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Question: Let \(N = 6 + 66 + 666 + 666...............\) where there are hundreds sixes in the last term in the...

Let N=6+66+666+666...............N = 6 + 66 + 666 + 666............... where there are hundreds sixes in the last term in the sum. How many times does the digit 77 occur in the number N?

Explanation

Solution

Try to find out the number by adding it , by using the Summation of GP if we take common 66 and multiple by 99 series become
⇒\Rightarrow N=69(9+99+999+9999.............)N = \dfrac{6}{9}(9 + 99 + 999 + 9999.............) or
⇒ \Rightarrow $N = \dfrac{6}{9}\left\{ {({{10}^1} - 1) + ({{10}^2} - 1) + ({{10}^3} - 1) + .............({{10}^{100}} - 1)} \right\}nowapplysummationofGPandgetthenumberafterconcludeseveninthedigitasnow apply summation of GP and get the number after conclude seven in the digit as7$ 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\}Nowsubtract Now subtract100 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 77 is repeated 3333 times in this digit.

Note: In the summation of series of if the common ratio r is in between 00 and 11 then it represent an infinite GP with summation formula is Sn=a1−r{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 arm−na{r^{m - n}}.