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MAXImum [283]
3 years ago
11

Cn someone help me plz

Computers and Technology
1 answer:
3241004551 [841]3 years ago
8 0

Answer:

A

Explanation:

if we put x=7 then the the condition will be true because after that x  which is 7 will be greater than 5

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Answer:

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

Explanation:

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