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Powers of 2 are all you need

Posted on: July 26th, 2013 by
1

Show that given any combination of digits, one can find a number 2^n that starts with the given combination of digits.


One Response to Powers of 2 are all you need

  1. Dinesh had this to say about that:

    Let the number given be a. If a itself is a power of 2, nothing more to be done. Else, we need to show that there exists some k such that there exists a power of 2 between a*10^k and (a+1)*10^k - 1. Taking log to the base 2, we need to show that there is some k such that there exists an integer between k \log(10) + \log(a) and  \log(10^k*(a+1)-1). The gap between these two numbers is an increasing function of k and so is lower bounded by the value when k=1. If the fractional part of the first number k \log(10) + \log(a) is less than this lower bound, the interval must include an integer. But this follows from the denseness of the fractional part of k \log(10) since \log_2(10) is irrational (the proof of this was part of another recent puzzle post also from Russian Olympiad).


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