I heard from a video that there is a theorem that proves that exists a power of two wghose leading or last digits can be any sequence of digits for example your telephone number. Can anyone point me to this theorem please?
Asked
Active
Viewed 115 times
1
-
1Not the last digits! For instance, the last digit must always be $2,4,6$, or $8$. – TonyK Mar 08 '14 at 20:14
1 Answers
1
You can stipulate the leading decimal digits of a number $n$ by setting limits on the fractional part of $\log_{10}n$. For instance, the leading digits of $n$ are $142857$ if
$\lfloor\log_{10}142857\rfloor \le \lfloor\log_{10}n\rfloor < \lfloor\log_{10}142858\rfloor$
To find an integer $k$ such that the leading digits of $2^k$ are $142857$, we want
$\lfloor\log_{10}142857\rfloor \le \lfloor k\log_{10}2\rfloor < \lfloor\log_{10}142858\rfloor$
$\log_{10}2$ is irrational (see e.g. this StackExchange answer), so such a $k$ always exists $-$ see this Wikipedia article.
