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 22:51, 9 February 2009

Problem

According to the standard convention for exponentiation, \[2^{2^{2^{2}}} = 2^{(2^{(2^2)})} = 2^{16} = 65536.\]

If the order in which the exponentiations are performed is changed, how many other values are possible?

$\mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } 4$


Solution

$Note that 2^{2^2} has a unique value of 16, because 2^4 = 4^2 = 16$

$So 2^{2^{2^2}} can be perenthesized as either 2^({2^2^2))=2^16 or (2^2^2)^2=16^2$ (Error compiling LaTeX. Unknown error_msg)

$Therefore, there is one other possible value of 2^2^2^2 \Rightarrow \mathrm {(B)}$ (Error compiling LaTeX. Unknown error_msg)

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
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