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Difference between revisions of "2002 AMC 12A Problems/Problem 3"
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{{duplicate|[[2002 AMC 12A Problems|2002 AMC 12A #3]] and [[2002 AMC 10A Problems|2002 AMC 10A #3]]}}
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==Problem==
 
==Problem==
 
According to the standard convention for exponentiation,  
 
According to the standard convention for exponentiation,  
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If the order in which the exponentiations are performed is changed, how many other values are possible?
 
If the order in which the exponentiations are performed is changed, how many other values are possible?
  
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<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } 4 </math>
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<math> \textbf{(A) } 0\qquad \textbf{(B) } 1\qquad \textbf{(C) } 2\qquad \textbf{(D) } 3\qquad \textbf{(E) } 4 </math>
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==Solution 1==
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The best way to solve this problem is by simple brute force.
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It is convenient to drop the usual way how exponentiation is denoted, and to write the formula as <math>2\uparrow 2\uparrow 2\uparrow 2</math>, where <math>\uparrow</math> denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:
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# <math>2\uparrow (2\uparrow (2\uparrow 2))</math>
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# <math>2\uparrow ((2\uparrow 2)\uparrow 2)</math>
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# <math>((2\uparrow 2)\uparrow 2)\uparrow 2</math>
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# <math>(2\uparrow (2\uparrow 2))\uparrow 2</math>
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# <math>(2\uparrow 2)\uparrow (2\uparrow 2)</math>
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We can note that <math>2\uparrow (2\uparrow 2) = (2\uparrow 2)\uparrow 2 =16</math>. Therefore options 1 and 2 are equal, and options 3 and 4 are equal.
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Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.
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<math>((2\uparrow 2)\uparrow 2)\uparrow 2 = 16\uparrow 2 = 256</math>
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<math>(2\uparrow 2)\uparrow (2\uparrow 2) = 4 \uparrow 4 = 256</math>
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Thus the only other result is <math>256</math>, and our answer is <math>\boxed{\textbf{(B) } 1}</math>.
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==Solution 2 (Recursive Method)==
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We will proceed by recursion using the same notation as Solution 1. Note that we can either have <math>(2 \uparrow 2 \uparrow 2) \uparrow 2</math> or <math>2 \uparrow (2 \uparrow 2 \uparrow 2)</math>.
  
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However, the expression in the parentheses can either be <math>(2 \uparrow 2) \uparrow 2</math> or <math>2 \uparrow (2 \uparrow 2)</math>, which correspond to the same value of <math>16</math>. Therefore, we can either have <math>16^2</math> or <math>2^{16}</math>, so our answer is <math>\boxed{\textbf{(B) }1}</math>.
  
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==Solution==
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~TPColor
  
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<math>Note that 2^{2^2} has a unique value of 16, because 2^4 = 4^2 = 16</math>
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==Video Solution by Daily Dose of Math==
  
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<math>So 2^{2^{2^2}} can be perenthesized as either 2^({2^2^2))=2^16 or (2^2^2)^2=16^2</math>
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https://youtu.be/fJndjYHWBrU
  
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<math>Therefore, there is one other possible value of 2^2^2^2 \Rightarrow \mathrm {(B)}</math>
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~Thesmartgreekmathdude
  
 
==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}}
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{{AMC10 box|year=2002|ab=A|num-b=2|num-a=4}}
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{{MAA Notice}}

Latest revision as of 07:31, 16 June 2025

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:

  1. $2\uparrow (2\uparrow (2\uparrow 2))$
  2. $2\uparrow ((2\uparrow 2)\uparrow 2)$
  3. $((2\uparrow 2)\uparrow 2)\uparrow 2$
  4. $(2\uparrow (2\uparrow 2))\uparrow 2$
  5. $(2\uparrow 2)\uparrow (2\uparrow 2)$

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

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
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. AMC Logo.png

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