Second Problem for Floors

Since $x = c^2 - 1 , \implies \lfloor\sqrt{x}\rfloor = c - 1$

$\large {\implies c - 1 + \lfloor\sqrt[4]{x}\rfloor + \lfloor\sqrt[8]{x}\rfloor + \lfloor\sqrt[16]{x}\rfloor + \lfloor\sqrt[32]{x}\rfloor = 100 \\ \lfloor\sqrt[4]{x}\rfloor + \lfloor\sqrt[8]{x}\rfloor + \lfloor\sqrt[16]{x}\rfloor + \lfloor\sqrt[32]{x}\rfloor = 101 - c \\ \text{Since} \ 0 < c < 101 \implies 0 < \lfloor\sqrt{c}\rfloor < 10 \\ \implies \max \ (\lfloor\sqrt{c}\rfloor) = 9}$

$\large {\implies \lfloor\sqrt[4]{x}\rfloor = \lfloor\sqrt[4]{c^2 - 1}\rfloor = \lfloor\sqrt{c}\rfloor = 9 \\ \lfloor\sqrt[8]{x}\rfloor = \lfloor\sqrt[8]{c^2 - 1}\rfloor = \lfloor\sqrt[4]{c}\rfloor = \lfloor\sqrt{\sqrt{c}}\rfloor = 3 \\ \lfloor\sqrt[16]{x}\rfloor = \lfloor\sqrt[16]{c^2 - 1}\rfloor = \lfloor\sqrt[8]{c}\rfloor = \lfloor\sqrt{\sqrt[4]{c}}\rfloor = 1 \\ \lfloor\sqrt[32]{x}\rfloor = \lfloor\sqrt[32]{c^2 - 1}\rfloor = \lfloor\sqrt[16]{c}\rfloor = \lfloor\sqrt{\sqrt[8]{c}}\rfloor = 1 }$

Hence,

$\large {\lfloor\sqrt{x}\rfloor + \lfloor\sqrt[4]{x}\rfloor + \lfloor\sqrt[8]{x}\rfloor + \lfloor\sqrt[16]{x}\rfloor + \lfloor\sqrt[32]{x}\rfloor = c - 1 + 9 + 3 + 1 + 1 = 100 \\ \boxed {c = 87} \implies \text{max (x)} = 87^2 - 1 .}$