pepeprepa wrote:Just one remark about Saule.
I think I see where your error is.
You say "Eliminate 3 from both numerator and denominator, and you'll be left with 2 in the denominator."
I think you cannot do that when you want to find a remainder.
Counter-exemple:
15 divided by 6 gives you a remainder of 3
15/6=(3*5)/(3*2)=5/2
5 divide by 2 gives you a remainder of 1
Am I right about this fact?
OK, the remainder should be 1, not 5. When an odd number is divided by 2, the remainder's always 1, my mistake. But the answer is still None of these, since 1 isn't an option either.
9^1 + 9^2 + 9^3 + ...... + 9^9 = 3x3 (9^0 + 9^1 + 9^2 + ...... + 9^8 )
this is our numerator.
the denominator is 6 = 3x2
cancel 3 in the numerator and denominator, and you're left with
3 (9^0 + 9^1 + 9^2 + ...... + 9^8 ) divided by 2.
Since the numerator is odd, the remainder is 1.
