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Evil powers (Posted on 2006-08-31) Difficulty: 4 of 5
Is there a power of 666 such that its decimal notation starts with the digits 123456789?

One may replace "starts" by "ends" for a much easier challenge ;-)

  Submitted by JLo    
Rating: 4.0000 (3 votes)
Solution: (Hide)
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.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Puzzle AnswerK Sengupta2022-06-30 23:08:44
re(2): WowJLo2006-09-04 17:25:35
re: WowDej Mar2006-09-02 22:47:53
WowJLo2006-09-01 13:16:43
Solutiona solutionDej Mar2006-09-01 00:15:17
re: solutionRichard2006-08-31 15:05:06
SolutionsolutionCharlie2006-08-31 13:51:05
SolutionAnswersFederico Kereki2006-08-31 13:29:26
Some ThoughtsHaving it both ways -- Possible spoilerSteve Herman2006-08-31 12:59:12
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