A problem on moderately fast algorithms

As already suggested, you were supposed to solve without WA.Here is one way,

$n\times lg\left(n\right)$ = $1000$ {solve for the equality to determine the largest $n$ and then take its floor value}

$n$ = $2^{\dfrac{1000}{n}}$

$1000$ = $\dfrac{1000}{n} \times 2^{\dfrac{1000}{n}}$

Let $\dfrac{1000}{n}$ = $z$ ................ $\left(*\right)$

Let $f\left(z\right)$ = $z \times 2^{z} -1000$ , $f'\left(z\right)$ = $2^{z}\left(zln(2) +1\right)$ and we are required to find roots of $f\left(z\right)$

Using $\color{#20A900}{\text{Newton's method}}$ ,

$z_{n+1}$ = $z_{n} - \dfrac{f\left(z_{n}\right)}{f'\left(z_{n}\right)}$

We know that $7\times 2^{7} < 1000 < 8\times 2^{8}$

$\therefore$ we can guess the initial root as ~ 7.25 {i.e. $z_{0}= 7.25$ }and proceed as per Newton's method to obtain the approx. root $z = 7.131561875$ .

$n = \dfrac{1000}{7.131561875}$ ......................from $\left(*\right)$

$\boxed{ \lfloor n \rfloor= 140}$