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Difference between revisions of "2002 AMC 12A Problems/Problem 3"
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{{duplicate|[[2002 AMC 12A Problems|2009 AMC 12A #3]] and [[2002 AMC 10A Problems|2009 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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==Solution==
 
==Solution==
  
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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>
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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 <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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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>
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<math>(2\uparrow 2)\uparrow (2\uparrow 2) = 4 \uparrow 4 = 256</math>
  
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Therefore, there is one other possible value of <math>2^{2^{2^2}} \Rightarrow \mathrm {(B)}</math>
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Thus the only other result is <math>256</math>, and our answer is <math>\boxed{\text{(B)}\ 1}</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}}
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{{AMC10 box|year=2002|ab=A|num-b=2|num-a=4}}

Revision as of 07:44, 18 February 2009

The following problem is from both the 2009 AMC 12A #3 and 2009 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?

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


Solution

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 $\boxed{\text{(B)}\ 1}$.

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
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