NEW HERE? LEARN MORE
closeclose
Last visit was: 23 Apr 2024, 16:34 It is currently 23 Apr 2024, 16:34
Decision Tracker
My Rewards
New posts
New comers' posts

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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

The operation x#n for all positive integers greater than 1

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
Financier
Manager
Manager
Joined: 18 Jun 2010
Posts: 230
Own Kudos [?]: 656 [101]
Given Kudos: 194
Schools:Chicago Booth Class of 2013
Send PM
Reviews Badge
The operation x#n for all positive integers greater than 1 [#permalink] New post   13 Aug 2010, 10:51
12
Kudos
88
Bookmarks
00:00
A
B
C
D
E

Difficulty:

95% (hard)

Question Stats:

42% (02:45) correct 58% (02:38) wrong based on 749 sessions
Hide Show timer Statistics
The operation x#n for all positive integers greater than 1 is defined in the following manner: x#n = x to the power of x#(n-1)

If x#1 = x, which of the following expressions has the greatest value?

A. (3#2)#2
B. 3#(1#3)
C. (2#3)#2
D. 2#(2#3)
E. (2#2)#3
ShowHide Answer
Official Answer
D
Most Helpful Reply
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92883
Own Kudos [?]: 618587 [35]
Given Kudos: 81563
Send PM
CAT Tests Forum Quiz Mode
The operation x#n for all positive integers greater than 1 [#permalink] New post   13 Aug 2010, 11:33
17
Kudos
18
Bookmarks
Expert Reply
Financier wrote:
The operation x#n for all positive integers greater than 1 is defined in the following manner:
x#n = x to the power of x#(n-1)

If x#1 = x,
which of the following expressions has the greatest value?


A. (3#2)#2
B. 3#(1#3)
C. (2#3)#2
D. 2#(2#3)
E. (2#2)#3


Couple of things before solving:

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
\(a^{m^n}=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).

So:
\((a^m)^n=a^{mn}\);

\(a^{m^n}=a^{(m^n)}\) and not \((a^m)^n\).

Back to the original question:
Let's replace # by AS # looks like the symbol "not equal to" and it might confuse someone.

Given: \(x@n=x^{(x@(n-1))}\) and \(x@1=x\):
______________________________________

\(x@2=x^{(x@1)}=x^x\), as \(x@1=x\);
\(x@3=x^{(x@2)}=x^{(x^x)}=x^{x^x}\);
\(x@4=x^{(x@3)}=x^{(x^{x^x})}=x^{x^{x^x}}\);
...
Basically n in x@n represents the # of stacked x-es.

A. \((3@2)@2=(3^3)@2=(27)@2=27^{27}=3^{81}\)

B. \(3@(1@3)=3@(1^{1^1})=3@1=3\)

C. \((2@3)@2=(2^{2^2})@2=16@2=16^{16}=2^{64}\)

D. \(2@(2@3)=2@(2^{2^2})=2@16\) this will be huge number:

    2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2=
    =2^2^2^2^2^2^2^2^2^2^2^2^2^2^4=
    =2^2^2^2^2^2^2^2^2^2^2^2^2^16=....;

E. \((2@2)@3=(2^2)@3=4@3=4^{4^4}=4^{256}=2^{512}\)

Option D (2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2) will be much bigger number than numbers from other answer choices.

Answer: D.
General Discussion
User avatar
B
YourDreamTheater
Manager
Manager
Joined: 22 Oct 2009
Posts: 223
Own Kudos [?]: 211 [2]
Given Kudos: 1
GMAT 1: 760 Q49 V44
GPA: 3.88
Send PM
Re: Just 800 Level question [#permalink] New post   15 Aug 2010, 22:39
1
Kudos
1
Bookmarks
Where'd this question come from?

And as always -- well done Bunuel!
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92883
Own Kudos [?]: 618587 [2]
Given Kudos: 81563
Send PM
CAT Tests Forum Quiz Mode
Re: The operation x#n for all positive integers greater than 1 [#permalink] New post   21 Jun 2013, 03:46
2
Bookmarks
Expert Reply
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Functions and Custom Characters questions: search.php?search_id=tag&tag_id=40
All PS Functions and Custom Characters questions: search.php?search_id=tag&tag_id=61
avatar
kpali
Intern
Intern
Joined: 22 May 2013
Posts: 35
Own Kudos [?]: 34 [2]
Given Kudos: 10
Concentration: General Management, Technology
Schools: Tippie '16 Sloan '16 Tepper '16 Duke '16
GPA: 3.9
WE:Information Technology (Computer Software)
Send PM
GMAT ToolKit User
Re: The operation x#n for all positive integers greater than 1 [#permalink] New post   21 Jun 2013, 22:29
2
Kudos
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Functions and Custom Characters questions: search.php?search_id=tag&tag_id=40
All PS Functions and Custom Characters questions: search.php?search_id=tag&tag_id=61


Well originally, i got this question wrong.
Bu then i figured an alternative approach which would have been very easy to start with, please let me know if there is something wrong with it.

First of all to look at the options, everything looks to be in the form of either,

a@(b@c) or (a@b)@c

Now lets quickly try to see the diff between the magnitude of these different forms, lets take a,b,c = 2 for comparison purpose.

2@(2@2) = 2@4 = 2^(2^(2^2)) = 2^16
while, (2@2)@2 = (2^2)@2 = 4@2 = 4^4 = 2^8

So well, there is a huge magnitude diff between the post, and option 1 def has much more wightage.

now, lets late at our options in the question : (all the options have either 2 or 3 as the values for a, b and c in a@(b@c) and (a@b)@c, so we can make a direct comparison.

We have B and D in a@(b@c) form, however since B has 1 as one of the numbers ( 1 and 0 are special no's with special properties),

so we can easily see that 3@(1@3) = 3@1 = 3 (which is most definitely the smallest)

Which leaves to D (2@16) now which is the answer.
avatar
kpali
Intern
Intern
Joined: 22 May 2013
Posts: 35
Own Kudos [?]: 34 [0]
Given Kudos: 10
Concentration: General Management, Technology
Schools: Tippie '16 Sloan '16 Tepper '16 Duke '16
GPA: 3.9
WE:Information Technology (Computer Software)
Send PM
GMAT ToolKit User
Re: The operation x#n for all positive integers greater than 1 [#permalink] New post   23 Jun 2013, 06:51
Bunuel wrote:
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE

All DS Functions and Custom Characters questions: search.php?search_id=tag&tag_id=40
All PS Functions and Custom Characters questions: search.php?search_id=tag&tag_id=61


Hi Bunuel,

Could you please let me know if the above approach is fine? or if there is something wrong with it?
avatar
Bhagavantha
Intern
Intern
Joined: 23 Jun 2013
Posts: 1
Own Kudos [?]: [0]
Given Kudos: 0
Send PM
Re: The operation x#n for all positive integers greater than 1 [#permalink] New post   24 Jun 2013, 02:12
Sir, I am totally confused with wat I have to study and on wat questions I should practice in this website. There are so many navigation from 1 page to other. Please provide me the solution where I can improve myself with enough practice.
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92883
Own Kudos [?]: 618587 [0]
Given Kudos: 81563
Send PM
CAT Tests Forum Quiz Mode
Re: The operation x#n for all positive integers greater than 1 [#permalink] New post   24 Jun 2013, 02:35
Expert Reply
Bhagavantha wrote:
Sir, I am totally confused with wat I have to study and on wat questions I should practice in this website. There are so many navigation from 1 page to other. Please provide me the solution where I can improve myself with enough practice.


New to the Math Forum? Please read this first: new-to-the-math-forum-please-read-this-first-140445.html

Hope it helps.
avatar
kartboybo
Intern
Intern
Joined: 16 Oct 2013
Posts: 6
Own Kudos [?]: 51 [0]
Given Kudos: 0
Send PM
Re: Just 800 Level question [#permalink] New post   22 Oct 2013, 23:46
Bunuel wrote:
Financier wrote:
The operation x#n for all positive integers greater than 1 is defined in the following manner:
x#n = x to the power of x#(n-1)

If x#1 = x,
which of the following expressions has the greatest value?


A. (3#2)#2
B. 3#(1#3)
C. (2#3)#2
D. 2#(2#3)
E. (2#2)#3


Couple of things before solving:

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).

So:
\((a^m)^n=a^{mn}\);

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\).

Back to the original question:
Let's replace # by @ as # looks like the symbol "not equal to" and it might confuse someone.

Given: \(x@n=x^{(x@(n-1))}\) and \(x@1=x\):
______________________________________

\(x@2=x^{(x@1)}=x^x\), as \(x@1=x\);
\(x@3=x^{(x@2)}=x^{(x^x)}=x^{x^x}\);
\(x@4=x^{(x@3)}=x^{(x^{x^x})}=x^{x^{x^x}}\);
...
Basically n in x@n represents the # of stacked x-es.

A. \((3@2)@2=(3^3)@2=(27)@2=27^{27}=3^{81}\)

B. \(3@(1@3)=3@(1^{1^1})=3@1=3\)

C. \((2@3)@2=(2^{2^2})@2=16@2=16^{16}=2^{64}\)

D. \(2@(2@3)=2@(2^{2^2})=2@16\) this will be huge number 2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2=2^2^2^2^2^2^2^2^2^2^2^2^2^2^4=2^2^2^2^2^2^2^2^2^2^2^2^2^16=.... ;

E. \((2@2)@3=(2^2)@3=4@3=4^{4^4}=4^{256}=2^{512}\)

Option D (2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2) will be much bigger number than numbers from other answer choices.

Answer: D.


Sorry to bump this topic but I don't see what you did with x@3. Could you write out the complete calculation? :)
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92883
Own Kudos [?]: 618587 [0]
Given Kudos: 81563
Send PM
CAT Tests Forum Quiz Mode
Re: Just 800 Level question [#permalink] New post   23 Oct 2013, 00:16
Expert Reply
kartboybo wrote:
Bunuel wrote:
Financier wrote:
The operation x#n for all positive integers greater than 1 is defined in the following manner:
x#n = x to the power of x#(n-1)

If x#1 = x,
which of the following expressions has the greatest value?


A. (3#2)#2
B. 3#(1#3)
C. (2#3)#2
D. 2#(2#3)
E. (2#2)#3


Couple of things before solving:

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\), which on the other hand equals to \(a^{mn}\).

So:
\((a^m)^n=a^{mn}\);

\(a^m^n=a^{(m^n)}\) and not \((a^m)^n\).

Back to the original question:
Let's replace # by @ as # looks like the symbol "not equal to" and it might confuse someone.

Given: \(x@n=x^{(x@(n-1))}\) and \(x@1=x\):
______________________________________

\(x@2=x^{(x@1)}=x^x\), as \(x@1=x\);
\(x@3=x^{(x@2)}=x^{(x^x)}=x^{x^x}\);
\(x@4=x^{(x@3)}=x^{(x^{x^x})}=x^{x^{x^x}}\);
...
Basically n in x@n represents the # of stacked x-es.

A. \((3@2)@2=(3^3)@2=(27)@2=27^{27}=3^{81}\)

B. \(3@(1@3)=3@(1^{1^1})=3@1=3\)

C. \((2@3)@2=(2^{2^2})@2=16@2=16^{16}=2^{64}\)

D. \(2@(2@3)=2@(2^{2^2})=2@16\) this will be huge number 2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2=2^2^2^2^2^2^2^2^2^2^2^2^2^2^4=2^2^2^2^2^2^2^2^2^2^2^2^2^16=.... ;

E. \((2@2)@3=(2^2)@3=4@3=4^{4^4}=4^{256}=2^{512}\)

Option D (2^2^2^2^2^2^2^2^2^2^2^2^2^2^2^2) will be much bigger number than numbers from other answer choices.

Answer: D.


Sorry to bump this topic but I don't see what you did with x@3. Could you write out the complete calculation? :)


Don't know what more can I add:
We have that \(x@2=x^x\), thus \(x@3=x^{(x@2)}=x^{(x^x)}=x^{x^x}\).
User avatar
ygdrasil24
Manager
Manager
Joined: 07 Apr 2012
Posts: 71
Own Kudos [?]: 25 [1]
Given Kudos: 45
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE:Operations (Manufacturing)
Send PM
Re: The operation x#n for all positive integers greater than 1 [#permalink] New post   31 Oct 2013, 09:48
1
Kudos
It took me a while to figure out, but then I am glad in the end.

I just have a diff approach compared to insanely awesome Bunuel's, so thought to share.

We are given x#1 = 1 . Given x > 1, so # could be either of ^ , / , *

So if we start solving x#1, as per the original expression i.e x#n = .....

We arrive to x#0 = 1 in the second setup. What could be # then ?
The only option is # is exponential.

Now its school stuff to know which one is greater, Option D clearly stands out.

A. (3^2)^2 = 81
B. 3
C. 2^6
D. 2^8
E. 2^6.

Having given GMAT once, I think this is a 700 question, easy, but looks intimidating in beginning :)

Liked it ? Appreciate it :)
avatar
B
WheatyPie
Manager
Manager
Joined: 01 Jan 2019
Posts: 52
Own Kudos [?]: 27 [0]
Given Kudos: 14
Send PM
Re: The operation x#n for all positive integers greater than 1 [#permalink] New post   27 May 2020, 07:24
ygdrasil24 wrote:
We are given x#1 = 1 . Given x > 1, so # could be either of ^ , / , *


This isn't true. # isn't limited to a normal algebraic operator, it could be anything you can imagine.

Maybe # means "count the number of intersections of lines in the character to the left and multiply it by the character to the right", so x#1=1 because x has 1 intersection, multiplied by 1 gives 1.

This is just a silly example, but the point is, we have no idea what # means without more information than "x#1=1"
User avatar
bumpbot
Non-Human User
Joined: 09 Sep 2013
Posts: 32628
Own Kudos [?]: 821 [0]
Given Kudos: 0
Send PM
Re: The operation x#n for all positive integers greater than 1 [#permalink] New post   02 Nov 2023, 05:40
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
gmatclubot
Re: The operation x#n for all positive integers greater than 1 [#permalink]
02 Nov 2023, 05:40
Moderators:
Bunuel
Math Expert
92883 posts
yashikaaggarwal
Senior Moderator - Masters Forum
3137 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions