Question 1076171
a) I agree with your expressions for the the general term in each of the three sequences.
To show if any number belongs to one of them,
I would just add digits, using module 9 clock arithmetic. The number belongs in the sequence that contains the final sum.
If the ending sum is 0, 3, or 6, the number is divisible by 3, and belongs in R0.
If it is 1, 4, or 7, it belongs in R1; if 2,5, or 8, it belongs in R2.
For {{{1706836}}} ,
the final {{{36}}} part adds up to {{{3+69}}} , and that 9 is dropped.
The {{{1}}} and {{{8}}} add up to {{{1+8=9}}} , and that 9 is dropped.
Then we are left with  {{{7+0+6=13}}} , and since {{{1+3=4}}} , we conclude that
when {{{1706836}}} is divided by {{{9}}} , the remainder is {{{4}}} .
So, {{{1706836=9K+4=3*3K+6-2=3(3K+2)-2}}} for some integer {{{K}}} ,
and that shows that it belongs in R1.
 
HOW AND WHY CLOCK ARITHMETIC MOD 9 WORKS:
The sum of the digits of a number divisible by 9 is 9 or a multiple of 9.
As you are adding,
you drop partial sums that are 9, and keep adding the rest of the digits.
Similarly, for any partial two-digit sum,
you replace the sum of the digits for the two-digit partial sum.
At the end, you should have a one-digit final sum from 0 to 8.
 
b)The numbers in R1 and R2 are {{{1}}} away from a multiple of {{{3}}} .
We could write them all as {{{3p +- 1}}} for some non-negative integer {{{p}}} 
.
(If you - or your teacher - do not like that expression, you could use
{{{3p + 1}}} for terms of R1, with {{{p=n-1}}} ,
and {{{3n - 1}}} for terms of R2,
and write twice as much for your proof).
For {{{x=3p +- 1}}} ,
{{{x^6=(3p +- 1)^6=(3p)^6 +- 6*(1)(3p)^5}}}{{{"+"}}}{{{15(3p)^4(1)^2 +- 20(3p)^3(1)^3}}}{{{"+"}}}{{{15(3p)^2(1)^4 +- 6(3p)(1)^5+1^6}}}
{{{"="}}}{{{(3p)^6 +- 6(3p)^5}}}{{{"+"}}}{{{15(3p)^4 +- 20(3p)^3}}}{{{"+"}}}{{{15(3p)^2 +- 6(3p) }}} {{{"+ 1"}}}
It should be obvious that all terms of that sum are multiples of {{{3}}} ,
except that {{{" + 1"}}} at the end,
and that shows that {{{x^6}}} is a term in R1.
 
If your teacher insisted on seeing the {{{n}}} that gives {{{x^6}}} the right to belong to R1,
you would have to write more, starting with getting out {{{3}}} as a common factor:
{{{x^6 = 3}}}{{{"["}}}{{{p*(3p)^5 +- 2(3p)^5}}}{{{"+"}}}{{{5(3p)^4 +- 20p(3p)^2}}}{{{"+"}}}{{{5(3p)^2(1)^4 +- 2(3p)}}}{{{"]"}}} {{{"+ 1"}}} .