2009 AIME I Problems/Problem 6: Difference between revisions

Revision as of 22:41, 27 January 2015

Problem

How many positive integers $N$ less than $1000$ are there such that the equation $x^{\lfloor x\rfloor} = N$ has a solution for $x$ ? (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x$ .)

Solution

First, $x$ must be less than $5$ , since otherwise $x^{\lfloor x\rfloor}$ would be at least $3125$ which is greater than $1000$ .

Because ${\lfloor x\rfloor}$ must be an integer, we can do some simple case work:

For ${\lfloor x\rfloor}=0$ , $N=1$ no matter what $x$ is

For ${\lfloor x\rfloor}=1$ , $N$ can be anything between $1^1$ to $2^1$ excluding $2^1$

This gives us $2^1-1^1=1$ $N$ 's

For ${\lfloor x\rfloor}=2$ , $N$ can be anything between $2^2$ to $3^2$ excluding $3^2$

This gives us $3^2-2^2=5$ $N$ 's

For ${\lfloor x\rfloor}=3$ , $N$ can be anything between $3^3$ to $4^3$ excluding $4^3$

This gives us $4^3-3^3=37$ $N$ 's

For ${\lfloor x\rfloor}=4$ , $N$ can be anything between $4^4$ to $5^4$ excluding $5^4$

This gives us $5^4-4^4=369$ $N$ 's

Since $x$ must be less than $5$ , we can stop here and the answer is $\boxed {4}$ possible values for $N$ .

@@ Line 6: / Line 6: @@
 First, <math>x</math> must be less than <math>5</math>, since otherwise <math>x^{\lfloor x\rfloor}</math> would be at least <math>3125</math> which is greater than <math>1000</math>.
-Now, <math>{\lfloor x\rfloor}</math> must be an integer, so lets do case work:
+Because <math>{\lfloor x\rfloor}</math> must be an integer, we can do some simple case work:
 For <math>{\lfloor x\rfloor}=0</math>, <math>N=1</math> no matter what <math>x</math> is

2009 AIME I (Problems • Answer Key • Resources)
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2009 AIME I Problems/Problem 6: Difference between revisions

Revision as of 22:41, 27 January 2015

Problem

Solution

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