Is sum of digits of $3^{1000}$ divisible by $7$? I'm working on a little exercise I found in my high school book (printed in 2007) which is pretty complicated.
Is the sum of digits of $3^{1000}$ a multiple of $7$?
Do you have any advice to solve this type of problem (without programming of course!)
PS :
We are a group of 3 french people working on it since 2007.
The sum calculated with Python is 2142, this number is a multiple of 7 BUT we want a mathematical answer.
All the results below are mathematically proved !!
$3^{1000}$ has 478 digits and the sum of digits of $3^{1000}$ can't be superior to 4302 (9*478).
This sum is a multiple of 3 and 9.
The last digits of $3^{1000}$ are 0001 (math proof not a result of a computer calculation).
The one who created this exercise doesn't know the answer.
Please help us with any clue!

Cross-posted at https://mathoverflow.net/q/282035/22954 on MathOverflow.
 A: I am not answering the question but the post asks for clues so here it is a couple of ideas.
If $a_0,a_1,a_2, \cdots , a_{477} $ are the decimal digits of $3^{1000}$ then the numbers 
$$b_i=a_{6i}+a_{6i+1}\cdot 10+a_{6i+2}\cdot 10^2+a_{6i+3}\cdot 10^3+a_{6i+4}\cdot 10^4+a_{6i+5}\cdot 10^5$$
for $i=0,1,2,\cdots, 79$ are the digits of $3^{1000}$ in base $10^6$  ($80$ is the nearest integer above $477 \div 6$ so there are $80$ digits numbered $0,1,2,\cdots,79$)
In other words:
$$ 3^{1000} = b_0 + b_1 \cdot 10^6+ \cdots + b_{79} \cdot (10^6)^{79}$$
Now, if we resort to modular arithmetic we see that 
$$ 3^0=1, 3^1=3 , 3^2=2, 3^3=6, 3^4=4, 3^5=5, 3^6=1 $$ (all the equalities taken modulo $7$).
Also $$3^{1000}=(3^6)^{166}\cdot 3^4= 1 \cdot 4 = 4$$(all the equalities taken modulo $7$).
Now if we note that $10^6=1$ (modulo 7) the expression of $3^{1000}$ in base $10^6$ reads (modulo 7)
$$ 4=b_0+b_1+ \cdots +b_{79}$$
So we can assert that the sum of digits of $3^{1000}$ in base one milion gives a remainder of $4$ when divided by $7$.
Another partial result comes from the decimal expansion read modulo 7:
$$3^{1000}= a_0+ a_1 \cdot 10 + \cdots +a_{477} \cdot 10^{477} =1 = a_0+3 a_1+ 2a_2 + 6 a_3 + 4 a_4 + 5 a_ 5 + \cdots $$
So, given that $a_0=1$ we can say that this particular linear combination of the remaining digits is divisible by $7$.
