Given a number x, find the smallest prime number which is greater than x.

3 Answers on this Question

1. # Python3 implementation of the approach
   import math

   # Function that returns True if n
   # is prime else returns False
   def isPrime(n):

   # Corner cases
   if(n <= 1):
   return False
   if(n <= 3):
   return True

   # This is checked so that we can skip
   # middle five numbers in below loop
   if(n % 2 == 0 or n % 3 == 0):
   return False

   for i in range(5,int(math.sqrt(n) + 1), 6):
   if(n % i == 0 or n % (i + 2) == 0):
   return False

   return True

   # Function to return the smallest
   # prime number greater than N
   def nextPrime(N):

   # Base case
   if (N <= 1):
   return 2

   prime = N
   found = False

   # Loop continuously until isPrime returns
   # True for a number lower than n
   while(not found):
   prime = prime – 1 # if i use prime =prime+1 will get greater than n number.

   if(isPrime(prime) == True):
   found = True

   return prime

   # Driver code
   N = 20
   print(nextPrime(N))

   # This code is contributed by ajay

   #the output will be 19

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2. import java.util.*;
   public class Main

   {
   public static boolean isprime(int n)
   {
   int i,j=0;
   for(i=1;i<=n/2;i++)
   {
   if(n%i==0)
   {
   j++;

   }
   }
   if(j==1)
   {
   return true;

   }
   else
   {
   return false;
   }

   }
   public static void isprim(int x)
   {
   int n=x+1;
   while(true){
   if(isprime(n))
   {
   break;
   }
   n++;
   }
   System.out.println(n);
   }
   public static void main(String[] args) {
   Scanner sc =new Scanner(System.in);
   System.out.println("enter number x");
   int x=sc.nextInt();
   isprim(x);
   }
   }

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3. x=153
   y=x*2
   arr=[]
   for(i=x+1; i<y;i++){
   temp=true
   for(j=2; j<i;j++){

   if(i%j==0){
   temp=false
   }

   }
   if(temp==true){
   arr.push(i)
   }
   }
   console.log(arr[0]);

   Reply