Is the sum of digits of $3^{1000}$ divisible by $7$? 
Is the sum of digits of $3^{1000}$ a multiple of $7$?

The sum of the digits of $3^{1000}$ can be computed using a computer. It is equal to $2142$, so the answer is positive.
Is there a short proof that the sum of the digits of $3^{1000}$ is a multiple of $7$ without using a computer?
Do you have any advice to solve this type of problem (without programming of course!)?
The results below are known:


*

*$3^{1000}$ has $478$ digits, and so the sum is at most $4302$ ($9\cdot478$).

*This sum is a multiple of $9$.

*The last four digits of $3^{1000}$ are $0001$.

Context: We are a group of 3 French people working on it since 2007. It's a little exercise I found in my high school book (printed in 2007) which is pretty complicated. The one who created this exercise doesn't know the answer.
This question was previously asked on Math.SE (link).
 A: Not an answer, but a series of considerations.
One expects not only the digit sum of 3^n to be a multiple of nine (for integral n greater than 1) but also for the string of digits (in the decimal representation of 3^n) to be somewhat normal in distribution, having roughly the same number of occurrences of each decimal digit.  For the given example, the actual digit sum is not far from the expected sum of 2151.   
I have not observed the growth of the digit sum of powers of 3, but it should grow linearly with n, subject of course to being a multiple of 9 and not deviating far from the expected value.  (Using the posted example for interpolation, I expect the rate of growth of digit sum to average about 2.1 for every increment of n, or to increase by 9 about every 4 steps of n.) Because of this, I would expect 1/7th of the exponents n to yield a multiple of 7 for the digit sum of 3^n, and to occur in runs (or near runs) of length about 4. (So the digit sum may be a multiple of 7 for n=998 or n=1002 as well.) Indeed, if it weren't for the variation, I would expect the digit sums to be multiples of 7 near n=1006.
Gerhard "Not Ready For A Summary" Paseman, 2017.09.26.
A: Middle digits of the numbers $3^n$ are unpredictable. At least it is too hard for current techniques to say anything about them. It means that the their sum  is unpredictable as well. Some good random number generators are based "digital" ideas. 
If we take binary digits of $3^n$ then we immediately get generalization 
 of $(3/2)^n$-problem which is out of reach today.
This picture is taken from a New Kind of Science:
The pattern is very similar to "rule 30" picture from the same book:
It is expected to have very good pseudorandom properties, see discussion at A New Kind of Science: A 15-Year View.
