Last visit was: 26 Dec 2024, 11:47 It is currently 26 Dec 2024, 11:47
Decision Tracker
My Rewards
New comers' posts
New posts
Unanswered

Close

GRE Prep 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 GRE 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.

Go to My Error Log Learn more
Close

Request Expert Reply

Confirm Cancel
PREVIOUS TOPIC
NEXT TOPIC

n^^ = n! - n^2, where n is a positive integer. For how many

Tags :
Find Similar Topics  About  Add a Tag

Show Tags
Hide Tags

avatar
Bunuel
Senior Manager
Senior Manager
Joined: 20 May 2014
Posts: 285
Own Kudos [?]: 725 [0]
Given Kudos: 225
n^^ = n! - n^2, where n is a positive integer. For how many [#permalink] New post   01 Oct 2017, 04:43
00:00

Question Stats:

64% (01:01) correct 35% (01:53) wrong based on 17 sessions

Hide Show timer Statistics

n^^ = n! - n^2, where n is a positive integer. For how many values of n is n^^ less than zero?

A. 1^^
B. 1^^ – 2^^
C. 3^^
D. 3^^ + 4^^
E. 5^^ – 2^^

Kudos for correct solution.
ShowHide Answer
Official Answer
B
avatar
pranab223
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2281 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM

Top Member of the Month
Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink] New post   01 Oct 2017, 07:20
Bunuel wrote:
n^^ = n! - n^2, where n is a positive integer. For how many values of n is n^^ less than zero?

A. 1^^
B. 1^^ – 2^^
C. 3^^
D. 3^^ + 4^^
E. 5^^ – 2^^

Kudos for correct solution.


Not clear with the ques?

It has asked for how many values of n, n^^<0, or should it be; for what value of n, n^^<0. PLz clarify??
avatar
IlCreatore
Director
Director
Joined: 03 Sep 2017
Posts: 518
Own Kudos [?]: 707 [0]
Given Kudos: 0
Send PM
Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink] New post   01 Oct 2017, 07:59
2
I interpreted the first way.

Thus,
1^^ = 1! - 1 = 0
2^^ = 2! - 4 = -2
3^^ = 3! - 9 = -3
4^^ = 4! - 16 = 18

So, I would say there are two cases in which n^^<0. Now, we have to find which one of the choice has value 2.
Given our prior computations, it is easy to notice that 1^^-2^^ = 0 - (-2) = 2.

Thus, the answer is B!
avatar
pranab223
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2281 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM

Top Member of the Month
Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink] New post   01 Oct 2017, 08:23
IlCreatore wrote:
I interpreted the first way.

Thus,
1^^ = 1! - 1 = 0
2^^ = 2! - 4 = -2
3^^ = 3! - 9 = -3
4^^ = 4! - 16 = 18

So, I would say there are two cases in which n^^<0. Now, we have to find which one of the choice has value 2.
Given our prior computations, it is easy to notice that 1^^-2^^ = 0 - (-2) = 2.

Thus, the answer is B!


you are right, i missed it :(
avatar
indiragre18
Intern
Intern
Joined: 27 Oct 2018
Posts: 49
Own Kudos [?]: 37 [0]
Given Kudos: 0
Send PM
Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink] New post   02 Nov 2018, 23:51
C & E
avatar
chetan2u
Supreme Moderator
Joined: 01 Nov 2017
Posts: 371
Own Kudos [?]: 471 [0]
Given Kudos: 0
Send PM
Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink] New post   03 Nov 2018, 04:50
1
Expert Reply
Bunuel wrote:
n^^ = n! - n^2, where n is a positive integer. For how many values of n is n^^ less than zero?

A. 1^^
B. 1^^ – 2^^
C. 3^^
D. 3^^ + 4^^
E. 5^^ – 2^^

Kudos for correct solution.



n^^ = \(n! - n^2\).... n^^ < 0 means \(n!-n^2<0...........n^2>n!......n^2>n(n-1)!.......n>(n-1)!\)
so the number must be greater than the product of numbers less than itself.
3>2*1 but 4 is NOT > 3*2 so values are 1, 2 and 3 but 1 will give us a 0, thus 2 values...
1^^ is 0, so cannot be the answer.
3^^ is negative, so cannot be the answer
so let us check 1^^-2^^ = 0-(2!-2^2)=0-(2-4)=2...answer

B
Prep Club for GRE Bot
gmatclubot
Re: n^^ = n! - n^2, where n is a positive integer. For how many [#permalink]
03 Nov 2018, 04:50
Moderators:
adewale223
GRE Instructor
88 posts
bronaugust
GRE Forum Moderator
37 posts
BrushMyQuant
Moderator
1115 posts
VibhuAnurag
GRE Instructor
234 posts

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

Main navigation

Partners

GRE Resources

www.ets.org/gre | About | Terms and Conditions and Privacy Policy | Prep Club for GRE Rules | Contact