Difference between revisions of "2002 AMC 10A Problems/Problem 3"
(New page: ==Problem== According to the standard convention for exponentiation, <math>2^{2^{2^2}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{16} = 65,536</math>. If the order in which the expone...) |
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==Solution== | ==Solution== | ||
The best way to solve this problem is by simple brute force. We find that the only other value is <math>(2^2)^{2^2}=4^{2^2}=4^4=256</math>. Our answer is just <math>\boxed{\text{(B)}\ 1}</math>. | The best way to solve this problem is by simple brute force. We find that the only other value is <math>(2^2)^{2^2}=4^{2^2}=4^4=256</math>. Our answer is just <math>\boxed{\text{(B)}\ 1}</math>. | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2002|ab=A|num-b=2|num-a=4}} | ||
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| + | [[Category:Introductory Algebra Problems]] | ||
Revision as of 17:39, 26 December 2008
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
The best way to solve this problem is by simple brute force. We find that the only other value is 
. Our answer is just 
.
See Also
| 2002 AMC 10A (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 10 Problems and Solutions | ||
