Difference between revisions of "2002 AMC 12A Problems/Problem 3"

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==Solution==
 
==Solution==
  
<math>Note that 2^{2^2} has a unique value of 16, because 2^4 = 4^2 = 16</math>
+
Note that <math>2^{2^2}</math> has a unique value of <math>16</math>, because <math>2^4 = 4^2 = 16</math>
  
<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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So <math>2^{2^{2^2}}</math> can be perenthesized as either <math>2^({2^2^2))=2^16</math> or <math>(2^2^2)^2=16^2</math>
  
<math>Therefore, there is one other possible value of 2^2^2^2 \Rightarrow \mathrm {(B)}</math>
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Therefore, there is one other possible value of <math>2^2^2^2 \Rightarrow \mathrm {(B)}</math>
  
 
==See Also==
 
==See Also==
  
 
{{AMC12 box|year=2002|ab=A|num-b=2|num-a=4}}
 
{{AMC12 box|year=2002|ab=A|num-b=2|num-a=4}}

Revision as of 21:52, 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 $2^({2^2^2))=2^16$ (Error compiling LaTeX. ! Double superscript.) 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 (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
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