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Difference between revisions of "2012 AIME II Problems/Problem 10"

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== Solution ==
 
== Solution ==
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We know that x cannot be irrational because the product of a rational number and an irrational number is irrational (but n is an integer). Therefore x is rational.
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We know that <math>x</math> cannot be irrational because the product of a rational number and an irrational number is irrational (but <math>n</math> is an integer). Therefore <math>x</math> is rational.
  
  
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a = 0 --> nothing because n is positive
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<math>a = 0 \implies</math> nothing because n is positive
  
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a = 1 --> b/c = 0/1
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<math>a = 1 \implies \frac{b}{c} = \frac{0}{1}</math>
  
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a = 2 --> b/c = 0/2, 1/2
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<math>a = 2 \implies \frac{b}{c} = \frac{0}{2},\frac{1}{2}</math>
  
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a = 3 --> b/c = 0/3, 1/3, 2/3
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<math>a = 3 \implies\frac{b}{c} =\frac{0}{3},\frac{1}{3},\frac{1}{3}</math>
  
  
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The pattern continues up to a = 31. Note that if a = 32, then n > 1000. However if a = 31, the largest possible x is 31 + 30/31, in which n is still less than 1000. Therefore the number of positive integers for n is equal to <math>1+2+3+...+31 = \frac{31*32}{2} = \boxed{496.}</math>
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The pattern continues up to <math>a = 31</math>. Note that if <math>a = 32</math>, then <math>n > 1000</math>. However if <math>a = 31</math>, the largest possible x is <math>31 + 30/31</math>, in which <math>n</math> is still less than <math>1000</math>. Therefore the number of positive integers for n is equal to <math>1+2+3+...+31 = \frac{31*32}{2} = \boxed{496.}</math>
  
 
== See Also ==
 
== See Also ==
 
{{AIME box|year=2012|n=II|num-b=9|num-a=11}}
 
{{AIME box|year=2012|n=II|num-b=9|num-a=11}}

Revision as of 01:48, 24 February 2013

Problem 10

Find the number of positive integers $n$ less than $1000$ for which there exists a positive real number $x$ such that $n=x\lfloor x \rfloor$.

Note: $\lfloor x \rfloor$ is the greatest integer less than or equal to $x$.


Solution

We know that $x$ cannot be irrational because the product of a rational number and an irrational number is irrational (but $n$ is an integer). Therefore $x$ is rational.


Let $x = a + \frac{b}{c}$ where a,b,c are nonnegative integers and $0 \le b < c$ (essentially, x is a mixed number). Then,


$n = (a + \frac{b}{c}) \lfloor a +\frac{b}{c} \rfloor \Rightarrow n = (a + \frac{b}{c})a = a^2 + \frac{ab}{c}$ Here it is sufficient for $\frac{ab}{c}$ to be an integer. We can use casework to find values of n based on the value of a:


$a = 0 \implies$ nothing because n is positive

$a = 1 \implies \frac{b}{c} = \frac{0}{1}$

$a = 2 \implies \frac{b}{c} = \frac{0}{2},\frac{1}{2}$

$a = 3 \implies\frac{b}{c} =\frac{0}{3},\frac{1}{3},\frac{1}{3}$


The pattern continues up to $a = 31$. Note that if $a = 32$, then $n > 1000$. However if $a = 31$, the largest possible x is $31 + 30/31$, in which $n$ is still less than $1000$. Therefore the number of positive integers for n is equal to $1+2+3+...+31 = \frac{31*32}{2} = \boxed{496.}$

See Also

2012 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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