How many zeroes do we write when we write all the integers from 1 to 256 in binary?
+373 Hey there, Guest!
So...
An n number is a 1 followed by 2^(n-1) combinations of n-1 0's and ones, symmetrically distributed, hence (n-1)2^(n-2) zeros.
In total,
1+2*2+3*4+4*8+5*16+6*32+7*64+8=777.
so that for numbers between 0 and 2^n, the number of zeroes is
(n−2)2^(n−1)+n+1
Hope this helped! :)
( ﾟдﾟ)つ Bye
