During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

Difference between revisions of "2002 AMC 12A Problems/Problem 3"

(Solution)
 
(One intermediate revision by the same user not shown)
Line 7: Line 7:
 
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?
  
<math> \mathrm{(A) \ } 0\qquad \mathrm{(B) \ } 1\qquad \mathrm{(C) \ } 2\qquad \mathrm{(D) \ } 3\qquad \mathrm{(E) \ } 4 </math>
+
<math> \textbf{(A) } 0\qquad \textbf{(B) } 1\qquad \textbf{(C) } 2\qquad \textbf{(D) } 3\qquad \textbf{(E) } 4 </math>
 
 
  
 
==Solution==
 
==Solution==
Line 28: Line 27:
 
<math>(2\uparrow 2)\uparrow (2\uparrow 2) = 4 \uparrow 4 = 256</math>
 
<math>(2\uparrow 2)\uparrow (2\uparrow 2) = 4 \uparrow 4 = 256</math>
  
Thus the only other result is <math>256</math>, and our answer is <math>\boxed{\text{(B)}\ 1}</math>.
+
Thus the only other result is <math>256</math>, and our answer is <math>\boxed{\textbf{(B) } 1}</math>.
  
 
==See Also==
 
==See Also==

Latest revision as of 11:28, 8 November 2021

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

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

See Also

2002 AMC 12A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS