Difference between revisions of "2002 AMC 12A Problems/Problem 3"
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Note that <math>2^{2^2}</math> has a unique value of <math>16</math>, because <math>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> | ||
− | So <math>2^{2^{2^2}}</math> can be perenthesized as either <math>2^({2^2^2 | + | 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> | + | 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 21:53, 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
Note that has a unique value of , because
So can be perenthesized as either or
Therefore, there is one other possible value of
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 |