Given: $m1$ and two positive numbers each containing $m$ digits, $a_{m-1}{a}_{m-2}dots{a}_0$

and $b_{m-1}{b}_{m-2}dots{b}_0$

Wanted: $c_m{c}_{m-1}{c}_{m-2}dots{c}_0,$ where

$b_{m-2}dots{b}_0$

Algorithm:

Step $1$ Set the value of carry to $0$

Step $2$ Set the value of $i$ to $0$

Step $3$ While the value of $i$ is less than or equal to $m-1,$ repeat the instruc$-$

tions in Steps $4$ through $6$

Step $4$ Add the two digits $a_i$ and $b_i$ to the current value of carry to get $c_i$

Step $5$ If $c_{i}10,$ then reset $c_i$ to $(c_i-10)$ and reset the value of carry to $1$ ;

otherwise, set the new value of carry to $0$

Step $6$ Add $1$ to $i,$ effectively moving one column to the left

Step $7$ Set $c_m$ to the value of carry

Step $8$ Print out the final answer, $c_m{c}_{m-1}{c}_{m-2}dots{c}_0$

Step $9$ Stop