Question

If the fractional part of the number $\dfrac{{{2}^{403}}}{15}$ is found as $\dfrac{k}{15}$, then find the value of k?
(a) 14
(b) 6
(c) 4
(d) 8

Answer 1

We start solving the problem by recalling the definition of fractional part and greatest integer function. We use binomial expansion to convert the given number ${{2}^{403}}$ in terms of multiple of 15. Once we convert the given number ${{2}^{403}}$ into multiples of 15, we find the $\left[ \dfrac{{{2}^{403}}}{15} \right]$ and then subtract this with $\dfrac{{{2}^{403}}}{15}$ to get the required value.

Complete step-by-step answer:
According to the problem, we have found the fractional part of the number $\dfrac{{{2}^{403}}}{15}$ as $\dfrac{k}{15}$. We need to find the value of k.
We know that the fractional part of x is defined as \left\\{ x \right\\}=x-\left[ x \right], where $\left[ . \right]$ is the greatest integer function.
Let us find the total value of $\dfrac{{{2}^{403}}}{15}$.
$\Rightarrow \dfrac{{{2}^{403}}}{15}=\dfrac{{{2}^{3}}{{.2}^{400}}}{15}$.
$\Rightarrow \dfrac{{{2}^{403}}}{15}=\dfrac{{{2}^{3}}.{{\left( {{2}^{4}} \right)}^{100}}}{15}$.
$\Rightarrow \dfrac{{{2}^{403}}}{15}=\dfrac{{{2}^{3}}.{{\left( 16 \right)}^{100}}}{15}$.
$\Rightarrow \dfrac{{{2}^{403}}}{15}=\dfrac{{{2}^{3}}.{{\left( 1+15 \right)}^{100}}}{15}$ -(1).
We know that binomial expansion of ${{\left( 1+x \right)}^{n}}$ is $1+{}^{n}{{C}_{1}}x+{}^{n}{{C}_{2}}{{x}^{2}}+......+{}^{n}{{C}_{n}}{{x}^{n}}$. We apply this in equation (1).
$\Rightarrow \dfrac{{{2}^{403}}}{15}=\dfrac{{{2}^{3}}.\left( 1+{}^{100}{{C}_{1}}15+{}^{100}{{C}_{2}}{{15}^{2}}+......+{}^{100}{{C}_{100}}{{15}^{100}} \right)}{15}$ -(2).
We can see in the binomial expansion present in the numerator every term is multiple of 15 except the first term. So, we assume the sum of all the other terms as a multiple of 15. We take ${}^{100}{{C}_{1}}15+{}^{100}{{C}_{2}}{{15}^{2}}+......+{}^{100}{{C}_{100}}{{15}^{100}}$ as $15a$. We substitute this in equation (2).
$\Rightarrow \dfrac{{{2}^{403}}}{15}=\dfrac{8.\left( 1+15a \right)}{15}$ here a is a positive integer.
$\Rightarrow \dfrac{{{2}^{403}}}{15}=\dfrac{8}{15}+\dfrac{120a}{15}$.
$\Rightarrow \dfrac{{{2}^{403}}}{15}=\dfrac{8}{15}+8a$.
We need to find the value of \left\\{ \dfrac{{{2}^{403}}}{15} \right\\}.
\Rightarrow \left\\{ \dfrac{{{2}^{403}}}{15} \right\\}=\left\\{ \dfrac{8}{15}+8a \right\\}.
\Rightarrow \left\\{ \dfrac{{{2}^{403}}}{15} \right\\}=\dfrac{8}{15}+8a-\left[ \dfrac{8}{15}+8a \right].
We know that the greatest integer function takes the value of the integer that is less than or equal to the given number. We have got 8a as a number and $\dfrac{8}{15}$ which is less than 1 which makes 8a as the least integer.
\Rightarrow \left\\{ \dfrac{{{2}^{403}}}{15} \right\\}=\dfrac{8}{15}+8a-8a
\Rightarrow \left\\{ \dfrac{{{2}^{403}}}{15} \right\\}=\dfrac{8}{15}.
So, we have found the value of k as 8.

So, the correct answer is “Option d”.

Note: We can see all other terms in binomial expansion expect 1 is divisible by 15. So, we take 15a to make our calculations easy for finding the nearest integer. We should know that the fractional part of the x lies between 0 and 1. Whenever we get this type of problem, we need to take binomial expansion and convert it into multiples of denominators to make calculations easy.