problem about number and power of 2

Asked Feb 02 '12 at 12:55

Active Jun 20 '12 at 20:53

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Possible Duplicate:
Starting digits of 2^n

can anyone give me a hint ?

Prove that any finite sequence of digits is a starting sequence of digits for some power of $2$.

This is my attempt : suppose the random finite sequence is $a_1a_2...a_i = A_i$ (its decimal value) need to prove that for every $A_i$ there exist $b_1b_2...b_i = B_i$ and $n$ such that $A_iB_i = 2^n$ or $A_i\times 10^k + B_i = 2^n$ ( $k$ is number of digits in $B_i$) we need $0 < Bi < 10^k$ the problem becomes, prove that for every $A_i$, there exist $n$ and $k$ such that $Ai\times 10^k < 2^n < A_{i+1}\times 10^k$. I'm still working on it... perhaps give me a hint on how to prove this or maybe there's a better way

edited Apr 13 '17 at 12:20

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asked Feb 02 '12 at 12:55

Geralt of Rivia

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You could use the Weyl Equidistribution Theorem: If $\alpha$ is irrational, then the set ${[n\alpha] : n\in\Bbb N}$ is uniformly distributed in $[0,1]$. – David Mitra Feb 02 '12 at 13:08
sorry i haven't learnt that theorem, the solution will definitely not involve such theorem. Unless, its proof is really simple – Geralt of Rivia Feb 02 '12 at 13:20
oh isee i will look it up – Geralt of Rivia Feb 02 '12 at 13:24

problem about number and power of 2

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