Convert it! (Problem 6)

$1020021$ in base 3 converted to base 10 is $1 \times 1 + 3 \times 2 + 9 \times 0 + 27 \times 0 + 81 \times 2 + 243 \times 0 + 729 \times 1 = \boxed{898_{10}}$ .

$898$ in base 10 converted to base 6 is $216 \times 4 + 36 \times 0 + 6 \times 5 + 1 \times 4= \boxed{4054_6}$

$\begin{aligned} 1020021_3 & = \left(1\times 3^6 + 2 \times 3^4 + 2 \times 3^1 + 1 \times 3^0 \right)_{10} \\ & = \left(729+ 162 + 6 + 1 \right)_{10} \\ & = 898_{10} \\ & = 4 \times \underbrace{216}_{=6^3} + \blue{34} & \small \blue{\text{34 = remainder when 898 is divided by }6^3} \\ & = 4 \times 6^3 + 0 \times 6^2 + 5 \times 6^1 + 4 \times 6^0 \\ & = \boxed{4054}_6 \end{aligned}$