Question

If the fractional part of the number \[\dfrac{{{2^{403}}}}{{15}}\] is \[\dfrac{k}{{15}}\] , then k is equal to

Answer 1

Hint: We will use binomial expansion for the given number to find the fractional part of the number and then we will compare it with the given value to get the value of k.
The binomial theorem describes the algebraic expansion of powers of a binomial.

Complete step by step solution:
The given number is \[\dfrac{{{2^{403}}}}{{15}}\]
And we are given that :
\[\dfrac{{{2^{403}}}}{{15}} = \dfrac{k}{{15}}..........\left( 1 \right)\]
According to the property of exponents if numbers with same and different powers are multiplied then their powers get added:
\[{a^b} \times {a^c} = {a^{\left( {b + c} \right)}}\]
Therefore, the numerator \[{2^{403}}\] can be further written as:
\[{2^{403}} = {2^3} \times {2^{400}}............\left( 2 \right)\]
According to the property of exponents if numbers with same and different powers are multiplied then their powers get added:
\[{a^b} \times {a^c} = {a^{\left( {b + c} \right)}}\]
Now according to another property of exponents :
\[{\left( {{a^b}} \right)^c} = {\left( a \right)^{bc}}\]
Applying this property in equation1 we get:
\[{2^{403}} = {2^3} \times {\left( {{2^4}} \right)^{100}}\]
Further solving it we get:
\[
  {2^{403}} = {2^3} \times {\left( {16} \right)^{100}} \\
  {2^{403}} = 8 \times {\left( {1 + 15} \right)^{100}} \\
 \]
Now using binomial expansion we get:
\[
  {2^{403}} = 8 \times \left[ {{}^{100}C0{{\left( {15} \right)}^0}{{\left( 1 \right)}^{100}} + {}^{100}C1{{\left( {15} \right)}^1}{{\left( 1 \right)}^{99}} + {}^{100}C2{{\left( {15} \right)}^2}{{\left( 1 \right)}^{98}} + ............... + {}^{100}C100{{\left( {15} \right)}^{100}}{{\left( 1 \right)}^0}} \right] \\
  {2^{403}} = 8 \times \left[ {1 + {}^{100}C1\left( {15} \right) + {}^{100}C2{{\left( {15} \right)}^2} + ................ + {}^{100}C100{{\left( {15} \right)}^{100}}} \right] \\
  {2^{403}} = 8 + 8 \times \left( {15} \right) + 8 \times {}^{100}C2{\left( {15} \right)^2} + ................ + 8 \times {}^{100}C100{\left( {15} \right)^{100}} \\
  {2^{403}} = 8 + 8 \times \left( {15} \right)\left[ \lambda \right] \\
 \]
Where \[\lambda = {}^{100}C2\left( {15} \right) + ................ + {}^{100}C100{\left( {15} \right)^{99}}\]
Now dividing the above equation by 15 we get:
 \[
  \dfrac{{{2^{403}}}}{{15}} = \dfrac{{8 + 8 \times \left( {15} \right)\left[ \lambda \right]}}{{15}} \\
  \dfrac{{{2^{403}}}}{{15}} = \dfrac{8}{{15}} + \dfrac{{8 \times \left( {15} \right)\left[ \lambda \right]}}{{15}} \\
  \dfrac{{{2^{403}}}}{{15}} = \dfrac{8}{{15}} + 8\lambda \\
 \]
Now the since fractional part in the above equation is \[\dfrac{8}{{15}}\]
Hence on comparing it with the value given we get:
\[
  \dfrac{8}{{15}} = \dfrac{k}{{15}} \\
  k = 8 \\
 \]
Hence the value of k is 8.

Therefore (B) is the correct option.

Note:
The binomial expansion of two numbers is given by:
\[{\left( {a + b} \right)^n} = {}^nC0{\left( a \right)^0}{\left( b \right)^n} + {}^nC1{\left( a \right)^1}{\left( b \right)^{n - 1}} + {}^nC2{\left( a \right)^2}{\left( b \right)^{n - 2}} + ............... + {}^nC100{\left( a \right)^{100}}{\left( b \right)^0}\]