# An inequality in numbers

Which number is larger? $\underbrace{888\cdots8}_\text{19 digits}\times\underbrace{333\cdots3}_\text{68 digits}$ or $\underbrace{444\cdots4}_\text{19 digits}\times\underbrace{666\cdots67}_\text{68 digits}$? Why? How much is it larger?

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Let those four numbers be $a,b,c,d$ respectively. Then $a=2c$ and $d=2b+1$. So $cd-ab=c$.

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note that $$\underbrace{888\cdots8}_\text{19 digits}=2*\underbrace{444\cdots4}_\text{19 digits}$$ and $$2*\underbrace{333\cdots3}_\text{68 digits}=\underbrace{666\cdots6}_\text{68 digits}$$ further $$\underbrace{666\cdots6}_\text{68 digits}+1=\underbrace{666\cdots7}_\text{68 digits}$$ Thus $$\underbrace{888\cdots8}_\text{19 digits}*\underbrace{333\cdots3}_\text{68 digits}=\underbrace{444\cdots4}_\text{19 digits}*2*\underbrace{333\cdots3}_\text{68 digits}= \underbrace{444\cdots4}_\text{19 digits}*\underbrace{666\cdots6}_\text{68 digits}$$ which is by $\underbrace{444\cdots4}_\text{19 digits}$ smaller than $$\underbrace{444\cdots4}_\text{19 digits}*\underbrace{666\cdots7}_\text{68 digits}=\underbrace{444\cdots4}_\text{19 digits}*\underbrace{666\cdots6}_\text{68 digits}+\underbrace{444\cdots4}_\text{19 digits}$$

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$\underbrace{888\cdots8}_\text{19 digits}\not=2\times\underbrace{444\cdots4}_\text{19 digits}$. Why did you write it???/ – bigli Jul 17 '14 at 20:46
it most truly does. $4*2=8$, $44*2=88$, $444*2=888$ etc. – cirpis Jul 17 '14 at 20:48
Excuse me. You are right. – bigli Jul 17 '14 at 20:52
No problem, looking at my solution i can tell that i shouldve represented those big numbers using letters for a more elegant and less confusing proof. – cirpis Jul 17 '14 at 20:53
In your answer: How did you get the last equality? – bigli Jul 17 '14 at 20:55

You have
$\underbrace{888\cdots8}_\text{19 digits}\times\underbrace{333\cdots3}_\text{68 digits} = 8\cdot 3 \cdot (\underbrace{111\cdots1}_\text{19 digits}\times \underbrace{111\cdots1}_\text{68 digits}) = 24 \cdot (\underbrace{111\cdots1}_\text{19 digits}\times \underbrace{111\cdots1}_\text{68 digits})$.
Similarly we get
$\underbrace{444\cdots4}_\text{19 digits}\times\underbrace{666\cdots6}_\text{68 digits} = 4\cdot 6 \cdot (\underbrace{111\cdots1}_\text{19 digits}\times \underbrace{111\cdots1}_\text{68 digits}) = 24 \cdot (\underbrace{111\cdots1}_\text{19 digits}\times \underbrace{111\cdots1}_\text{68 digits})$.
So these two numbers are equal.

It is clear that $$\underbrace{444\cdots4}_\text{19 digits}\times\underbrace{666\cdots6}_\text{68 digits} \le \underbrace{444\cdots4}_\text{19 digits}\times\underbrace{666\cdots7}_\text{68 digits}.$$ Since the multiplier is increased by one, the difference is exactly $\underbrace{444\cdots4}_\text{19 digits}$.

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