Answer
\[{{250}_{\text{six}}}\].
Work Step by Step
To multiply numerals of same bases other than base ten,multiply unit digits first according to base 10 then convert them to their respective base. Then continue the same process for other digits.Solve the provided numerals as follows:
\[\begin{align}
& {{4}_{\text{six}}}\,\times {{3}_{\text{six}}}\,={{12}_{\text{ten}}} \\
& \,={{\left( 2\times 6 \right)}_{{}}}+\left( 0\times 1 \right) \\
& ={{20}_{\text{six}}}
\end{align}\]
Base ten product of \[4\times 3=12\]which is larger than base six. So, can be written as \[2\] times six and \[0\]times one.
\[\begin{align}
& \underline{\begin{align}
& \overset{2}{\mathop{5}}\,{{4}_{\text{six}}} \\
& \times {{3}_{\text{six}}}
\end{align}} \\
& \text{ }0 \\
\end{align}\]
Now,
\[\begin{align}
& {{5}_{\text{six}}}\,\times {{3}_{\text{six}}}\,+{{2}_{\text{six}}}\,={{17}_{\text{ten}}} \\
& \,={{\left( 2\times 6 \right)}_{{}}}+\left( 5\times 1 \right) \\
& ={{25}_{\text{six}}}
\end{align}\]
Base ten product of \[5\times 3+2=17\]which is larger than base six.So, can be written as \[2\] times six and \[5\] times one.
\[\begin{align}
& \underline{\begin{align}
& \overset{2}{\mathop{5}}\,{{4}_{\text{six}}} \\
& \times {{3}_{\text{six}}}
\end{align}} \\
& 25{{0}_{\text{six}}}
\end{align}\]
Hence, the result is\[{{250}_{\text{six}}}\].
