Longarithms!

Algebra Level 4

Find the value of $\lfloor \log_{2}{1}\rfloor + \lfloor \log_{2}{2}\rfloor + \lfloor \log_{2}{3}\rfloor..........+ \lfloor \log_{2}{66}\rfloor.$

2 solutions

Daniel Liu
Jun 8, 2014

Note that $\lfloor \log_2 (2^n+k)\rfloor = n$ for all $k=0\to 2^n-1$ , which is a total of $2^n$ values of $k$ .

Thus, the sum we want is $(1\cdot 0+2\cdot 1+4\cdot 2+8\cdot 3+16\cdot 4+32\cdot 5)+6+6+6=\boxed{276}$ .

Same way but I found the rule after trying the long way !!

Niranjan Khanderia - 6 years, 11 months ago

The symbol is not for greatest integer function it is for smallest integer function

Aman Singh - 6 years, 12 months ago

Daniel, please answer me.

$k=0\rightarrow{}2^n-1$

is a formal notation or just something you made up to represent your idea?

I am really bad with demonstrations and proofs and to me yours just seem to be pretty solid. Just wanted to know if that notation is a standard.

Bernardo Sulzbach - 6 years, 11 months ago

the same I did :D

Harshit Joshi - 6 years, 9 months ago

Paola Ramírez
Jan 13, 2015

$1\times0+2\times1+2\times4+3\times8+4\times16+5\times32+6\times3=\boxed{276}$