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16 Sep 2014, 01:51
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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:
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Question Stats:
26% (01:04) correct
74%
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The operation \(x N\) for all positive integers n greater than 1 is defined in the following manner:
\(x N = x\) raised 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. \((2 @ 2) @ 3\)
\(x N = x\) raised 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. \((2 @ 2) @ 3\)
16 Sep 2014, 01:51
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Official Solution:
The operation \(x [b]N\) for all positive integers n greater than 1 is defined in the following manner:
\(x N = x\) raised to the power of \(x @ (n-1)\).
If \(x 1 = x\), which of the following expressions has the greatest value?
[/b]
A. \((3 @ 2) @ 2\)
B. \(3 @ (1 @ 3)\)
C. \((2 @ 3) @ 2\)
D. \(2 @ (2 @ 3)\)
E. \((2 @ 2) @ 3\)
First, we need to figure out what this strange operation means for a few small integers n. Let's build upward from 1:
\(x 1 = x\)
\(x @ 2 = x\) raised to the power of \(x 1\) (which is just \(x\)), so \(x @ 2 = x^x\)
\(x @ 3 = x\) raised to the power of \(x @ 2\), so \(x @ 3 = x^{x^x}\)
\(x @ 4 = x^{x^{x^x}}\)
So the number after the @ sign tells you how many \(x\)'s are in the exponential expression. Now we can translate the answer choices. As always, do the operation inside the parentheses first.
(A) \((3 @ 2) @ 2\)
\(3 @ 2 = 3^3 = 27\)
\(27 @ 2 = 27^{27} = (3^3)^{27} = 3^{81}\)
(B) \(3 @ (1 @ 3)\)
\(1 @ 3 = 1^{1^1} = 1^1 = 1\)
\(3@1 = 3\)
(C) \((2 @ 3) @ 2\)
\(2 @ 3 = 2^{2^2} = 2^4 = 16\)
\(16 @ 2 = 16^{16} = (2^4)^{16} = 2^{64}\)
Because both the base and the exponent of this answer choice are smaller, we can tell that choice A is still the winner at this point.
(D) \(2 @ (2 @ 3)\)
\(2 @ 3 = 2^{2^2} = 2^4 = 16\)
\(2 16 = 2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^2}}}}}}}}}}}}}}\)
There are sixteen 2's in this "tower of powers"! This number is incredibly large, far larger than \(3^{81}\). Let's start to collapse the layers to see why.
\(2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^2}}}}}}}}}}}}}}\)
\(= 2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^4}}}}}}}}}}}}}\)
\(= 2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{16}}}}}}}}}}}}}\)
\(2^{16} = 65536\) You aren't expected to know that, of course, but now imagine 2 raised to that power. This number has thousands of digits.
Now imagine 2 raised to THAT power.
Then 2 raised to THAT power.
And so on, over 10 more times!
This number is the winner by far among the first four answer choices.
(E) \((2 @ 2) @ 3\)
\(2 @ 2 = 2^2 = 4\)
\(4 @ 3 = 4^{4^4} = 4^{256} = 2^{512}\)
While enormous, this number is still far smaller than answer choice (D).
By the way, the operation represented by the @ sign in this problem is sometimes called "tetration." The reason is that just as multiplication is repeated addition, and exponentiation is repeated multiplication, so-called "tetration" is repeated exponentiation. ("Tetra-" means "four," and this operation is fourth in line: addition, multiplication, exponentiation, tetration.) Tetration is also called superexponentiation, ultraexponentiation, hyper-4, and power tower.
Answer: D
The operation \(x [b]N\) for all positive integers n greater than 1 is defined in the following manner:
\(x N = x\) raised to the power of \(x @ (n-1)\).
If \(x 1 = x\), which of the following expressions has the greatest value?
[/b]
A. \((3 @ 2) @ 2\)
B. \(3 @ (1 @ 3)\)
C. \((2 @ 3) @ 2\)
D. \(2 @ (2 @ 3)\)
E. \((2 @ 2) @ 3\)
First, we need to figure out what this strange operation means for a few small integers n. Let's build upward from 1:
\(x 1 = x\)
\(x @ 2 = x\) raised to the power of \(x 1\) (which is just \(x\)), so \(x @ 2 = x^x\)
\(x @ 3 = x\) raised to the power of \(x @ 2\), so \(x @ 3 = x^{x^x}\)
\(x @ 4 = x^{x^{x^x}}\)
So the number after the @ sign tells you how many \(x\)'s are in the exponential expression. Now we can translate the answer choices. As always, do the operation inside the parentheses first.
(A) \((3 @ 2) @ 2\)
\(3 @ 2 = 3^3 = 27\)
\(27 @ 2 = 27^{27} = (3^3)^{27} = 3^{81}\)
(B) \(3 @ (1 @ 3)\)
\(1 @ 3 = 1^{1^1} = 1^1 = 1\)
\(3@1 = 3\)
(C) \((2 @ 3) @ 2\)
\(2 @ 3 = 2^{2^2} = 2^4 = 16\)
\(16 @ 2 = 16^{16} = (2^4)^{16} = 2^{64}\)
Because both the base and the exponent of this answer choice are smaller, we can tell that choice A is still the winner at this point.
(D) \(2 @ (2 @ 3)\)
\(2 @ 3 = 2^{2^2} = 2^4 = 16\)
\(2 16 = 2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^2}}}}}}}}}}}}}}\)
There are sixteen 2's in this "tower of powers"! This number is incredibly large, far larger than \(3^{81}\). Let's start to collapse the layers to see why.
\(2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^2}}}}}}}}}}}}}}\)
\(= 2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^4}}}}}}}}}}}}}\)
\(= 2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{2^{16}}}}}}}}}}}}}\)
\(2^{16} = 65536\) You aren't expected to know that, of course, but now imagine 2 raised to that power. This number has thousands of digits.
Now imagine 2 raised to THAT power.
Then 2 raised to THAT power.
And so on, over 10 more times!
This number is the winner by far among the first four answer choices.
(E) \((2 @ 2) @ 3\)
\(2 @ 2 = 2^2 = 4\)
\(4 @ 3 = 4^{4^4} = 4^{256} = 2^{512}\)
While enormous, this number is still far smaller than answer choice (D).
By the way, the operation represented by the @ sign in this problem is sometimes called "tetration." The reason is that just as multiplication is repeated addition, and exponentiation is repeated multiplication, so-called "tetration" is repeated exponentiation. ("Tetra-" means "four," and this operation is fourth in line: addition, multiplication, exponentiation, tetration.) Tetration is also called superexponentiation, ultraexponentiation, hyper-4, and power tower.
Answer: D
14 Mar 2016, 19:38
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Awesome explanation and cool question. I read about Graham's number recently and that talks about tetration and even higher powers of these kinds of operations. The numbers get so large that all the atoms in the universe aren't even close to come near an approximation. It's mind blowing. Here is the article: https://waitbutwhy.com/2014/11/1000000-grahams-number.html
