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

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Let <math>x = a + \frac{b}{c}</math> where a,b,c are nonnegative integers and <math>0 \le b < c</math> (essentially, x is a mixed number). Then,
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Let <math>x = a + \frac{b}{c}</math> where <math>a,b,c</math> are nonnegative integers and <math>0 \le b < c</math> (essentially, <math>x</math> is a mixed number). Then,
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<cmath>n = (a + \frac{b}{c}) \lfloor a +\frac{b}{c} \rfloor \Rightarrow n = (a + \frac{b}{c})a = a^2 + \frac{ab}{c}</cmath>  
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<math>n = (a + \frac{b}{c}) \lfloor a +\frac{b}{c} \rfloor \Rightarrow n = (a + \frac{b}{c})a = a^2 + \frac{ab}{c}</math> Here it is sufficient for <math>\frac{ab}{c}</math> to be an integer. We can use casework to find values of n based on the value of a:
 
  
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Here it is sufficient for <math>\frac{ab}{c}</math> to be an integer. We can use casework to find values of n based on the value of a:
  
 
<math>a = 0 \implies</math> nothing because n is positive
 
<math>a = 0 \implies</math> nothing because n is positive
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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>
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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>
  
 
== Solution 2==
 
== Solution 2==

Revision as of 12:34, 6 August 2020

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

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

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

$a = 3 \implies\frac{b}{c} =\frac{0}{3},\frac{1}{3},\frac{2}{3}$


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 $1+2+3+...+31 = \frac{31*32}{2} = \boxed{496}.$

Solution 2

Notice that $x\lfloor x\rfloor$ is continuous over the region $x \in [k, k+1)$ for any integer $k$. Therefore, it takes all values in the range $[k\lfloor k\rfloor, (k+1)\lfloor k+1\rfloor) = [k^2, (k+1)k)$ over that interval. Note that if $k>32$ then $k^2 > 1000$ and if $k=31$, the maximum value attained is $31*32 < 1000$. It follows that the answer is $\sum_{k=1}^{31} (k+1)k-k^2 = \sum_{k=1}^{31} k = \frac{31\cdot 32}{2} = \boxed{496}.$

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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