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!
Best Answer
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.