Difference between revisions of "2002 AMC 12A Problems/Problem 3"
| Line 10: | Line 10: | ||
==Solution== | ==Solution== | ||
| − | <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 <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> |
| − | + | 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,
![\[2^{2^{2^{2}}} = 2^{(2^{(2^2)})} = 2^{16} = 65536.\]](http://latex.artofproblemsolving.com/2/9/d/29dee8eac6c7812722f0013ad002044b30d8f220.png)
If the order in which the exponentiations are performed is changed, how many other values are possible?
Solution
Note that 
has a unique value of 
, because 
So 
can be perenthesized as either $2^({2^2^2))=2^16$ (Error compiling LaTeX. Unknown error_msg) 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 (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 | |
