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

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by Anurag@Gurome » Tue Jul 10, 2012 3:19 am
gmatter2012 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?
x@1 = x
x@2 = x^(x@(2 - 1)) = x^(x@1) = x^x
x@3 = x^(x@(3 - 1)) = x^(x@2) = x^(x^x)
Hence, x@n = x^(x^(x^... n times)))

Let us check the options,
  • A. (3@2)@2 = (3^3)@2 = 27@2 = 27^27 = (3^3)^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 = (2^4)^16 = 2^64
    D. 2@(2@3) = 2@(2^(2^2)) = 2@16 ---> This will be a huge number 2(2^(2^... 16 times)))
    E. (2@2)@3 = (2^2)@3 = 4@3 = 4^(4^4) = (2^2)^(256) = 2^512
The correct answer is D.
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by gmatter2012 » Tue Jul 10, 2012 3:36 am
Anurag@Gurome wrote:
gmatter2012 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?
x@1 = x
x@2 = x^(x@(2 - 1)) = x^(x@1) = x^x
x@3 = x^(x@(3 - 1)) = x^(x@2) = x^(x^x)
Hence, x@n = x^(x^(x^... n times)))

Let us check the options,
  • A. (3@2)@2 = (3^3)@2 = 27@2 = 27^27 = (3^3)^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 = (2^4)^16 = 2^64
    D. 2@(2@3) = 2@(2^(2^2)) = 2@16 ---> This will be a huge number 2(2^(2^... 16 times)))
    E. (2@2)@3 = (2^2)@3 = 4@3 = 4^(4^4) = (2^2)^(256) = 2^512
The correct answer is D.
Great !
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