What is the remainder when $9^3+9^4+9^5+9^6+...+9^{100}$ is divided by 6?

The correct option is 0.

To find the remainder when the given sum is divided by 6, let us analyze the behavior of the terms modulo 6.

First, notice the expression we want to evaluate:

$S=9^3+9^4+9^5+9^6+...+9^{100}$

Let us look at the remainder of $9$ when divided by 6:

$9\equiv 3\text{ (mod 6)}$

Now let us compute higher powers of 9 modulo 6:

$9^2=81\equiv 3\text{ (mod 6)}$

$9^3=9\times 9^2\equiv 3\times 3=9\equiv 3\text{ (mod 6)}$

In general, for any positive integer power $k\ge 1$:

$9^k\equiv 3\text{ (mod 6)}$

Therefore, every term in the sum $9^3+9^4+...+9^{100}$ leaves a remainder of 3 when divided by 6.

Next, let us determine the total number of terms in the sum. The powers range from $3$ to $100$ inclusive:

$\text{Number of terms}=100-3+1=98$

Now, substituting the modular value for each term into the sum $S$:

$S\equiv 3+3+...+3\text{ (98 times)}\text{ (mod 6)}$

$S\equiv 98\times 3\text{ (mod 6)}$

Calculating $98\times 3$:

$98\times 3=294$

Finally, we find the remainder of $294$ when divided by 6:

$294=6\times 49+0$

Since 294 is completely divisible by 6, the remainder is 0.