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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]

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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]

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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).

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

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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.
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