jmc

Using the definition of base numbers, $100111011_6=6^8+6^5+6^4+6^3+6+1$. Let $x=6$, so the number equals $x^8+x^5+x^4+x^3+x+1$. By using the Rational Root Theorem, $x+1$ is a factor of $x^8+x^5+x^4+x^3+x+1$, so the polynomial factors into $(x+1)(x^7-x^6+x^5+x^3+1)$. The first three terms share a common factor of $x^5$, and the last two terms is a sum of cubes, so the expression can be grouped and factored as $(x+1)(x^5(x^2-x+1)+(x+1)(x^2-x+1)=(x+1)(x^2-x+1)(x^5+x+1)$. To factor the quintic polynomial, add and subtract $x^2$ to get $x^5-x^2+x^2+x+1$. Factoring out $x^2$ in the first two terms results in $x^2(x^3-1)+x^2+x+1=x^2(x-1)(x^2+x+1)+x^2+x+1$, and factoring by grouping results in $(x^2+x+1)(x^3-x^2+1)$. Thus, the polynomial can be factored into $(x+1)(x^2-x+1)(x^2+x+1)(x^3-x^2+1)$, and substituting $x=6$ results in $7\cdot 31\cdot 43\cdot 181$. A prime test shows that $\boxed{181}$ is the largest prime factor of $100111011_6$ in decimal form.