Question

111……1 (91 times) is a
A. Prime Number
B. Composite Number
C. Not an integer
D. Integer

Answer 1

Hint: A prime number is one which has at least one factor other than 1 and itself. As such, what needs to be done is that you need to convert this number into a multiple of two numbers. Also this question will use the formula of sum of G.P. (geometric progression).

Formulas used:
${S_{G.P.}} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}}$
Where ‘a’ is the first term, ‘r’ is the common ratio and ‘n’ is the number of terms.

Complete Step by Step Solution:
Looking at the question, we see this number and observe that it doesn’t have any number after the decimal place and as such it is an integer so the option D is correct and option C is wrong.
Now moving forward, we check if it is a prime number or a composite number. A prime number is one which has at least one factor other than 1 and itself. Thus, we will try to convert the number into a multiple of two numbers.
For this, first of all we will expand this number, say N.
\[\
  N = 111.....1\left( {91\,times} \right) \\
   = {10^0} + {10^1} + {10^2} + {10^3} + ....... + {10^{90}} \\
\ \]
Now, we need to make this a G.P. so we will create a common ratio. 1 cannot be taken as the same and so we will take it to be 10. Thus, we will be putting $10 - 1 = 9$ in the place of ‘r-1’ in the sum of G.P. formula which demands that we multiply the number N by 9. As such,
$\
  N \times 9 = 999......9\left( {91\,times} \right) \\
   = {10^{91}} - 1 \\
\ $
So, sum of G.P. whose result will be N can be given by
$\
  {S_{G.P.}} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}} = N \\
   = \dfrac{{1\left( {{{10}^{91}} - 1} \right)}}{{10 - 1}} \\
\ $
Now we multiply and divide this equation by a number which is one less than 10 raised to a multiple of 91 where 7 and 13 are both its multiples. We will be going with 13, so
$\
  {S_{G.P.}} = \dfrac{{\left( {{{10}^{91}} - 1} \right)}}{{10 - 1}} \times \dfrac{{{{10}^{13}} - 1}}{{{{10}^{13}} - 1}} \\
   = \dfrac{{\left( {{{10}^{91}} - 1} \right)}}{{{{10}^{13}} - 1}} \times \dfrac{{{{10}^{13}} - 1}}{{10 - 1}} \\
   = \dfrac{{{{\left( {{{10}^{13}}} \right)}^7} - 1}}{{{{10}^{13}} - 1}} \times \dfrac{{{{10}^{13}} - 1}}{{10 - 1}} \\
\ $
Now, we expand the G.P. which gives us two multiples.
\[\
  {S_{G.P.}} = N \\
   = \dfrac{{{{\left( {{{10}^{13}}} \right)}^7} - 1}}{{{{10}^{13}} - 1}} \times \dfrac{{{{10}^{13}} - 1}}{{10 - 1}} \\
   = \left[ {{{\left( {{{10}^{13}}} \right)}^0} + {{\left( {{{10}^{13}}} \right)}^1} + {{\left( {{{10}^{13}}} \right)}^2} + {{\left( {{{10}^{13}}} \right)}^3} + {{\left( {{{10}^{13}}} \right)}^4} + {{\left( {{{10}^{13}}} \right)}^5} + {{\left( {{{10}^{13}}} \right)}^6}} \right] \cdot \left[ {{{10}^0} + {{10}^1} + {{10}^2} + {{10}^3} + {{10}^4} + {{10}^5} + {{10}^6}} \right] \\
\ \]
Since it has two multiples, it is a composite number.

Note:
In such types of questions, making it a G.P. form is the most effective method to solve it. If it were anything like 222…. or 333….. or something like that, first take the number common and make it in 111…. Form to use G.P. format. For example,
$\
  222..... = 2 \times 111.... \\
  333..... = 3 \times 111.... \\
\ $
Secondly, about being an integer, if any number has something other than zero on the right of its decimal point, it is not an integer.