Answer
The quotient of the provided expression is, \[{{21}_{\text{five}}}\].
Work Step by Step
Consider the expression, \[{{4}_{\text{five}}}\overline{\left){{{134}_{\text{five}}}}\right.}\]
Start to divide the expression same as base ten but in this take base as five.
So, starts divide\[13\]by\[4\],
Use the multiplication table that is given above, find the column represented by \[2\] and the row represented by \[4\]that is less than or equal to\[13\]. As \[{{4}_{\text{five}}}\times {{2}_{\text{five}}}={{13}_{\text{five}}}\] from the table, then it is purely divisible not getting any remainder, further its quotient is \[{{2}_{\text{five}}}\]; the division is represented below:
\[{{4}_{\text{five}}}\overset{2}{\overline{\left){\begin{align}
& \ \ {{134}_{\text{five}}} \\
& \underline{-13} \\
& \ \ \ 0 \\
\end{align}}\right.}}\]
Take down the further digit in the dividend, that is\[{{4}_{\text{five}}}\]
\[{{4}_{\text{five}}}\overset{2}{\overline{\left){\begin{align}
& \ \ {{134}_{\text{five}}} \\
& \underline{-13} \\
& \ \ \ \ 04 \\
\end{align}}\right.}}\]
Then, again use the table and find the column represented by \[1\] and row represented by \[4\]that is less than or equal to\[4\]. As \[{{4}_{\text{five}}}\times {{1}_{\text{five}}}={{4}_{\text{five}}}\]from the table, then it is purely divisible not getting any remainder, further its quotient is \[{{1}_{\text{five}}}\]; the division is represented below:
\[{{4}_{\text{five}}}\overset{{{21}_{\text{five}}}}{\overline{\left){\begin{align}
& \ \ {{134}_{\text{five}}} \\
& \underline{-13} \\
& \ \ \ \ 04 \\
& \underline{\ -04} \\
& \ \ \ \ \ 0 \\
\end{align}}\right.}}\]
So, \[{{21}_{\text{five}}}\] is the quotient of the \[{{4}_{\text{five}}}\overline{\left){{{134}_{\text{five}}}}\right.}\]
Hence, the quotient of the provided expression is, \[{{21}_{\text{five}}}\].