], given: 112

Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

Show Tags

New post 23 Aug 2017, 19:07
1
This post received
KUDOS
Expert's post
DharLog wrote:
chetan2u wrote:
gary391 wrote:
If n >2 and n^4-5n^2+4 = 20h, Is h an integer?
\(\frac{n}{4}\) is an integer..
let n/4 = s where s is the integer.. n = 4s
so \(h=\frac{n^4-5n^2+4}{20}.......\frac{(4s)^4-5*(4s)^2+4}{20}.....\)
so if s is 1, ans is 180/20=9 yes h is an integer..
but when s is multiple of 5, clearly numerator will be multiple of 5+4, which will not be multiple of 5 and thus 20.. so h is NOT an integer

insufficient


Chetan2u,

can you, please explain this moment? It seems, I miss some properties of numbers.
but when s is multiple of 5, clearly numerator will be multiple of 5+4 - from where does it come from?

I solved in next manner:
(n-2)(n-1)(n+1)(n+2) = 20h

|. n=4 -> divisable. n=20 -> not divisable ---> insufficient The disadvantage of this method - I can miss option n = 20. And in your solution it seems impossible.
||. n is divisable by 5 -> n-2, n-1, n+1, n+2 - none of them is divisable by 5 ---> so the answer is: never divisable ---> not integer. Sufficient.



hi..
equation is \((4s)^4-5s^2+4\)..
now if s is a multiple of 5, (4s)^4 will become (4*5)^4=20^4, which will be div by 5
similarly 5s^2 will also be multiple of 5.
left is 4, so the entire equation becomes MULTIPLE of 5 +4, which will not be div by 5
_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html

Kudos [?]: 5472 [1], given: 112

Manager
Manager
avatar
S
Joined: 26 Jun 2017
Posts: 104

Kudos [?]: 27 [0], given: 176

Location: Russian Federation
Concentration: General Management, Strategy
WE: Information Technology (Other)
Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ? [#permalink]

Show Tags

New post 24 Aug 2017, 03:15
hi..
equation is \((4s)^4-5s^2+4\)..
now if s is a multiple of 5, (4s)^4 will become (4*5)^4=20^4, which will be div by 5
similarly 5s^2 will also be multiple of 5.
left is 4, so the entire equation becomes MULTIPLE of 5 +4, which will not be div by 5
=============================

Fuf, it seems to me, I wanted to sleep too much =)
(MULTIPLE of 5) + 4 - how could I understand it in another way?

Thank you very much!

Kudos [?]: 27 [0], given: 176

Re: If n >2 and n^4-5n^2+4 = 20h, Is h an integer ?   [#permalink] 24 Aug 2017, 03:15
Display posts from previous: Sort by

If n >2 and n^4-5n^2+4 = 20h, Is h an integer ?

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


GMAT Club MBA Forum Home| About| Terms and Conditions| 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®.