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Prove that number $\underbrace{11 \ldots1}_{100}$$\underbrace{22 \ldots2}_{100}$ is product of two consecutive numbers

$\begin{align}\underbrace{11 \ldots1}_{100} \underbrace{22 \ldots2}_{100}&=10^{199}+10^{198}+\ldots+10^{100}+2(10^{99}+10^{98}+\ldots+10+1)\\&=(10^{100}+2)(10^{99}+10^{98}+\ldots+10+1)=(10^{100}+2)\frac{10^{100}-1}{10-1}\end{align}$.

Is this good path or not?

Yadati Kiran
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Marko Škorić
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    You need a backslash before underbrace to get what you want. – Ross Millikan Nov 16 '18 at 15:43
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    I don't see any advantage in re-writing the way you have (though of course I might be missing something). I'd start small. Note that $12=3\times 4$. What about $1122$? Well, that's $33\times 34$. Maybe there's a pattern.... – lulu Nov 16 '18 at 15:45

3 Answers3

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You are not done because the two factors you exhibited are not consecutive. For three digits, you have shown $111222=1002\cdot 111$. You can, however, make one more step and be there. lulu has given a good hint.

Ross Millikan
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Hint $\ $ Apply the result below, with $\,a = 10^{100}\!-1,\ n = 3,\,$ using $\,n^2\mid a$

$$\begin{align} &\dfrac{a+n}{1}\ \dfrac{a}{n^2}\\[.3em] =\ &\dfrac{a+n}{n}\ \dfrac{a}{n}\\[.3em] =\ &(b+1)\, b \end{align}$$

Bill Dubuque
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Insert $1=\frac{1}{3}*3$

$$ (10^{100}+2)\frac{1}{3} 3\frac{10^{100}-1}{10-1}\\ x= \frac{10^{100}+2}{3}\\ y=3\frac{10^{100}-1}{10-1} = \frac{10^{100}-1}{3}\\\\ $$

AHusain
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