AIME 2012 II · 第 10 题

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 $\sum_{k=1}^{31} (k+1)k-k^2 = \sum_{k=1}^{31} k = \frac{31\cdot 32}{2} = \boxed{496}.$

Solution 3

Bounding gives $x^2\le n. Thus there are a total of$x$possible values for$n$, for each value of$x^2$. Checking, we see$31^2+31=992<1000$, so there are

$$1+2+3+...+31= \boxed{496}$$ such values for $n$.

Here writer's x and question's x are different. Here writer means [x] everywhere he writes x so you can let [x] = k and then use the inequality $k^2\\le n. this inequality comes from [x] = k$k\le x $k[x]\le x[x]<(k+1)[x]$ $k^2\le n<(k+1)k$ further you can proceed.

Solution 4

After a bit of experimenting, we let $n=l^2+s, s < 2n+1$. We claim that I (the integer part of $x$) = $l$ . (Prove it yourself using contradiction !) so now we get that $x=l+\frac{s}{l}$. This implies that solutions exist iff $s, or for all natural numbers of the form$l^2+s$where$s. Hence, 1 solution exists for $l=1$! 2 for $l=2$ and so on. Therefore our final answer is $31+30+\dots+1= \boxed{496}$

Solution 5

Notice we can write $\lfloor x \rfloor$ as $x - \{x\}$. Since $n$ is a positive integer less than $1000$,

$$1 \leq x(x - \{x\}) \leq 999.$$ We now have

$$1 \leq x^{2} - x\{x\} \leq 999.$$ Since $x\{x\}$ ranges from $0$ to $x$, we can rewrite as

$$1 \leq x^{2} \leq 999 + x.$$ Notice all $x$ that satisfy this inequality range from $1$ to $32$.

Now every range from $y$ to $y+1$ for $x$ results in exactly $y$ solutions that give a positive integer solution for $n$. The first range is $1$ to $2$ and the last range is $31$ to $32$.

Thus, by using our above lemma, we can find the total number of solutions:

$$1+2+\cdots+31 = \frac{31 \cdot 32}{2} = \boxed{496}.$$ ~ilikemath247365

Video Solution

2012 AIME II #10

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