Difference between revisions of "2004 AMC 12A Problems/Problem 16"
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== Solution 2 == | == Solution 2 == | ||
− | + | Let | |
− | 2001^a= | + | <cmath>\begin{align*} |
− | + | x &= 2001^a, \\ | |
− | 2002^b = | + | a &= 2002^b, \\ |
− | + | b &= 2003^c, \\ | |
− | 2003^c = | + | c &= 2004^d. |
− | + | \end{align*}</cmath> | |
− | 2004^d | + | It follows that <cmath>x = 2001^{2002^{2003^{2004^d}}}.</cmath> |
− | + | The smallest value of <math>x</math> occurs when <math>d\rightarrow -\infty,</math> so this expression becomes | |
− | + | <cmath>x = 2001^{2002^{2003^0}} = 2001^{2002^1} = \boxed{\textbf {(B) }2001^{2002}}.</cmath> | |
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==Video Solution (Logical Thinking)== | ==Video Solution (Logical Thinking)== |
Latest revision as of 02:01, 23 January 2023
Problem
The set of all real numbers for which
is defined is . What is the value of
?
Solution 1
For all real numbers and
such that
note that:
is defined if and only if
if and only if
Therefore, we have
from which
~Azjps ~MRENTHUSIASM
Solution 2
Let
It follows that
The smallest value of
occurs when
so this expression becomes
Video Solution (Logical Thinking)
~Education, the Study of Everything
See also
2004 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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 |