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Difference between revisions of "2009 AIME I Problems/Problem 6"
(→‎Solution)
Line 10: Line 10:
 
So let do case work:
 
So let do case work:
  
−
For <math>{\lfloor x\rfloor}=0</math> N=<math>1</math> no matter what x is
+
For <math>{\lfloor x\rfloor}=0</math>, N=<math>1</math> no matter what x is
  
−
For <math>{\lfloor x\rfloor}=1</math> N can be anything between <math>1^1</math> to <math>2^1</math> excluding <math>2^1</math>
+
For <math>{\lfloor x\rfloor}=1</math>, N can be anything between <math>1^1</math> to <math>2^1</math> excluding <math>2^1</math>
  
 
This gives us <math>2^1-1^1=1</math> N's
 
This gives us <math>2^1-1^1=1</math> N's
  
−
For <math>{\lfloor x\rfloor}=2</math> N can be anything between <math>2^2</math> to <math>3^2</math> excluding <math>3^2</math>
+
For <math>{\lfloor x\rfloor}=2</math>, N can be anything between <math>2^2</math> to <math>3^2</math> excluding <math>3^2</math>
  
 
This gives us <math>3^2-2^2=5</math> N's
 
This gives us <math>3^2-2^2=5</math> N's
  
−
For <math>{\lfloor x\rfloor}=3</math> N can be anything between <math>3^3</math> to <math>4^3</math> excluding <math>4^3</math>
+
For <math>{\lfloor x\rfloor}=3</math>, N can be anything between <math>3^3</math> to <math>4^3</math> excluding <math>4^3</math>
  
 
This gives us <math>4^3-3^3=37</math> N's
 
This gives us <math>4^3-3^3=37</math> N's
  
−
For <math>{\lfloor x\rfloor}=4</math> N can be anything between <math>4^4</math> to <math>5^4</math> excluding <math>5^4</math>
+
For <math>{\lfloor x\rfloor}=4</math>, N can be anything between <math>4^4</math> to <math>5^4</math> excluding <math>5^4</math>
  
 
This gives us <math>5^4-4^4=369</math> N's
 
This gives us <math>5^4-4^4=369</math> N's

Revision as of 23:53, 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 $= 1+5+37+369= \boxed {412}$

See also

2009 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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