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What is the product of $\underbrace{333...3}_\text{$100$ digits of $3$} \times2019$?

I am trying to solve this problem

$\underbrace{333...3}_\text{$100$ digits of $3$}$ $×$ $2019$ = $\underbrace{333...3}_\text{$100$ digits of $3$}$ $×$ $(2020-1)$

=$\underbrace{333...3}_\text{$100$ digits of $3$}$ $×$ $(2000 + 20-1)$

I am not sure if I am at the right process.

Any help will be much appreciated. Thank you!

AYA
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    It's all good, looks good, please go on with your solution :) – Matti P. Jul 13 '20 at 09:03
  • Or one could try $2019\times3=6057$, then $6057\times111\cdots 1111$ and see the pattern here. – Chrystomath Jul 13 '20 at 09:06
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    $2$ answrs, although it was not clarified which kind of solution is expected. Do you want a description of the decimal expansion of this product or just a handy expression for this product ? – Peter Jul 13 '20 at 09:09

2 Answers2

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For $n\ge1,$

$$\underbrace{33\cdots33}_{ n \text{ digits}}=\dfrac{10^n-1}3$$

Now $\underbrace{33\cdots33}_{ n \text{ digits}}\cdot2019=\dfrac{10^n-1}3\cdot2019=(10^n-1)673$

lab bhattacharjee
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Your process is correct. Following it gives:

$$\quad 666 \ \underbrace{666 \cdots 666}_\text{$97$ times} \ 000$$ $$+ \ \ \ \ \ 6 \ \underbrace{666 \cdots 666}_\text{$97$ times} \ 660$$ $$- \quad \ \ \ \underbrace{333 \cdots 333}_\text{$97$ times} \ 333$$

and from here it is straightforward.

Toby Mak
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