Question

Consider the number $N = {\left( {106} \right)^{85}} - {\left( {85} \right)^{106}}$ then:
A. $N$ is divisible by 7
B. $N$ is not divisible by 7
C. $N$ is prime
D. None of these

Answer 1

First write the number $N$ in the form of ${\left( {105 + 1} \right)^{85}} - {\left( {84 + 1} \right)^{106}}$. Then expand both the terms binomially by using the formula ${\left( {x + 1} \right)^n} = {}^n{C_0}{x^n} + {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} + .... + {}^n{C_n}{x^0}$. Then simplify the number further and check whether 7 can be taken out from it or not.

Complete step-by-step answer:
According to the question, the given number is:
$\Rightarrow N = {\left( {106} \right)^{85}} - {\left( {85} \right)^{106}}$
First, we will rewrite the above number as:
$\Rightarrow N = {\left( {105 + 1} \right)^{85}} - {\left( {84 + 1} \right)^{106}}$
Now, we use binomial expansion of ${\left( {1 + x} \right)^n}$. As we know that:
$\Rightarrow {\left( {x + 1} \right)^n} = {}^n{C_0}{x^n} + {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} + .... + {}^n{C_n}{x^0}$
Applying this expansion for $N$, we’ll get:
$\Rightarrow N = \left[ {{}^{85}{C_0}{{\left( {105} \right)}^{85}} + {}^{85}{C_1}{{\left( {105} \right)}^{84}} + ... + {}^{85}{C_{84}}} \right] - \left[ {{}^{106}{C_0}{{\left( {84} \right)}^{106}} + ....{}^{106}{C_{105}}84 + {}^{106}{C_{106}}} \right]$
We know that ${}^n{C_n} = 1$. From this we have ${}^{85}{C_{85}} = {}^{106}{C_{106}} = 1$. Using these values, we’ll get:
$\Rightarrow N = \left[ {{}^{85}{C_0}{{\left( {105} \right)}^{85}} + {}^{85}{C_1}{{\left( {105} \right)}^{84}} + ... + 1} \right] - \left[ {{}^{106}{C_0}{{\left( {84} \right)}^{106}} + ....{}^{106}{C_{105}}84 + 1} \right]$
This can be further simplified as:

$$\Rightarrow N = \left[ {{}^{85}{C_0}{{\left( {105} \right)}^{85}} + {}^{85}{C_1}{{\left( {105} \right)}^{84}} + ...} \right] + 1 - \left[ {{}^{106}{C_0}{{\left( {84} \right)}^{106}} + ....{}^{106}{C_{105}}84} \right] - 1 \\\ \Rightarrow N = \left[ {{}^{85}{C_0}{{\left( {105} \right)}^{85}} + {}^{85}{C_1}{{\left( {105} \right)}^{84}} + ...} \right] - \left[ {{}^{106}{C_0}{{\left( {84} \right)}^{106}} + ....{}^{106}{C_{105}}84} \right] \\\$$

In the above number, 105 can be taken as common from the first term and 84 can be taken as common from the second term. So we have:
$\Rightarrow N = 105\left[ {{}^{85}{C_0}{{\left( {105} \right)}^{84}} + {}^{85}{C_1}{{\left( {105} \right)}^{83}} + ...} \right] - 84\left[ {{}^{106}{C_0}{{\left( {84} \right)}^{105}} + ....{}^{106}{C_{105}}} \right]$
Taking 7 outside from the complete number as both 105 and 84 are multiple of 7, we’ll get:
$\Rightarrow N = 7\left( {15\left[ {{}^{85}{C_0}{{\left( {105} \right)}^{84}} + {}^{85}{C_1}{{\left( {105} \right)}^{83}} + ...} \right] - 12\left[ {{}^{106}{C_0}{{\left( {84} \right)}^{105}} + ....{}^{106}{C_{105}}} \right]} \right)$
Now, let $15\left[ {{}^{85}{C_0}{{\left( {105} \right)}^{84}} + {}^{85}{C_1}{{\left( {105} \right)}^{83}} + ...} \right] - 12\left[ {{}^{106}{C_0}{{\left( {84} \right)}^{105}} + ....{}^{106}{C_{105}}} \right] = k$, where $k$ is a natural number. Then we have:
$\Rightarrow N = 7k$
Therefore we can see that the number $N$ is divisible by 7.

(A) is the correct option.

Note: For checking divisibility, binomial theorem comes in handy if the number can be transformed in the form of ${\left( {x \pm 1} \right)^n}$ because in such cases the last term becomes ${\left( { - 1} \right)^n}$ which can be easily managed.
The expansion of ${\left( {x \pm 1} \right)^n}$ is done by using the formula:

$$\Rightarrow {\left( {x + 1} \right)^n} = {}^n{C_0}{x^n} + {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} + .... + {}^n{C_n}{x^0} \\\ \Rightarrow {\left( {x - 1} \right)^n} = {}^n{C_0}{x^n} - {}^n{C_1}{x^{n - 1}} + {}^n{C_2}{x^{n - 2}} + .... + {\left( { - 1} \right)^n}{}^n{C_n}{x^0} \\\$$