Problem:
Convert $3321_5$ to base 16.
Answer:
\begin{array}{l}
 
\begin{array}{l}
 \text{Convert the following to base }16: \\
 3321_5 \\
\end{array}
 \\
\hline
 
\begin{array}{l}
 \text{First, }\text{convert }3321_5 \text{to }\text{base }10: \\
 \text{Find }\text{the }\text{base-10 }\text{representation }\text{of }3321_5 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Split }3321_5 \text{into }\text{its }\text{component }\text{digits}: \\
 
\begin{array}{c}
 \text{Base-5 digit} \\
 3 \\
 3 \\
 2 \\
 1 \\
\end{array}
 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Label }\text{each }\text{digit }\text{of }3321_5 \text{with }\text{its }\text{place}: \\
 
\begin{array}{cc}
 \text{Base-5 digit value} & \text{Place} \\
 3 & 5^3 \\
 3 & 5^2 \\
 2 & 5^1 \\
 1 & 5^0 \\
\end{array}
 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Write }\text{the }\text{product }\text{that }\text{each }\text{base-}5 \text{digit }\text{represents }\text{in }\text{base }10: \\
 
\begin{array}{cccc}
 \text{Base-5 digit value} & \text{Place} & \rightarrow  & \text{Base-10 equivalent} \\
 3 & 5^3 & \rightarrow  & 3\ 5^3 \\
 3 & 5^2 & \rightarrow  & 3\ 5^2 \\
 2 & 5^1 & \rightarrow  & 2\ 5^1 \\
 1 & 5^0 & \rightarrow  & 1\ 5^0 \\
\end{array}
 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Compute }\text{the }\text{base-10 }\text{equivalent }\text{of }3321_5: \\
 
\begin{array}{cccccc}
\hline
 \text{Base-5 digit value} & \text{Place} & \rightarrow  & \text{Base-10 equivalent} &   & \text{Value} \\
\hline
 3 & 5^3 & \rightarrow  & 3\ 5^3 &   & 375 \\
\hline
 3 & 5^2 & \rightarrow  & 3\ 5^2 & + & 75 \\
\hline
 2 & 5^1 & \rightarrow  & 2\ 5^1 & + & 10 \\
\hline
 1 & 5^0 & \rightarrow  & 1\ 5^0 & + & 1 \\
\hline
   &   &   &   & = & 461 \\
\hline
\end{array}
 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{number }3321_5 \text{is }\text{equivalent }\text{to }461_{10} \text{in }\text{base }10. \\
 3321_5=461_{10} \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Now }\text{convert }\text{461 }\text{to }\text{base }16: \\
 \text{Now, }\text{find }\text{the }\text{base-16 }\text{representation }\text{of }461 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Determine }\text{the }\text{powers }\text{of }\text{16 }\text{that }\text{will }\text{be }\text{used }\text{as }\text{the }\text{places }\text{of }\text{the }\text{digits }\text{in }\text{the }\text{base-16 }\text{representation }\text{of }461: \\
 
\begin{array}{|c|c|c}
\hline
 \text{Power} & \text{Base}^{\text{Power}} & \text{Place value} \\
\hline
 3 & 16^3 & 4096 \\
\hline
 2 & 16^2 & 256 \\
 1 & 16^1 & 16 \\
 0 & 16^0 & 1 \\
\end{array}
 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Label }\text{each }\text{place }\text{of }\text{the }\text{base-16 }\text{representation }\text{of }\text{461 }\text{with }\text{the }\text{appropriate }\text{power }\text{of }16: \\
 
\begin{array}{ccccccc}
 \text{Place} &   &   & 16^2 & 16^1 & 16^0 &   \\
   &   &   & \downarrow  & \downarrow  & \downarrow  &   \\
 461_{10} & = & ( & \_\_ & \_\_ & \_\_ & )_{\text{}_{16}} \\
\end{array}
 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Determine }\text{the }\text{value }\text{of }\text{the }\text{first }\text{digit }\text{from }\text{the }\text{right }\text{of }\text{461 }\text{in }\text{base }16: \\
 
\begin{array}{l}
 \frac{461}{16}=28 \text{with remainder} \fbox{$13$} \\
 
\begin{array}{ccccccc}
 \text{Place} &   &   & 16^2 & 16^1 & 16^0 &   \\
   &   &   & \downarrow  & \downarrow  & \downarrow  &   \\
 461_{10} & = & ( & \_\_ & \_\_ & 13 & )_{\text{}_{16}} \\
\end{array}
 \\
\end{array}
 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Determine }\text{the }\text{value }\text{of }\text{the }\text{next }\text{digit }\text{from }\text{the }\text{right }\text{of }\text{461 }\text{in }\text{base }16: \\
 
\begin{array}{l}
 \frac{28}{16}=1 \text{with remainder} \fbox{$12$} \\
 
\begin{array}{ccccccc}
 \text{Place} &   &   & 16^2 & 16^1 & 16^0 &   \\
   &   &   & \downarrow  & \downarrow  & \downarrow  &   \\
 461_{10} & = & ( & \_\_ & 12 & 13 & )_{\text{}_{16}} \\
\end{array}
 \\
\end{array}
 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Determine }\text{the }\text{value }\text{of }\text{the }\text{last }\text{remaining }\text{digit }\text{of }\text{461 }\text{in }\text{base }16: \\
 
\begin{array}{l}
 \frac{1}{16}=0 \text{with remainder} \fbox{$1$} \\
 
\begin{array}{ccccccc}
 \text{Place} &   &   & 16^2 & 16^1 & 16^0 &   \\
   &   &   & \downarrow  & \downarrow  & \downarrow  &   \\
 461_{10} & = & ( & 1 & 12 & 13 & )_{\text{}_{16}} \\
\end{array}
 \\
\end{array}
 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{Convert }\text{all }\text{digits }\text{that }\text{are }\text{greater }\text{than }\text{9 }\text{to }\text{their }\text{base }\text{16 }\text{alphabetical }\text{equivalent}: \\
 
\begin{array}{ccc}
 \text{Base-16 digit value} &   & \text{Base-16 digit} \\
 1 & \rightarrow  & 1 \\
 12 & \rightarrow  & \text{c} \\
 13 & \rightarrow  & \text{d} \\
\end{array}
 \\
\end{array}
 \\
 
\begin{array}{l}
 \text{The }\text{number }461_{10}, \text{and }\text{therefore }\text{the }\text{number }3321_5, \text{is }\text{equivalent }\text{to }\text{1cd}_{16} \text{in }\text{base }16. \\
 \fbox{$
\begin{array}{ll}
 \text{Answer:} &  \\
 \text{} & 3321_5=461_{10}=\text{1cd}_{16} \\
\end{array}
$} \\
\end{array}
 \\
\end{array}

{'source': 'AMPS', 'file': 'amps/mathematica/number_theory/convert_base_w_steps/9251.txt'}