Math Problem Solver

$$3591 + 266=3857$$

$$\begin{aligned} \mathtt{3}\overset{\mathtt{1}}{\mathtt{5}}\mathtt{9}\mathtt{1}\\ \mathtt{+\phantom{0}}\phantom{0}\mathtt{2}\mathtt{6}\mathtt{6}\\ \hline \mathtt{3}\mathtt{8}\mathtt{5}\mathtt{7} \end{aligned}$$

$7$ is the digit in the $10^{0}$ place. $1 \times 10^{0} + 6 \times 10^{0} = 7 \times 10^{0}$.

$5$ is the digit in the $10^{1}$ place. $9 \times 10^{1} + 6 \times 10^{1} = 15 \times 10^{1} = 1 \times 10^{2} + 5 \times 10^{1}$.$1 \times 10^{2}$ carries to the $10^{2}$ place.

$8$ is the digit in the $10^{2}$ place. $1 \times 10^{2} + 5 \times 10^{2} + 2 \times 10^{2} = 8 \times 10^{2}$.

$3$ is the digit in the $10^{3}$ place. $3 \times 10^{3} + 0 \times 10^{3} = 3 \times 10^{3}$.