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# Digits Problem | PRMO - 2018 | Question 19

Try this beautiful Digits Problem from Number theorm from PRMO 2018, Question 19.

## Digits Problem - 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 $N ?$

• $30$
• $33$
• $36$
• $39$
• $42$

### Key Concepts

Number theorm

Digits Problem

integer

Answer:$33$

PRMO-2018, Problem 19

Pre College Mathematics

## Try with Hints

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 ..........

$\mathrm{N}=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}]$
$=\frac{6}{9}\left[(10-1)+\left(10^{2}-1\right)+.......+\left(10^{100}-1\right)\right]$
$=\frac{6}{9}\left[\left(10+10^{2}+.....+10^{100}\right)-100\right]$

Can you finish the problem........

$=\frac{6}{9}\left[\left(10^{2}+10^{3}+\ldots \ldots \ldots+10^{100}\right)-90\right]$
$=\frac{6}{9}\left(10^{2} \frac{\left(10^{99}-1\right)}{9}\right)-60$
$=\frac{200}{27}\left(10^{99}-1\right)-60$
$=\frac{200}{27}\underbrace{(999....99)}_{99 \text{times}}-60$
$=\frac{1}{3}\underbrace{(222.....200)}_{99 \mathrm{times}}-60$

$=\underbrace{740740 \ldots \ldots .7400-60}_{740 \text { comes } 33 \text { times }}$ $=\underbrace{740740 \ldots \ldots .740}_{32 \text { times }}+340$
$\Rightarrow 7$ comes 33 times

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Try this beautiful Digits Problem from Number theorm from PRMO 2018, Question 19.

## Digits Problem - 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 $N ?$

• $30$
• $33$
• $36$
• $39$
• $42$

### Key Concepts

Number theorm

Digits Problem

integer

Answer:$33$

PRMO-2018, Problem 19

Pre College Mathematics

## Try with Hints

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 ..........

$\mathrm{N}=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}]$
$=\frac{6}{9}\left[(10-1)+\left(10^{2}-1\right)+.......+\left(10^{100}-1\right)\right]$
$=\frac{6}{9}\left[\left(10+10^{2}+.....+10^{100}\right)-100\right]$

Can you finish the problem........

$=\frac{6}{9}\left[\left(10^{2}+10^{3}+\ldots \ldots \ldots+10^{100}\right)-90\right]$
$=\frac{6}{9}\left(10^{2} \frac{\left(10^{99}-1\right)}{9}\right)-60$
$=\frac{200}{27}\left(10^{99}-1\right)-60$
$=\frac{200}{27}\underbrace{(999....99)}_{99 \text{times}}-60$
$=\frac{1}{3}\underbrace{(222.....200)}_{99 \mathrm{times}}-60$

$=\underbrace{740740 \ldots \ldots .7400-60}_{740 \text { comes } 33 \text { times }}$ $=\underbrace{740740 \ldots \ldots .740}_{32 \text { times }}+340$
$\Rightarrow 7$ comes 33 times

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