Question

If $\left\{x\right\}$ represents the fractional part of $x$, then $\left\{\frac{5^{200}}{8}\right\}$ is

• A.
• B.
• C.
• D.

Answer 1

Answer: (B)

$\frac{5^{200}}{8}$

$=\frac{(5^2{)}^{100}}{8}$

$=\frac{(25{)}^{100}}{8}$

$=\frac{(1+24{)}^{100}}{8}$
$=\frac{1{+}^{100}{C}_1.24{+}^{100}{C}_2(24{)}^2+....{+}^{100}{C}_{100}(24{)}^{100}}{8}$

[$\because (1+x{)}^n=1{+}^n{C}_1.x{+}^n{C}_2{x}^2+....{+}^n{C}_n{x}^n$]

$=\frac{1}{8}+\frac{{}^{100}{C}_1.24{+}^{100}{C}_2(24{)}^2+....{+}^{100}{C}_{100}(24{)}^{100}}{8}$

$=\frac{1}{8}+\frac{24{(}^{100}{C}_1.{+}^{100}{C}_2\left(24\right)+....{+}^{100}{C}_{100}\left(24{)}^{99}\right)}{8}$

$=\frac{1}{8}+3{(}^{100}{C}_1.{+}^{100}{C}_2\left(24\right)+....{+}^{100}{C}_{100}\left(24{)}^{99}\right)$

Since, the value of ${}^{100}{C}_1.{+}^{100}{C}_2\left(24\right)+....{+}^{100}{C}_{100}(24{)}^{99}$ is always an integer.
i.e., fractional part of $x$ is $\frac{1}{8}$.

Hence, Option A is correct.