Try this beautiful Digits Problem from Number theorm from PRMO 2018, Question 19.
Let $N=6+66+666+\ldots \ldots+666 \ldots .66,$ where there are hundred 6 's in the last term in the sum. How many times does the digit 7 occur in the number $N ?$
Number theorm
Digits Problem
integer
But try the problem first...
Answer:$33$
PRMO-2018, Problem 19
Pre College Mathematics
First hint
Given that $\mathrm{N}=6+66+666+....... \underbrace{6666 .....66}_{100 \text { times }}$
If you notice then we can see there are so many large terms. but we have to find out the sum of the digits. but since the number of digits are large so we can not calculate so eassily . we have to find out a symmetry or arrange the number so that we can use any formula taht we can calculate so eassily. if we multiply \(\frac{6}{9}\) then it becomes $=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}]$
Can you now finish the problem ..........
Second Hint
$\mathrm{N}=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}]$
$=\frac{6}{9}\left[(10-1)+\left(10^{2}-1\right)+.......+\left(10^{100}-1\right)\right]$
$=\frac{6}{9}\left[\left(10+10^{2}+.....+10^{100}\right)-100\right]$
Can you finish the problem........
Final Step
$=\frac{6}{9}\left[\left(10^{2}+10^{3}+\ldots \ldots \ldots+10^{100}\right)-90\right]$
$=\frac{6}{9}\left(10^{2} \frac{\left(10^{99}-1\right)}{9}\right)-60$
$=\frac{200}{27}\left(10^{99}-1\right)-60$
$=\frac{200}{27}\underbrace{(999....99)}_{99 \text{times}}-60$
$=\frac{1}{3}\underbrace{(222.....200)}_{99 \mathrm{times}}-60$
$=\underbrace{740740 \ldots \ldots .7400-60}_{740 \text { comes } 33 \text { times }}$ $=\underbrace{740740 \ldots \ldots .740}_{32 \text { times }}+340$
$\Rightarrow 7$ comes 33 times
Try this beautiful Digits Problem from Number theorm from PRMO 2018, Question 19.
Let $N=6+66+666+\ldots \ldots+666 \ldots .66,$ where there are hundred 6 's in the last term in the sum. How many times does the digit 7 occur in the number $N ?$
Number theorm
Digits Problem
integer
But try the problem first...
Answer:$33$
PRMO-2018, Problem 19
Pre College Mathematics
First hint
Given that $\mathrm{N}=6+66+666+....... \underbrace{6666 .....66}_{100 \text { times }}$
If you notice then we can see there are so many large terms. but we have to find out the sum of the digits. but since the number of digits are large so we can not calculate so eassily . we have to find out a symmetry or arrange the number so that we can use any formula taht we can calculate so eassily. if we multiply \(\frac{6}{9}\) then it becomes $=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}]$
Can you now finish the problem ..........
Second Hint
$\mathrm{N}=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}]$
$=\frac{6}{9}\left[(10-1)+\left(10^{2}-1\right)+.......+\left(10^{100}-1\right)\right]$
$=\frac{6}{9}\left[\left(10+10^{2}+.....+10^{100}\right)-100\right]$
Can you finish the problem........
Final Step
$=\frac{6}{9}\left[\left(10^{2}+10^{3}+\ldots \ldots \ldots+10^{100}\right)-90\right]$
$=\frac{6}{9}\left(10^{2} \frac{\left(10^{99}-1\right)}{9}\right)-60$
$=\frac{200}{27}\left(10^{99}-1\right)-60$
$=\frac{200}{27}\underbrace{(999....99)}_{99 \text{times}}-60$
$=\frac{1}{3}\underbrace{(222.....200)}_{99 \mathrm{times}}-60$
$=\underbrace{740740 \ldots \ldots .7400-60}_{740 \text { comes } 33 \text { times }}$ $=\underbrace{740740 \ldots \ldots .740}_{32 \text { times }}+340$
$\Rightarrow 7$ comes 33 times