2002 AMC 12A Problems/Problem 3

The following problem is from both the 2002 AMC 12A #3 and 2002 AMC 10A #3, so both problems redirect to this page.

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?

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

Solution 1

The best way to solve this problem is by simple brute force.

It is convenient to drop the usual way how exponentiation is denoted, and to write the formula as $2\uparrow 2\uparrow 2\uparrow 2$ , where $\uparrow$ denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:

$2\uparrow (2\uparrow (2\uparrow 2))$
$2\uparrow ((2\uparrow 2)\uparrow 2)$
$((2\uparrow 2)\uparrow 2)\uparrow 2$
$(2\uparrow (2\uparrow 2))\uparrow 2$
$(2\uparrow 2)\uparrow (2\uparrow 2)$

We can note that $2\uparrow (2\uparrow 2) = (2\uparrow 2)\uparrow 2 =16$ . Therefore options 1 and 2 are equal, and options 3 and 4 are equal. Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.

$((2\uparrow 2)\uparrow 2)\uparrow 2 = 16\uparrow 2 = 256$

$(2\uparrow 2)\uparrow (2\uparrow 2) = 4 \uparrow 4 = 256$

Thus the only other result is $256$ , and our answer is $\boxed{\textbf{(B) } 1}$ .

Solution 2 (Recursive Method)

We will proceed by recursion using the same notation as Solution 1. Note that we can either have $(2 \uparrow 2 \uparrow 2) \uparrow 2$ or $2 \uparrow (2 \uparrow 2 \uparrow 2)$ .

However, the expression in the parentheses can either be $(2 \uparrow 2) \uparrow 2$ or $2 \uparrow (2 \uparrow 2)$ , which correspond to the same value of $16$ . Therefore, we can either have $16^2$ or $2^{16}$ , so our answer is $\boxed{\textbf{(B) }1}$ .

~TPColor

Video Solution by Daily Dose of Math

https://youtu.be/fJndjYHWBrU

~Thesmartgreekmathdude

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

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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2002 AMC 12A Problems/Problem 3

Contents

Problem

Solution 1

Solution 2 (Recursive Method)

Video Solution by Daily Dose of Math

See Also