Question

Choose the correct option. ${{\left( 115 \right)}^{96}}-{{\left( 96 \right)}^{115}}$ is divisible by
A) 15
B) 17
C) 19
D) 21

Answer 1

In the given question, we have been asked that the given expression is divisible by which number. In order to solve the question, first we need to rewrite the equation to make it possible to have a common factor. Later by applying the combination, we will need to expand the given expression by using the commutation formula i.e. ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. Then, by taking out the common factor we will observe that the given expression is divisible by any number which is given as the option.

Complete step-by-step solution:
We have given that,
$\Rightarrow {{\left( 115 \right)}^{96}}-{{\left( 96 \right)}^{115}}$
Rewritten the above expression as,
$\Rightarrow {{\left( 1+114 \right)}^{96}}-{{\left( 1+95 \right)}^{115}}$
Now, applying the combinations, we get
$\Rightarrow \left( 1+{}^{96}{{C}_{1}}\left( 114 \right)+{}^{96}{{C}_{2}}{{\left( 114 \right)}^{2}}+..... \right)-\left( 1+{}^{115}{{C}_{1}}\left( 95 \right)+{}^{115}{{C}_{2}}{{\left( 95 \right)}^{2}}+.... \right)$
As we know that,
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
Therefore,
Solving the above, we obtained
Taking out the common term from both the brackets, we get
$\Rightarrow 114\left( {}^{96}{{C}_{1}}+{}^{96}{{C}_{2}}\left( 114 \right)+..... \right)-95\left( {}^{115}{{C}_{1}}+{}^{115}{{C}_{2}}\left( 95 \right)+.... \right)$
$\Rightarrow 114\left( {}^{96}{{C}_{1}}+{}^{96}{{C}_{2}}\left( 114 \right)+..... \right)-95\left( {}^{115}{{C}_{1}}+{}^{115}{{C}_{2}}\left( 95 \right)+.... \right)$
As we can see that 19 is the common factor of both the numbers i.e. 114 and 95.
Thus,
Taking out 19 as a common factor, we get
$\Rightarrow 19\left( 6\left( {}^{96}{{C}_{1}}+{}^{96}{{C}_{2}}\left( 114 \right)+..... \right)-5\left( {}^{115}{{C}_{1}}+{}^{115}{{C}_{2}}\left( 95 \right)+.... \right) \right)$
Thus, the given expression is divisible by 19.

Hence, the option (C) is the correct answer.

Note: In mathematics, a permutation is known as the choice of ‘r’ things from the given set of ‘n’ things without the replacement. In permutation, order matters.
The formula of permutation is given by;
$\Rightarrow {}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$
And a combination is known as the choice of ‘r’ things from the given set of ‘n’ things without the replacement and where the order does not matter.
The formula of combination is given by;
$\Rightarrow {}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.