2012 AIME II Problems/Problem 10

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

Solution 1

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, \[n = \left(a + \frac{b}{c}\right) \left\lfloor a +\frac{b}{c} \right\rfloor \Rightarrow n = \left(a + \frac{b}{c}\right)a = a^2 + \frac{ab}{c}\]

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 $31 + \frac{30}{31}$, in which $n$ is still less than $1000$. Therefore the number of positive integers for $n$ is equal to $1+2+3+...+31 = \frac{31 \cdot 32}{2} = \boxed{496}.$

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

2009 AIME I Problems/Problem 6

2012 AIME II (ProblemsAnswer KeyResources)
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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