Question

If $111....1\left( {91\,{\text{times}}} \right)$ is a
(The question has multiple correct options)
(A) Prime number
(B) Composite number
(C) not a integer
(D) Integer

Answer 1

Hint: In this question we can write $111...1$ as the sum of increasing powers of $10$ and then from it we can find the sum of the series which is a geometric series with the help of the formula of sum of geometric series. Now, from the result of this sum we can decide what $111...1$ is.

Formula used:
${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}$ , where ${S_n}$ is the sum of $n$ terms, $a$ is the first term of the series and $r$ is the common ratio.

Complete step-by-step answer:
We can write $111....1\left( {91\,{\text{times}}} \right)$ as the sum of increasing powers of $10$
$ \Rightarrow 1 + {10^1} + {10^2} + {10^3} + .... + {10^{90}}$
We can say that the above expression is the geometric sequence; its common ratio in the above expression is $10$ , the first term is \[1\] and the number of terms are $91$ . Therefore, we can write $a = 1,\,n = 91$ and $r = 10$ .
Now, substitute all this values in the formula ${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{\left( {r - 1} \right)}}$
$
   \Rightarrow {S_n} = \dfrac{{1\left( {{{10}^{91}} - 1} \right)}}{{\left( {10 - 1} \right)}} \\
   \Rightarrow {S_n} = \dfrac{{\left( {{{10}^{91}} - 1} \right)}}{{\left( {10 - 1} \right)}} \\
 $
Now, the above equation can be written as
$ \Rightarrow {S_n} = \dfrac{{\left( {{{\left( {{{10}^{13}}} \right)}^7} - 1} \right)}}{{\left( {10 - 1} \right)}}$
Multiply and divide the above equation with ${10^{13}} - 1$ . Therefore, the equation can be written as follows:

 $ \Rightarrow {S_n} = \dfrac{{\left( {{{\left( {{{10}^{13}}} \right)}^7} - 1} \right)}}{{{{10}^{13}} - 1}} \times \dfrac{{{{10}^{13}} - 1}}{{10 - 1}}$
Now, we can observe from the above expression that the above expression is the multiplication of the sum of two geometric series. Therefore, the above expression can be written as follows:
$ \Rightarrow {S_n} = \left( {{{10}^{13}} + {{10}^{26}} + .... + {{10}^{91}}} \right) \times \left( {1 + 10 + {{10}^2} + ... + {{10}^{13}}} \right)$
Now, from the above observation we can write $111....1\left( {91\,{\text{times}}} \right)$ as multiplication of its two factor
$ \Rightarrow 111...1 = \left( {{{10}^{13}} + {{10}^{26}} + .... + {{10}^{91}}} \right) \times \left( {1 + 10 + {{10}^2} + ... + {{10}^{13}}} \right)$
Now, as $111....1\left( {91\,{\text{times}}} \right)$ can be expressed as multiplication of two factors it is a composite number and also we can say that it is not a prime number. It is also an integer.
Hence, the option (B) and (D) are correct.

Note: In this question the important thing is to convert the number given into the sum of geometric series and from that observing how it can be solved. The other important thing is the definition of composite numbers and also one should know how to say that the number is a prime number.