Question

Let $N = 6 + 66 + 666 + 666...............$ where there are hundreds sixes in the last term in the sum. How many times does the digit $7$ occur in the number N?

Answer 1

Hint: Try to find out the number by adding it , by using the Summation of GP if we take common $6$ and multiple by $9$ series become
\[ \Rightarrow \] $N = \dfrac{6}{9}(9 + 99 + 999 + 9999.............)$ or
\[ \Rightarrow \]$N = \dfrac{6}{9}\left\{ {({{10}^1} - 1) + ({{10}^2} - 1) + ({{10}^3} - 1) + .............({{10}^{100}} - 1)} \right\}$ now apply summation of GP and get the number after conclude seven in the digit as $7$ digits in number repeat in sequence.

Complete step-by-step answer:
First try to find out the number by adding it to the number or by use it summation formula of AP by making some changes as
\[ \Rightarrow \]$N = 6 + 66 + 666 + 666...............\underbrace {66666....66}_{100{\text{ times }}}$
Take $6$ common from this number ,
\[ \Rightarrow \]$N = 6(1 + 11 + 111 + 1111.............)$
Now multiple and divide by $9$ in this number ,
\[ \Rightarrow \]$N = \dfrac{6}{9}(9 + 99 + 999 + 9999.............)$
or it can be written as
\[ \Rightarrow \]$N = \dfrac{6}{9}\left\{ {(10 - 1) + (100 - 1) + (1000 - 1) + .............} \right\}$
\[ \Rightarrow \]$N = \dfrac{6}{9}\left\{ {({{10}^1} - 1) + ({{10}^2} - 1) + ({{10}^3} - 1) + .............({{10}^{100}} - 1)} \right\}$
\[ \Rightarrow \]$N = \dfrac{6}{9}\left\{ {({{10}^1} + {{10}^2} + {{10}^3} + {{..........10}^{100}}) - 100} \right\}$
As the given term ${10^1} + {10^2} + {10^3} + {..........10^{100}}$ are in GP with first term is $10$ and common ratio is $10$ total number of term is $100$
${S_n} = \dfrac{{a({r^n} - 1)}}{{r - 1}}$
Hence for this $a = 10,r = 10,n = 100$
\[ \Rightarrow \]${10^1} + {10^2} + {10^3} + {..........10^{100}}$ = $\dfrac{{10({{10}^{100}} - 1)}}{9}$
So ,
\[ \Rightarrow \]$N = \dfrac{2}{3}\left\{ {\dfrac{{10({{10}^{100}} - 1)}}{9} - 100} \right\}$
\[ \Rightarrow \]$N = \dfrac{2}{3}\left\{ {\dfrac{{10(\underbrace {9999999.......99}_{100times})}}{9} - 100} \right\}$
Multiple by $10$ in it and divide it in next step ,
\[ \Rightarrow \]$N = \dfrac{2}{3}\left\{ {\dfrac{{999999......9990}}{9} - 100} \right\}$
\[ \Rightarrow \]$N = \dfrac{2}{3}\left\{ {\underbrace {111111.....1110}_{1{\text{ is 100times}}} - 100} \right\}$
Now subtract $100$ from it ,
\[ \Rightarrow \]$N = \dfrac{2}{3}\left\{ {1111.....1010} \right\}$
\[ \Rightarrow \]$N = \dfrac{1}{3}\left\{ {2222.....2020} \right\}$
Now divide $3$ from number ,
\[ \Rightarrow \]$N = \underbrace {740740....7407}_{96digits}340$

Hence $7$ is repeated $33$ times in this digit.

Note: In the summation of series of if the common ratio r is in between $0$ and $1$ then it represent an infinite GP with summation formula is ${S_n} = \dfrac{a}{{1 - r}}$ where a is first term of GP. If a is the first term, r is the common ratio of a finite G.P. consisting of m terms, then the nth term from the end will be $a{r^{m - n}}$.