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Yes, there is such an integer power of 666. With some impressive programming Charlie even managed to calculate the power. This is something I haven't anticipated!
Below proof for a more general statement: For every sequence of digits and every integer p which is not a power of 10, there is a power of p starting with this exact sequence.
The proof follows Frederico's posting, only a little more detailed:
1. Statement: log(p) is irrational, where log is the logarithm with base 10
Otherwise log(p)=g/h with co-prime g and h. Then 10(g/h)=p, then 10g=ph. This is only possible if p is a power of 10.
2. Statement: Given any non-empty interval ]a,b[ within the unit interval [0,1] and given an irrational number x, then there are infinitely many integers n such that nx-[nx] lies in ]a,b[. (Here [x] denotes the largest integer smaller or equal than x) This is known as Kronecker's theorem as I learned from Frederico's post.
Proof: Can be found in number theory books. It is however not hard to prove and requires only elementary maths.
Now let the starting digits be represented by a fraction d, where 1<=d<10 (In our case we had d=1.23456789) So we are looking for k and n such that
d 10k <= pn < (d+1) 10k
<=>
log(d) + k <= n log(p) < log(d+1) + k
<=>
log(d) <= n log(p) - k < log(d+1)
Now we choose k:=[n log(p)] and get
log(d) <= n log(p) - [n log(p)] < log(d+1)
The interval [log(d),log(d+1)] fits the requirement of the Kronecker Theorem, therefore the n we are looking for must exist.
The easier "challenge" is of course much easier to answer: All powers of 666 end with 6 and can therefore not end with 123456789. |
