How can I solve this problem?

I am a bit lost with this Matlab problema that I have to do. Can anyone give me a hand? I would be very grateful!

" Write a function called circular_primes that finds the number of circular prime numbers smaller than n, where n is a positive integer scalar input argument. For example, the number, 197, is a circular prime because all rotations of its digits: 197, 971, and 719, are themselves prime. For instance, there are thirteen such primes below 100: 2, 3, 5, 7, 11, 13, 17, 31, 37, 71, 73, 79, and 97. It is important to emphasize that rotation means circular permutation not all possible permutations "

"2520 is the smallest number that can be divided by each of the numbers from 1 to 10 without any remainder. Write a function called smallest_multiple that returns a uint64, the smallest positive number that is evenly divisible by all of the numbers from 1 to n where n is a positive integer scalar and is the only input argument of the function. If the result would be greater than what can be represented as a uint64, the function returns 0 "

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/8026-best-way-s-to-master-matlab

http://www.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer

Answer 1

[Srishti Saha]

This should work perfectly:

function mul = smallest_multiple(n)
  facts = zeros(1,n);             % store the exponents of various factors
  for ii = 2:n
      f = factor(ii);             % get factors for current integer
      for jj = 2:ii
          k = sum(f == jj);       % what's the exponent of this factor?
          if k > facts(jj)        % if it is greater than what we have so far
              facts(jj) = k;      % update to this new value
          end
      end
  end
  % Compute the result with one command. 
  % The 'native' option tells MATLAB to work in uint64
  mul = prod(uint64((1:n).^facts),'native');   
  if mul == intmax('uint64')
     mul = uint64(0);
  end
end

Answer 2

[RAMAKANT SHAKYA]

function out=circular_primes(no)
prim=primes(no);% find the all prime number till the given number
pr=0;
nos=[];
po=[];
for p=1:length(prim)
    n=prim(p); % picking up each prime no one by one
    n=num2str(n);% change into string for rotation of no
    d=length(n); % length of string
    if d>1          % take nos greater than 10 beacuase below 10 no need for rotation
        for h=1:d
            a=n(1);
            for r=1:d % for rotation right to left

                if r==d  5 % for the last element of string
                    n(d)=a;
                else
                n(r)=n(r+1); %shifting
                end

            end
            
        s=str2num(n); % string to number
        nos=[nos,s]; % store rotated elements in array
        end
        if nos(end)==no   %if given no is also a circular prime we need smaller
            break;
        end
        for gr=1:length(nos) % checking rotated nos are prime or not
            p1=isprime(nos(gr));
            po=[po,p1];    %storing logical result in array
        end
            if sum(po(:))==length(nos) %if all are prime the length and sum are must be equal
                
                pr=pr+1;
                out=pr;
            else
                out=pr;
            end
        po=[];
        nos=[];
    else 
        s=str2num(n); %numbers less than 10
        f=isprime(s);
        if f==1
            pr=pr+1;
            out=pr;
        else
            out=pr;
        end
    end
        
       end
end