% Given a string 's1' representing a non-negative integer s and a character 'c'
% representing a digit, find s*c 
% Assume the output string is named s3
%
% Algorithm: 
%   convert 'c' to a number c
%   repeat the following, starting with the least significant digit: 
%      convert the ith digit of s1 to a number x, find m = x*c+carry
%      convert the least significant digit of m to a character
%      and insert into the answer as the ith digit.  Save the most 
%      significant digit of m as the carry.  If a carry exists at the
%      and, stick it on the front of s3.
function s3 = multiplyLine(s1, c);

if length(s1) < 1 s3 = '0';           % if s1 == [] answer = 0
else
   carry = 0;                         % define the carry
   s3 = '';                           % define the answer

   % Change c to a number
   c = c-'0';
   
   % From least significant digit, change digit chars to numbers, add with
   % carry to get new carry and new digit, convert new digit back to char
   % and stick into solution.  If a number runs out of digits, use 0.
   for i=1:length(s1)
      m = (s1(length(s1)+1-i)-'0')*c + carry;
      carry = floor(m/10);            % save the carry
      s3 = [char(mod(m,10)+'0') s3];  % compute the answer digit
   end
   
   % If there is a carry at the end, change to char and stick it at front 
   if carry > 0 s3 = [char(carry+'0') s3]; end
end