## Thinking Mathematically (6th Edition)

The result is${{1212}_{\text{five}}}$.
To add numerals of same bases other than base ten, add unit digits first according to base 10 then convert them to their respective base. Then continue the same process for other digits. Solve the given numerals as follows: \begin{align} & {{4}_{\text{five}}}+{{3}_{\text{five}}}={{7}_{\text{ten}}} \\ & ={{\left( 1\times 5 \right)}_{\text{five}}}+{{\left( 2\times 1 \right)}_{\text{five}}} \\ & ={{12}_{\text{five}}} \end{align} Base ten sum of $4+3=7$which is larger than base five. So, it can be written as $1$ time five and $2$times one. \begin{align} & \underline{\begin{align} & \overset{{}}{\mathop{\text{ }2}}\,\overset{1}{\mathop{3}}\,{{\overset{{}}{\mathop{4}}\,}_{\text{five}}} \\ & +{{423}_{five}} \\ \end{align}} \\ & \text{ }2 \end{align} Now, solve further as: \begin{align} & {{1}_{\text{five}}}\,+{{3}_{\text{five}}}+{{2}_{\text{five}}}\,={{6}_{\text{ten}}} \\ & \,={{\left( 1\times 5 \right)}_{\text{five}}}+{{\left( 1\times 1 \right)}_{\text{five}}} \\ & ={{11}_{\text{five}}} \end{align} Base ten sum of $1+3+2=6$which is larger than base five.So, it can be written as $1$ time five and $1$ time one. \begin{align} & \underline{\begin{align} & {{\overset{\text{ }1}{\mathop{\text{ 2}34}}\,}_{\text{five}}} \\ & +{{423}_{five}} \end{align}} \\ & \text{ }12 \end{align} Again, \begin{align} & {{1}_{\text{five}}}\,+{{2}_{\text{five}}}+{{4}_{\text{five}}}\,={{7}_{\text{ten}}} \\ & \,={{\left( 1\times 5 \right)}_{\text{five}}}+{{\left( 2\times 1 \right)}_{\text{five}}} \\ & ={{12}_{\text{five}}} \end{align} Base ten sum of $1+2+4=7$which is larger than base five. So, can be written as $1$ time five and $2$times one. \begin{align} & \underline{\begin{align} & \,\overset{1}{\mathop{\,\,2}}\,\overset{1}{\mathop{3}}\,{{\overset{{}}{\mathop{4}}\,}_{\text{five}}} \\ & +{{423}_{five}} \end{align}} \\ & \,\,\,1\,\,2\,1\,2 \end{align} Hence, the result is${{1212}_{\text{five}}}$.