Prove that the decimal expansion of (sqrt(35) + 6)^1995 starts with at least 1000 nines after the decimal...

arithmetic/digits/near.integer New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

Prove that the decimal expansion of (sqrt(35) + 6)^1995 starts with at
least 1000 nines after the decimal point.

There must be other digits following somewhere far away, since
sqrt(35) + 6 is clearly no integer.

What about (sqrt(37) + 6)^1995 and (sqrt(37) + 6)^1994?

[ Editor: posted by Norbert Petsch <petsch@mi.uni-erlangen.de>, who
  added: "It is more a mathematical question than a puzzle, but I think
  it can be done with school math.  I don't know if this is the number
  closest to the next integer, but I don't think so." ]

Show solution

HINT: Think of (6 - sqrt(35))^1995.

BIGGER HINT (in fact SOLUTION): Look at

     (sqrt(35) + 6)^1995 + (6 - sqrt(35))^1995.

-- Norbert Petsch <petsch@mi.uni-erlangen.de>

[ Editor: no fuller solution, and no answer for sqrt(37), was found in
  the collected postings. ]