Anty solutions written

$(100)_{6} = (1\times6^{2}+0\times6^{1}+0\times6^{0}) = (1\times36+0+0) = \boxed{36}$

Haha, this was Level 5 (rating 2900) for a while since the answer was typo'ed :P

Anyway, in decimal notation, $100_6$ is written as $1 \cdot 6^2 + 0 \cdot 6 + 0 \cdot 1 = \boxed{36}.$

$1 \times 6^2 + 0 \times 6^1 + 0 \times 6^0$

Using the Number Base Representation as a guide, we know that to convert to our decimal system we can do

$(100)_{6}=(1\times 6^{2})+(0\times 6^{1})+(0\times 6^{0})=36$

The answer is $\boxed{36}.$

(100)_6 is equivalent to, in base 10,

$(1 \times 6^2) + (0 \times 6^1) + (0 \times 6^0$ )

$= (1 \times 36) + 0 + 0$

$= \boxed{36}$

1 6^2+0 6^1+0*6^0 =36