Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 24 Aug 2016, 05:15

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

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

Hide Tags

Senior Manager
Joined: 18 Jun 2010
Posts: 302
Schools: Chicago Booth Class of 2013
Followers: 24

Kudos [?]: 203 [8] , given: 194

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

Show Tags

13 Aug 2010, 10:51
8
KUDOS
14
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

42% (08:40) correct 58% (02:39) wrong based on 319 sessions

HideShow 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
[Reveal] Spoiler: OA
Math Expert
Joined: 02 Sep 2009
Posts: 34409
Followers: 6248

Kudos [?]: 79362 [10] , given: 10016

Re: Just 800 Level question [#permalink]

Show Tags

13 Aug 2010, 11:33
10
KUDOS
Expert's post
2
This post was
BOOKMARKED
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.

_________________
Manager
Joined: 22 Oct 2009
Posts: 242
GMAT 1: 760 Q49 V44
GPA: 3.88
Followers: 6

Kudos [?]: 78 [0], given: 1

Re: Just 800 Level question [#permalink]

Show Tags

15 Aug 2010, 22:39
Where'd this question come from?

And as always -- well done Bunuel!
Math Expert
Joined: 02 Sep 2009
Posts: 34409
Followers: 6248

Kudos [?]: 79362 [0], given: 10016

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

Show Tags

21 Jun 2013, 03:46
Expert's post
2
This post was
BOOKMARKED
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

_________________
Intern
Joined: 22 May 2013
Posts: 49
Concentration: General Management, Technology
GPA: 3.9
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 16 [1] , given: 10

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

Show Tags

21 Jun 2013, 22:29
1
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.
_________________

Intern
Joined: 22 May 2013
Posts: 49
Concentration: General Management, Technology
GPA: 3.9
WE: Information Technology (Computer Software)
Followers: 0

Kudos [?]: 16 [0], given: 10

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

Show Tags

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?
_________________

Intern
Joined: 23 Jun 2013
Posts: 1
Followers: 0

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

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

Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 34409
Followers: 6248

Kudos [?]: 79362 [0], given: 10016

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

Show Tags

24 Jun 2013, 02:35
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.

Hope it helps.
_________________
Intern
Joined: 16 Oct 2013
Posts: 8
Followers: 0

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

Re: Just 800 Level question [#permalink]

Show Tags

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.

Sorry to bump this topic but I don't see what you did with x@3. Could you write out the complete calculation?
Math Expert
Joined: 02 Sep 2009
Posts: 34409
Followers: 6248

Kudos [?]: 79362 [0], given: 10016

Re: Just 800 Level question [#permalink]

Show Tags

23 Oct 2013, 00:16
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.

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 $$x@3=x^{(x@2)}=x^{(x^x)}=x^{x^x}$$.
_________________
Manager
Joined: 07 Apr 2012
Posts: 126
Location: United States
Concentration: Entrepreneurship, Operations
Schools: ISB '15
GMAT 1: 590 Q48 V23
GPA: 3.9
WE: Operations (Manufacturing)
Followers: 0

Kudos [?]: 10 [0], given: 45

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

Show Tags

31 Oct 2013, 09:48
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
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11033
Followers: 509

Kudos [?]: 133 [0], given: 0

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

Show Tags

01 Nov 2014, 19:04
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.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 11033
Followers: 509

Kudos [?]: 133 [0], given: 0

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

Show Tags

12 Nov 2015, 23:42
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] 12 Nov 2015, 23:42
Similar topics Replies Last post
Similar
Topics:
2 X1=1, for all positive integers, Xn=1+7Xn-1 , what is the units digit 4 08 Feb 2016, 18:45
9 If M is the product of all positive integers greater than 59 8 29 Jun 2014, 02:44
7 If n is a positive integer greater than 1, then 2^{n-1} + 2^ 7 07 Sep 2013, 00:46
12 If n is a positive integer greater than 1, then p(n) represe 5 22 Dec 2012, 07:24
6 If x and y are positive integers greater than 1 such that x 9 14 Dec 2012, 23:53
Display posts from previous: Sort by