2002 AMC 10A Problems/Problem 3

Problem

According to the standard convention for exponentiation,

$2^{2^{2^2}} = 2^{\left(2^{\left(2^2\right)}\right)} = 2^{16} = 65,536$ .

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

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

Solution

The best way to solve this problem is by simple brute force. We find that the only other value is $(2^2)^{2^2}=4^{2^2}=4^4=256$ . Our answer is just $\boxed{\text{(B)}\ 1}$ .

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
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2002 AMC 10A Problems/Problem 3

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