Question

If $\left\{ p \right\}$ denotes the fractional part of the number $p$, then what is the value of $\left\{ {\dfrac{{{3^{200}}}}{8}} \right\}$?
A. $\dfrac{5}{8}$
B. $\dfrac{1}{8}$
C. $\dfrac{7}{8}$
D. $\dfrac{3}{8}$

Answer 1

Hint: Write ${3^{200}}$ as ${9^{100}}$ and $9$ as $\left( {8 + 1} \right)$. After that use the binomial theorem and separate the fractional part from the expression obtained. The last term represents the fractional part. All other terms are the whole number.

Formula Used:
${\left( {a + b} \right)^n} = {a^n} + {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + ..... + {}^n{C_{n - 1}}a{b^{n - 1}} + {b^n}$
${\left( {{a^m}} \right)^n} = {a^{mn}}$
$\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$

Complete step by step solution:
The given number is $\dfrac{{{3^{200}}}}{8}$
Use the formula ${\left( {{a^m}} \right)^n} = {a^{mn}}$
${3^{200}} = {\left( {{3^2}} \right)^{100}} = {9^{100}}$
We can write $9 = 8 + 1$
So, ${9^{100}} = {\left( {8 + 1} \right)^{100}}$
Use binomial theorem ${\left( {a + b} \right)^n} = {a^n} + {}^n{C_1}{a^{n - 1}}b + {}^n{C_2}{a^{n - 2}}{b^2} + ..... + {}^n{C_{n - 1}}a{b^{n - 1}} + {b^n}$
Putting $a = 8,b = 1,n = 100$, we get
${\left( {8 + 1} \right)^{100}} = {\left( 8 \right)^{100}} + {}^{100}{C_1}{\left( 8 \right)^{100 - 1}}\left( 1 \right) + {}^{100}{C_2}{\left( 8 \right)^{100 - 2}}{\left( 1 \right)^2} + ..... + {}^{100}{C_{99}}\left( 8 \right){\left( 1 \right)^{100 - 1}} + {\left( 1 \right)^{100}}$
$ = {8^{100}} + 100 \times {8^{99}} + 4950 \times {8^{98}} + ...100 \times 8 + 1$
Dividing both sides by $8$, we get
$ \Rightarrow \dfrac{{{{\left( {8 + 1} \right)}^{100}}}}{8} = \dfrac{{{8^{100}} + 100 \times {8^{99}} + 4950 \times {8^{98}} + ... + 100 \times 8 + 1}}{8}$
Separate each term on right-hand side
$ \Rightarrow \dfrac{{{9^{100}}}}{8} = \dfrac{{{8^{100}}}}{8} + 100 \times \dfrac{{{8^{99}}}}{8} + 4950 \times \dfrac{{{8^{98}}}}{8} + ... + 100 \times \dfrac{8}{8} + \dfrac{1}{8}$
Use the formula $\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}$
$ \Rightarrow \dfrac{{{9^{100}}}}{8} = {8^{99}} + 100 \times {8^{98}} + 4950 \times {8^{97}} + ... + 100 + \dfrac{1}{8}$
Clearly, the fractional part of the above expression is $\dfrac{1}{8}$

Option ‘B’ is correct

Note: Fractional part of a fraction is a fraction. So, it is obvious that fractional part of every proper fraction is itself but for improper fraction it is not. We can express every improper fraction as a mixed fraction writing the whole part and the fractional part separately.