Difference between revisions of "2009 AIME I Problems/Problem 6"

Revision as of 23:45, 19 March 2009

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 $1000$ .

Now in order for $x^{\lfloor x\rfloor}$ to be an integer, $x$ must be an integral root of an integer,

So let do 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

Answer $= 1+5+37+369= \boxed {412}$

@@ Line 31: / Line 31: @@
 Answer <math>= 1+5+37+369= \boxed {412}</math>
+== See also ==
+{{AIME box|year=2009|n=I|num-b=5|num-a=7}}

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

Revision as of 23:45, 19 March 2009

Problem

Solution

See also