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| − | ==Problem==
| + | #redirect [[2002 AMC 12A Problems/Problem 3]] |
| − | According to the standard convention for exponentiation,
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| − | <math>2^{2^{2^2}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{16} = 65,536</math>.
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| − | If the order in which the exponentiations are performed is changed, how many other values are possible?
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| − | <math>\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 3 \qquad \text{(E)}\ 4</math>
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| − | ==Solution==
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| − | The best way to solve this problem is by simple brute force.
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| − | It is convenient to drop the usual way how exponentiation is denoted, and to write the formula as <math>2\uparrow 2\uparrow 2\uparrow 2</math>, where <math>\uparrow</math> denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:
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| − | # <math>2\uparrow (2\uparrow (2\uparrow 2))</math>
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| − | # <math>2\uparrow ((2\uparrow 2)\uparrow 2)</math>
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| − | # <math>((2\uparrow 2)\uparrow 2)\uparrow 2</math> | |
| − | # <math>(2\uparrow (2\uparrow 2))\uparrow 2</math>
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| − | # <math>(2\uparrow 2)\uparrow (2\uparrow 2)</math>
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| − | We can note that <math>2\uparrow (2\uparrow 2) = (2\uparrow 2)\uparrow 2 =16</math>. Therefore options 1 and 2 are equal, and options 3 and 4 are equal.
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| − | Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.
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| − | <math>((2\uparrow 2)\uparrow 2)\uparrow 2 = 16\uparrow 2 = 256</math>
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| − | <math>(2\uparrow 2)\uparrow (2\uparrow 2) = 4 \uparrow 4 = 256</math>
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| − | Thus the only other result is <math>256</math>, and our answer is <math>\boxed{\text{(B)}\ 1}</math>.
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| − | ==See Also==
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| − | {{AMC10 box|year=2002|ab=A|num-b=2|num-a=4}}
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| − | [[Category:Introductory Algebra Problems]] | |
