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Find $\overbrace{999\dots 9}^{2^{n+2}}\over 99$. Would you help me to find out exact answer?

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  • $\begingroup$ What about $n$? Is it $n \geq 0$? $\endgroup$ – Alex Silva Dec 30 '14 at 11:54
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    $\begingroup$ Try some induction. It seems to be $$\;\overbrace{101010....101}^{2^{n+2}-1\;\text{digits}}$$ $\endgroup$ – Timbuc Dec 30 '14 at 11:54
  • $\begingroup$ Seems like, $\dfrac{10^{2^{\large{n+2}}}-1}{99}$ $\endgroup$ – Mann Dec 30 '14 at 12:18
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$$\underbrace{99...9}_{k\text{ -digits}}=10^{k}-1;\quad99=10^2-1$$ So you wish to seek: $$\frac{10^{2^{n+2}}-1}{10^2-1}=\sum_{k=0}^{2^{n+1}-1}10^{2k}=\underbrace{101010...101}_{2^{n+2}-1\quad \text{digits}}$$

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Let's denote $a_n = \underbrace{999\dots 9}_{n}, b = 99$. You have to find $q$ such that $a_{n+2} = qb$. Let's note that $$ a_n +1 = 10^{n-2}(b+1) = 10^{n-2}b + 10^{n-2}\\ \Rightarrow a_n = 10^{n-2}b + a_{n-2} $$ Now we can use induction and say that $a_n = (10^{n-2}+10^{n-4}+\dots)b$.

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Note that $$\overbrace{999\dots 9}^{k}=9(1+10+10^2+...+10^k)=10^{k+1}-1.$$ Also $$\overbrace{999\dots 9}^{2^k}=99\times10^{2^k-2}+99\times10^{2^k-4}+99\times10^{2^k-6}+...+99$$

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