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13 Aug 2010, 10:51
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
43% (02:43) correct
57%
(02:38)
wrong
based on 861
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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
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
13 Aug 2010, 11:33
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Financier
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
15 Aug 2010, 22:39
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Where'd this question come from?
And as always -- well done Bunuel!
And as always -- well done Bunuel!
