Broke the right Glass!

Let $\lambda$ be the least no. to be subtracted from $\color{#D61F06}{200}$ to make it a perfect cube.

$\therefore$

$200-\lambda=a^3$

Perfect cube numbers nearest neighbors to $\color{#D61F06}{200}$ are $216, 125$ .

$\Large \text{case 1 :}$

Let $a^3$ be $125$ for which value of $a$ will be $5$ .

$\implies 200-\lambda=125$

$\implies \lambda=75$

$\Large \text{case 2 :}$

Let $a^3$ be $216$ for which value of $a$ will be $6$ .

$\implies 200-\lambda=216$

$\implies \lambda=-16$

$\color{#20A900}{\bullet}$ If you take any other case than the above two, the value goes out of the range $(-100,100)$

Out of the above two values of $\lambda$ the least value is $\boxed{-16}$ . (which is not given in the options)

So, the correct option is None of the given.

enjoy !

The answer is -16 , which when subtracted from 200 , gives the answer as 216 , which is the nearest perfect cube to 200.

Let $n$ be the integer $-100 < n < 100$ and $m = 200-n$ .

• The smallest $m$ , $m_{min} = 200-100 = 100$ and the smallest perfect cubic $p_{min} > 100$ , therefore $p_{min} = \left \lceil \sqrt[3]{100} \right \rceil ^3 = 5^3 = 125$ . $200-n_1 = 125$ $\Longrightarrow n_1 = 75$ .
• The greatest $m$ , $m_{max} = 200-(-100) = 300$ and the largest perfect cubic $p_{max} < 300$ , therefore $p_{max} = \left \lfloor \sqrt[3]{300} \right \rfloor ^3 = 6^3 = 216$ . $200-n_2 = 216$ $\Longrightarrow n_2 = -16$ .

Therefore, the smallest $n_{min} = n_2 = \boxed{-16}$