Question

If $\left\{p\right\}$ denotes the fractional part of the number $p$, then $\left\{\frac{3^{200}}{8}\right\}$ is equal to:

• A. $\frac{5}{8}$
• B. $\frac{1}{8}$
• C. $\frac{7}{8}$
• D. $\frac{3}{8}$

Answer 1

The correct option is B

$\frac{1}{8}$

Explanation for the correct option

Step 1: Simplification of the given expression

The given factor can be rewritten as,

$\begin{array}{rcl}\left\{\frac{3^{200}}{8}\right\}& =& \left\{\frac{{\left(3^2\right)}^{100}}{8}\right\}\\ & =& \left\{\frac{{\left(9\right)}^{100}}{8}\right\}\\ & =& \left\{\frac{{\left(1+8\right)}^{100}}{8}\right\}\end{array}$

Step 2: Determination of the required fractional part

Solve the RHS separately,

Suppose in the numerator $x=1,y=8,\text{and}n=100$.

Then we know that the binomial expansion of ${\left(x+y\right)}^n$ is

${\left(x+y\right)}^{\mathbf{n}}\mathbf{=}{}^{\mathbf{n}}\mathit{C}_{\mathbf{0}}{\mathit{x}}^{\mathbf{n}}{\mathit{y}}^{\mathbf{0}}\mathbf{+}{}^{\mathbf{n}}\mathit{C}_{\mathbf{1}}{\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{1}}{\mathit{y}}^{\mathbf{1}}\mathbf{+}{}^{\mathbf{n}}\mathit{C}_{\mathbf{2}}{\mathit{x}}^{\mathbf{n}\mathbf{-}\mathbf{2}}{\mathit{y}}^{\mathbf{2}}\mathbf{+}\mathbf{\dots }\mathbf{+}{}^{\mathbf{n}}\mathit{C}_{\mathbf{n}}{\mathit{x}}^{\mathbf{0}}{\mathit{y}}^{\mathbf{n}}$

With the use of this, the numerator will be,

${\left(1+8\right)}^{100}={}^{100}C_0{1}^{100}{8}^0+{}^{100}C_1{1}^{99}{8}^1+{}^{100}C_2{1}^{98}{8}^2+\dots +{}^{100}C_{100}{1}^0{8}^{100}$

Now, we can write that,

$\begin{array}{rcl}\therefore \left\{\frac{{\left(1+8\right)}^{100}}{8}\right\}& =& \left\{\frac{{}^{100}C_0{1}^{100}{8}^0+{}^{100}C_1{1}^{99}{8}^1+{}^{100}C_2{1}^{98}{8}^2+\dots +{}^{100}C_{100}{1}^0{8}^{100}}{8}\right\}\\ & =& \left\{\frac{{}^{100}C_0{1}^{100}+8t}{8}\right\}\\ & =& \left\{\frac{1+8t}{8}\right\}\\ & =& \frac{1}{8}+t\end{array}$

Therefore, the fractional part of the number $p$ will be $\frac{1}{8}$.

Hence, the correct option is (B).