Difference between revisions of "2002 AMC 12A Problems/Problem 3"
(New page: ==Problem== According to the standard convention for exponentiation, <cmath> 2^{2^{2^{2}}} = 2^{(2^{(2^2)})} = 2^{16} = 65536. </cmath> If the order in which the exponentiations are perf...) |
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<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } 4 </math> | <math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } 4 </math> | ||
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+ | ==Solution== | ||
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+ | <math>Note that 2^{2^2} has a unique value of 16, because 2^4 = 4^2 = 16</math> | ||
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+ | <math>So 2^{2^{2^2}} can be perenthesized as either 2^({2^2^2))=2^16 or (2^2^2)^2=16^2</math> | ||
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+ | <math>Therefore, there is one other possible value of 2^2^2^2 \Rightarrow \mathrm {(B)}</math> | ||
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+ | ==See Also== | ||
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+ | {{AMC12 box|year=2002|ab=A|num-b=2|num-a=4}} |
Revision as of 21:51, 9 February 2009
Problem
According to the standard convention for exponentiation,
If the order in which the exponentiations are performed is changed, how many other values are possible?
Solution
$So 2^{2^{2^2}} can be perenthesized as either 2^({2^2^2))=2^16 or (2^2^2)^2=16^2$ (Error compiling LaTeX. ! Double superscript.)
$Therefore, there is one other possible value of 2^2^2^2 \Rightarrow \mathrm {(B)}$ (Error compiling LaTeX. ! Double superscript.)
See Also
2002 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 2 |
Followed by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |