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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. We find that the only other value is <math>(2^2)^{2^2}=4^{2^2}=4^4=256</math>. Our answer is just <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]] | |
