Prove that the number is the product of two successive positive integers Prove that the number $\underbrace{11\cdots 1}_{100\text{ digits}}\underbrace{22\cdots 2}_{100\text{ digits}}$ is the product of two successive positive integers.
What is the general method for this class of proof (big integers)?
 A: Hint:
$$
\begin{align}
\underbrace{11\cdots 1}_{100\text{ digits}}\underbrace{22\cdots 2}_{100\text{ digits}} & = 1 \cdot 10^{199} + 1 \cdot 10^{198} + \cdots + 1 \cdot 10^{100} + 2 \cdot 10^{99} + 2 \cdot 10^{98} + \cdots + 2 \cdot 10^0 \\
 & = 2 \sum_{k=0}^{99} 10^k + \sum_{k=100}^{199} 10^k \\
 & = \frac{(10^{100}-1)(10^{100}+2)}{9}
\end{align}
$$

[ EDIT ]  To conclude the proof, just note that $\frac{10^{100}-1}{3}$ is an integer, and $\frac{10^{100}+2}{3} = \frac{10^{100}-1}{3} + 1$.
A: Let $u=11\cdots1$ ($100$ digits). Then $10^{100}=9u+1$ and so
$$11\cdots 1 22\cdots 2 = u10^{100}+2u = u(10^{100}+2) = u(9u+3) = 3u(3u+1)$$
A: Note that:
$$\underbrace{1\dots1}_{n\text{ times }}\underbrace{2\dots2}_{n\text{ times }}=\Big(\sum\limits_{k=0}^{2n-1}10^k\Big)+\Big(\sum\limits_{k=0}^{n-1}10^k\Big)$$
So you can prove by induction that:
$$\Big(\sum\limits_{k=0}^{2n-1}10^k\Big)+\Big(\sum\limits_{k=0}^{n-1}10^k\Big)=\Big(3\sum\limits_{k=0}^{n-1}10^k\Big)\times\Big(1+3\sum\limits_{k=0}^{n-1}10^k\Big)$$

First, show that this is true for $n=1$:
$\Big(\sum\limits_{k=0}^{2-1}10^k\Big)+\Big(\sum\limits_{k=0}^{1-1}10^k\Big)=\Big(3\sum\limits_{k=0}^{1-1}10^k\Big)\times\Big(1+3\sum\limits_{k=0}^{1-1}10^k\Big)$
Second, assume that this is true for $n$:
$\Big(\sum\limits_{k=0}^{2n-1}10^k\Big)+\Big(\sum\limits_{k=0}^{n-1}10^k\Big)=\Big(3\sum\limits_{k=0}^{n-1}10^k\Big)\times\Big(1+3\sum\limits_{k=0}^{n-1}10^k\Big)$
Third, prove that this is true for $n+1$:
$\Big(\sum\limits_{k=0}^{2n+1}10^k\Big)+\Big(\sum\limits_{k=0}^{n}10^k\Big)=$
$\color\red{\Big(\sum\limits_{k=0}^{2n-1}10^k\Big)+\Big(\sum\limits_{k=0}^{n-1}10^k\Big)}+10^{2n+1}+10^{2n}+10^{n}=$
$\color\red{\Big(3\sum\limits_{k=0}^{n-1}10^k\Big)\times\Big(1+3\sum\limits_{k=0}^{n-1}10^k\Big)}+10^{2n+1}+10^{2n}+10^{n}=$
$\Big(3\sum\limits_{k=0}^{n}10^k\Big)\times\Big(1+3\sum\limits_{k=0}^{n}10^k\Big)$

Please note that the assumption is used only in the part marked red.
