New Year Problem (3)

Consider a number $A_n = \underbrace{11111\cdots11111}_{\text{Number of 1's}=n} = \displaystyle \sum_{k=0}^{n-1} 10^k = \frac {10^n-1}9$ . Then

$\begin{aligned} \underbrace{333\cdots3}_n5^2 & = (30A_n + 5)^2 \\ & = 900A_n^2 + 300A_n + 25 \\ & = 900 \left(\frac {10^n-1}9\right)A_n + 300A_n + 25 \\ & = 100 \left(10^n-1 \right)A_n + 300A_n + 25 \\ & = 10^{n+2}A_n + 200A_n + 25 \\ & = \underbrace{111\cdots 1}_n\underbrace{222\cdots2}_{n+1}5 \end{aligned}$

Put $n=2019$ and it is "Yes" .

$\underbrace{1111111 ... 1111111}_\text{Number of 1's=2019} \underbrace{2222222 ... 2222222}_\text{Number of 2's=2020} 5$

$= \underbrace{11111 ... 11111}_\text{2 * 2020} + \underbrace{11111 ... 11111}_\text{2021} + 3$

$= \frac{10^{2 \cdot 2020} - 1}{9} + \frac{10^{2021} - 1}{9} + 3$

$= \frac{10^{2 \cdot 2020} - 1}{9} + \frac{10^{2021} - 1}{9} - \frac{9}{9} + \frac{9}{9} + 3$

$= \frac{10^{2 \cdot 2020} - 1}{9} + \frac{10^{2021} - 10}{9} + 4$

$= \frac{10^{2 \cdot 2020} - 1}{9} + 10 \cdot \frac{10^{2020} - 1}{9} + 4$

$= \frac{10^{2 \cdot 2020} - 1}{9} - 2 \cdot \frac{10^{2020} - 1}{9} + 4 \cdot \frac{10^{2020} - 1}{3} + 4$

$= \frac{10^{2 \cdot 2020} - 1 - 2(10^{2020} - 1)}{9} + 4 \cdot \frac{10^{2020} - 1}{3} + 4$

$= \frac{10^{2 \cdot 2020} - 2 \cdot 10^{2020} + 1}{9} + 4 \cdot \frac{10^{2020} - 1}{3} + 4$

$= (\frac{10^{2020} - 1}{3})^2 + 4 \cdot \frac{10^{2020} - 1}{3} + 4$

$= (\frac{10^{2020} - 1}{3} + 2)^2$