Prove that the number is the product of two successive positive integers

Question

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)?

Answer 1

Hint:

$$ \begin{align} \underbrace{11\cdots 1}_{100\text{ digits}}\underbrace{22\cdots 2}_{100\text{ digits}} & = 1 \cdot 10^{199} + 1 \cdot 10^{198} + \cdots + 1 \cdot 10^{100} + 2 \cdot 10^{99} + 2 \cdot 10^{98} + \cdots + 2 \cdot 10^0 \\ & = 2 \sum_{k=0}^{99} 10^k + \sum_{k=100}^{199} 10^k \\ & = \frac{(10^{100}-1)(10^{100}+2)}{9} \end{align} $$

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[ EDIT ]  To conclude the proof, just note that $\frac{10^{100}-1}{3}$ is an integer, and $\frac{10^{100}+2}{3} = \frac{10^{100}-1}{3} + 1$.

Answer 2

Let $u=11\cdots1$ ($100$ digits). Then $10^{100}=9u+1$ and so

$$11\cdots 1 22\cdots 2 = u10^{100}+2u = u(10^{100}+2) = u(9u+3) = 3u(3u+1)$$

Answer 3

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:

$$\Big(\sum\limits_{k=0}^{2n-1}10^k\Big)+\Big(\sum\limits_{k=0}^{n-1}10^k\Big)=\Big(3\sum\limits_{k=0}^{n-1}10^k\Big)\times\Big(1+3\sum\limits_{k=0}^{n-1}10^k\Big)$$

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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 $n+1$:

$\Big(\sum\limits_{k=0}^{2n+1}10^k\Big)+\Big(\sum\limits_{k=0}^{n}10^k\Big)=$

$\color\red{\Big(\sum\limits_{k=0}^{2n-1}10^k\Big)+\Big(\sum\limits_{k=0}^{n-1}10^k\Big)}+10^{2n+1}+10^{2n}+10^{n}=$

$\color\red{\Big(3\sum\limits_{k=0}^{n-1}10^k\Big)\times\Big(1+3\sum\limits_{k=0}^{n-1}10^k\Big)}+10^{2n+1}+10^{2n}+10^{n}=$

$\Big(3\sum\limits_{k=0}^{n}10^k\Big)\times\Big(1+3\sum\limits_{k=0}^{n}10^k\Big)$

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