Find the positive integer $n$ which satisfy given condition.

Question

I want to know whether how the following inequality can be solved $$1993+n\ge \lfloor \log_2 n\rfloor $$

The original question was

Find the positive integer $n\,$ for which $$\lfloor\log_2{1}\rfloor+\lfloor\log_2{2}\rfloor+\lfloor\log_2{3}\rfloor+\cdots+\lfloor\log_2{n}\rfloor=1994$$ (For real $x\,$, $\lfloor x\rfloor\,$ is the greatest integer $\le x.\,$)

For this at first I used the fact that $\lfloor x\rfloor + \lfloor y\rfloor+1\ge \lfloor x+y\rfloor$

Answer 1

HINT: Regarding your original question, there is a neat pattern hiding in plain sight:

1. $\lfloor\log_2{1}\rfloor=0$ for one number
2. $\lfloor\log_2{2}\rfloor,\lfloor\log_2{3}\rfloor=1$ for two numbers
3. $\lfloor\log_2{4}\rfloor, \lfloor\log_2{5}\rfloor, ..., \lfloor\log_2{7}\rfloor=2$ for four numbers
4. $\lfloor\log_2{8}\rfloor, \lfloor\log_2{9}\rfloor, ..., \lfloor\log_2{15}\rfloor=3$ for eight numbers

Their corresponding sum can be written as:

$$0\cdot2^0+1\cdot2^1+2\cdot2^2+3\cdot2^3+...$$

which is an AGP sequence. Can you solve the question now?

Answer 2

Since the question was specifically about the inequality, it's easy to observe that $f(x)=x-\log_2{x}$ has $f'(x)=1-\frac{1}{x\ln{2}}\geq0$ for $x\geq 2$ which means $f(x)$ is ascending for $x\geq 2$. Particularly $f(x) \geq f(2)=1>0$, thus $$x>\log_2{x}>\left \lfloor \log_2{x} \right \rfloor \geq0, \forall x\geq2$$ This applies to $n\in \mathbb{N}\setminus\{0\}$ too, where $n=1$ can be validated "manually".