Find the product of ($100$ digits of $3$) $333...3 × 2019$.

Question

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!

Answer 1

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.

Answer 2

For $n\ge1,$

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

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