For what $n$ does $[\log_21]+[\log_22]+[\log_23]+\dotsb+[\log_2n] = 1538$?

Question

I just can't solve this problem in spite of doing a whole book on logs and inequalities

Where $[\dotsc]$ denotes the greatest integer function, what is the value of the natural number $n$ satisfying the equation $$ [\log_21]+[\log_22]+[\log_23]+\dotsb+[\log_2n] = 1538 $$

A. 255
B. 256
C. 254
D. 313

The answer is A.

Please give an easy solution as I am a high school student. And I do fully understand the meaning of the question but I can't solve it.

Since stack exchange forces me to provide an attempt here it goes- I wrote the equation as $log_2n!=1538$

after this I feel n! should be $2^{1538} $

and now i am confused...

I hope you got your attempt for the people who put the question on hold.

nice you pointed out chris.. yeah the question is same but the person who posted it has accepted the answer to it not me!....the answer maybe sufficient to understand for him.. but it may not be for me...

PS-whats the point of giving a fake attempt and not speaking the truth?!!

Answer 1

In fact 255 is the only value that does satisfy the equation. The way to make it easier is to recognize that the values you sum are $n$ for each bracket from $2^n$ to $2^{n+1}-1$, giving a total of $n2^n$. This gives $$\sum_{n=1}^{255} \lfloor \log_2 n \rfloor=\sum_{k=0}^7k2^k=1538$$ It was convenient that $255=2^8-1$. Adding or subtracting terms will change the result. If I were doing the problem from scratch, I would do this to get the sum up to $255$ (even if that were not one of the choices), then see what correction was needed to get another. So for example $$\sum_{n=1}^{313} \lfloor \log_2 n \rfloor=\sum_{n=1}^{255} \lfloor \log_2 n \rfloor+\sum_{n=256}^{313} \lfloor \log_2 n \rfloor=\sum_{k=0}^7k2^k+(313-255)*8=2002$$

Added: the reason you can't write the equation as $\log_2 (n!)=1538$ is because of the greatest integer function applied to each term. For example, $\log_2(100)\approx 6.644$. The greatest integer means you only add in $6$ for the term $\lfloor \log_2(100) \rfloor$. My argument above is based on the fact that all the terms from $\lfloor \log_2(64) \rfloor$ through $\lfloor \log_2(127) \rfloor$ each contribute $6$ to the sum, so they sum to $6 \cdot 64=384$

Answer 2

Since part of the question is, at least implicitly, "How do I attempt this?", I thought I would write about how I started looking at the question.

So I didn't worry too much about what the target number was to start with, just tried looking at a few terms of the sequence:

$$\{0,1,1,2,2,2,2,3,\ldots\}$$

with each number being represented twice as many times as the previous. So then, looking at the target number and seeing it was "quite big", I could get the sum within intervals where the value is constant, ie. from $2^{k}$ to $2^{k+1}-1$, as $k.2^k$:

$$ \{ 2,8,24,64,160,384,896,2048\}$$

(I dropped $k=0$) and clearly now I have bracketed the problem. As it happens, this is enough to solve the problem too, because the first $7$ terms actually add to $1538$, the target number, so that is reached at $2^8-1=255$.