Math Problem Solver

$$59 + 39=98$$

$$\begin{aligned} \overset{\mathtt{1}}{\mathtt{5}}\mathtt{9}\\ \mathtt{+\phantom{0}}\mathtt{3}\mathtt{9}\\ \hline \mathtt{9}\mathtt{8} \end{aligned}$$

$8$ is the digit in the $10^{0}$ place. $9 \times 10^{0} + 9 \times 10^{0} = 18 \times 10^{0} = 1 \times 10^{1} + 8 \times 10^{0}$.$1 \times 10^{1}$ carries to the $10^{1}$ place.

$9$ is the digit in the $10^{1}$ place. $1 \times 10^{1} + 5 \times 10^{1} + 3 \times 10^{1} = 9 \times 10^{1}$.