number system

Hi,
The remainder is 5, hence the answer is E, none of these.

Let me explain my logic.

9^1 + 9^2 + 9^3 + ...... + 9^9 = 9 (9^0 + 9^1 + 9^2 +.....+ 9^8 )
Divide this by 6.
Eliminate 3 from both numerator and denominator, and you'll be left with 2 in the denominator.
When you divide a number by 2 you can only have 0 in the remainder if the number is even or 5 if it's odd.
So is 3 (9^0 + 9^1 + 9^2 +.....+ 9^8 ) even or odd?
It's odd, because 9 to any power is always odd and you have it nine times in the brackets, so odd * 9 = odd, and then 3 * odd = odd.
So the remainder is 5.

My answer is 3.

let me know its right?

Ian Stewart wrote:

nikhilagrawal wrote:What is the reminder when 9^1 + 9^2 + 9^3 + ...... + 9^9 is divided by 6?

So the sum is an odd multiple of 3, and the remainder will be 3 when you divide by 6.

I have to disagree that the remainder is 3 when you divide an odd multiple of 3 by 6.
Divide for example 9 by 6, you get 1.5

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?

Sal

"I have to disagree that the remainder is 3 when you divide an odd multiple of 3 by 6"
That's wrong.

9 divided by 6 gives you a remainder of 3
81 (which is 3*27) divided by 6 gives you a remainder of 3

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.

But Saule what do you think about my counter-example?
I mean when you reduce numerator and denominator by the same factor, the remainder is not the same for both equations.

You're right! I just realized when you reduce a fraction by some number, the reminder is also reduced by the same number.
So in end we need to multiply 1 by 3 and get 3.

Thank you, pepeprepa!