Find the positive integer #n# such that #sum_(k=1)^n floor(log_2 k) = 2018#?
2 Answers
Explanation:
Completing what Oliver started...
Note that:
#8(256) = 2048 > 2018#
So we need less terms than:
#1+2+4+8+16+32+64+128+256 = 511#
and probably more than:
#1+2+4+8+16+32+64+128 = 255#
Start by summing:
#0(1)+1(2)+2(4)+3(8)+4(16)+5(32)+6(64)+7(128)#
#= 0+2+8+24+64+160+384+896 = 1538#
So we need an additional:
#(2018-1538)/8 = 480/8 = 60# terms
So
See below.
Explanation:
We have
We know also that
Now considering
and for
Solving for
so considering
Now