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.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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.

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.

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.
_________________

PS: Like my approach? Please Help me with some Kudos.

Re: The operation x#n for all positive integers greater than 1 [#permalink]

Show Tags

24 Jun 2013, 01:12

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.

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.

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?

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 \(x@3=x^{(x@2)}=x^{(x^x)}=x^{x^x}\).
_________________

Re: The operation x#n for all positive integers greater than 1 [#permalink]

Show Tags

01 Nov 2014, 18:04

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.
_________________

Re: The operation x#n for all positive integers greater than 1 [#permalink]

Show Tags

12 Nov 2015, 22:42

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.
_________________

Re: The operation x#n for all positive integers greater than 1 [#permalink]

Show Tags

10 Jan 2017, 13:58

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.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...