Find the positive integer $n$ $n$ such that $n \sum k = 1 ⌊ {log}_{2} k ⌋ = 2018$ $sum_(k=1)^n floor(log_2 k) = 2018$ ?

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Find the positive integer $n$ $n$ such that $n \sum k = 1 ⌊ {log}_{2} k ⌋ = 2018$ $sum_(k=1)^n floor(log_2 k) = 2018$ ?

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Answer 1 · 2017-10-13T20:26:58

$n = 315$ $n=315$

Explanation:

Completing what Oliver started...

Note that:

$8 (256) = 2048 > 2018$ $8(256) = 2048 > 2018$

So we need less terms than:

$1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 = 511$ $1+2+4+8+16+32+64+128+256 = 511$

and probably more than:

$1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 = 255$ $1+2+4+8+16+32+64+128 = 255$

Start by summing:

$0 (1) + 1 (2) + 2 (4) + 3 (8) + 4 (16) + 5 (32) + 6 (64) + 7 (128)$ $0(1)+1(2)+2(4)+3(8)+4(16)+5(32)+6(64)+7(128)$

$= 0 + 2 + 8 + 24 + 64 + 160 + 384 + 896 = 1538$ $= 0+2+8+24+64+160+384+896 = 1538$

So we need an additional:

$\frac{2018 - 1538}{8} = \frac{480}{8} = 60$ $(2018-1538)/8 = 480/8 = 60$ terms

So $n = 255 + 60 = 315$ $n = 255+60 = 315$

Answer 2 · 2017-10-13T23:23:08

See below.

Explanation:

We have

${log}_{2} k = ⌊ {log}_{2} k ⌋$ $log_2 k = floor(log_2 k)$ for

$k = 2^{j}$ $k = 2^j$

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

$m \sum k = 1 k 2^{k - 1} + k_{2} = 2018$ $sum_(k=1)^m k 2^(k-1) + k_2= 2018$

Now considering

$n \sum k = 1 k x^{k - 1} = \frac{d}{d x} n \sum k = 1 x^{k} - 1 = \frac{d}{d x} (\frac{x^{n + 1} - 1}{x - 1}) - 1 = \frac{(2 - x) x + (m (x - 1) - 1) x^{m}}{{(x - 1)}^{2}}$ $sum_(k=1)^n k x^(k-1) = d/(dx) sum_(k=1)^n x^k -1=d/(dx)((x^(n+1)-1)/(x-1)) -1= ((2-x) x + (m (x-1)-1) x^m)/(x-1)^2$

and for $x = 2$ $x = 2$

$m \sum k = 1 k 2^{k - 1} = 2^{m} (m - 1)$ $sum_(k=1)^m k 2^(k-1)=2^m (m-1)$

Solving for $m$ $m$

$2^{m} (m - 1) = 2018$ $2^m (m-1)=2018$ we obtain

$m = \frac{{log}_{e} 2 + W (1009 {log}_{e} 2)}{{log}_{e} 2} = 8.14232$ $m = (Log_e 2 +W(1009 Log_e 2))/Log_e 2 = 8.14232$

so considering $⌊ 8.14232 ⌋ = 8$ $floor(8.14232)= 8$ we have

$m_{1} = 2^{8} = 256$ $m_1 = 2^8 = 256$ and

$m_{1} \sum k = 1 ⌊ {log}_{2} k ⌋ = 1546$ $sum_(k=1)^(m_1) floor(log_2 k) = 1546$

Now $2018 - 1546 = k_{2} = 472 = 8 \times 59$ $2018-1546 = k_2=472 = 8 xx 59$ then finally

$n = m_{1} + 59 = 256 + 59 = 315$ $n = m_1 + 59 = 256+59=315$

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