Question

Find the positive integer `n` such that `sum_(k=1)^n floor(log_2 k) = 2018`?

Answer 1

`n=315`

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 `n = 255+60 = 315`

Answer 2

See below.

Explanation:

We have

`log_2 k = floor(log_2 k)` for

`k = 2^j`

We know also that `log_2 k` is a monotonically strict increasing function of `k` so from `j` to `j+1` we have `2^j` equal terms with value `j` so we are looking for

`sum_(k=1)^m k 2^(k-1) + k_2= 2018`

Now considering

`sum_(k=1)^n k x^(k-1) = d/(dx) sum_(k=1)^n x^k -1=d/(dx)((x^(n+1)-1)/(x-1)) -1= ((2-x) x + (m (x-1)-1) x^m)/(x-1)^2`

and for `x = 2`

`sum_(k=1)^m k 2^(k-1)=2^m (m-1)`

Solving for `m`

`2^m (m-1)=2018` we obtain

`m = (Log_e 2 +W(1009 Log_e 2))/Log_e 2 = 8.14232`

so considering `floor(8.14232)= 8` we have

`m_1 = 2^8 = 256` and

`sum_(k=1)^(m_1) floor(log_2 k) = 1546`

Now `2018-1546 = k_2=472 = 8 xx 59` then finally

`n = m_1 + 59 = 256+59=315`