Let $f(x) = \lfloor x \lfloor x \rfloor \rfloor$ for $x \ge 0.$
Find the number of possible values of $f(x)$ for $0 \le x \le 10.$
I split the problem into three cases, $1 \leq n < 10, n=10,$ and $0 \leq x < 1.$ I got the answer 47, but I'm not sure if it's correct. Could someone clarify for me? Also, is it correct to use $n$ in my cases or $x$? I'm confused a little bit, so I would be please by a hint/help!
On
$f(x) = \lfloor x \lfloor x \rfloor \rfloor$ can be rewritten as $$\lfloor x\rfloor^2 + \lfloor \{x\}\lfloor x \rfloor \rfloor$$
where $\{x\}$ denotes the fractional part of $x$. Let $n = \lfloor x \rfloor$. Then you are trying to find $$2+\sum_{n=1}^9 g(n)$$
where $g(n)$ denotes the number of values that $n^2 + \lfloor t n\rfloor$ reaches for $0 \le t < 1$ that weren't reached by an earlier $n$. The $+2$ at the beginning is for $n = 0$ and $n = 10$.
Because there would be no "overlap" in values between values of $n$, this definition can be simplified to the number of values that $\lfloor t \rfloor$ reaches for $0 \le x < n$, which is easily seen to be $n$. Then, the sum becomes $$2 + \sum_{n=1}^9 n = 2+ \frac{9\cdot 10}{2} = 47$$
which is exactly what you got.
Assuming that you had defined $n = \lfloor x \rfloor$ as I had, your notation looks correct.
For a real number $x$, please make sure that we understand its integer part $\lfloor x \rfloor$ and its fractional part $\{x\}$. We have the following identity $$ x=\lfloor x \rfloor+\{x\}.$$
Back to the problem, we consider 3 cases:
+Case 1: $0\leq x<1$. In this case, we have $f(x)=\lfloor x \lfloor x \rfloor \rfloor= \lfloor 0 \rfloor= 0$
+Case 2: $x=10$. Then $f(x)=100$.
+Case 3: $n\leq x<n+1$ where $n$ is some integer from $1$ to $9$. We further assume that $x\in \left[n+\frac{k}{n}, n+\frac{k+1}{n} \right)$ for some integer $0\leq k<n$, then we have $\lfloor x \rfloor=n$ and $\lfloor \{x\} \lfloor x \rfloor \rfloor=k$. Now we have $$f(x)=\lfloor \lfloor x \rfloor^2+ \{x\} \lfloor x \rfloor \rfloor=n^2+k.$$
Totally, we see that the set of all possible values of $f(x)$ for $0\leq x\leq 10$ is $$A=\{0, 100\}\cup \{ n^2+k: n=1,...9 \text{ and } 0\leq k\leq n\}.$$ This set has $47$ elements.