Answer
$110_{\text{two}}$
Work Step by Step
The place values of base two are: $,,,, 2^3, 2^2, 2^1, 1$.
The place value that are less then $6$ are $2^2, 2^1$ and $1$
Divide $6$ by $2^2$ or $4$ to obtain:
$6 \div 4 = 1$ remainder $2$
Divide the remainder $2$ by $2^1$ or $2$ to obtain:
$2 \div 2 = 1$ remainder $0$
Divide $0$ by $1$ to obtain
$0 \div 1= 0$
Thus,
$6=(1 \times 2^2) + (1 \times 2) + (0 \times 1)
\\5=110_{\text{two}}$
