To write a Java program to check if a number is a twisted prime or not, you first need to create a function that checks if the number is prime. You can do this by iterating from 2 to the square root of the number and checking if the number is divisible by any of these values. Once you have verified that the number is prime, you can then check if the number remains prime after twisting its digits (reversing the number and checking if the reversed number is also prime). If both conditions are met, then the number is a twisted prime.
121 is not a prime number because it has more factors than 1 and itself.
for s being prime find all s so that s² < 50 { 4 , 25 , 49...and one more }
Not everyone is agreed on this but Peano's axioms, which are the basis for the axiomatic structure of numbers defines 0 as a natural number and then all other natural numbers in terms of successors.0 is not a successor of any natural number.1 is the successor of 0 is 1 = S(0)2 = S(S(0))3 = S(S(S(0))) and so on.Accordingly, the first natural number is 0.
0,1,3,5,7,
No numbers, because 2 is the smallest prime number 0 and 1 aren't prime #'s.
9 0's so50,000,000,000
composite. prime #s can only be divided by 1 and itself
To write a Java program to check if a number is a twisted prime or not, you first need to create a function that checks if the number is prime. You can do this by iterating from 2 to the square root of the number and checking if the number is divisible by any of these values. Once you have verified that the number is prime, you can then check if the number remains prime after twisting its digits (reversing the number and checking if the reversed number is also prime). If both conditions are met, then the number is a twisted prime.
A prime number has only 2 factors which are 1 and itself. Composite numbers are everything else except 1 and 0. 1 and 0 are neither prime, nor composite.
Begin Read num for(i=2; i<num; i++) if(num%2==0) then print "the number is not a prime no."; else if print "the number is prime"; end if Stop
121 is not a prime number because it has more factors than 1 and itself.
for s being prime find all s so that s² < 50 { 4 , 25 , 49...and one more }
#include <iostream> void main() { using namespace std; int num, rem, sum=0; //Declaring variables cout<<"Enter a number :"<<endl; cin>>num; //Loop to calculate the sum of the digits of the given number. while(num!=0) { rem=num%10; num=num/10; sum=sum+rem; } cout<<"Sum of the digits is "<<sum<<endl; cin.get(); }
2 x 5 x 5 = 50
Prime 9 - 2009 Players of the 50's 4-2 was released on: USA: 2011
/** * @author BHARGAV MODI * */ class prime { public static void main(String p[]) { int i,j,s=0,d=0; for(i=1;i<100;i++) { s=0; for(j=1;j<=i;j++) { if(i%j==0) s=s+1; } if(s==2) { System.out.println(i); d=d+1; } } System.out.println("the no. of prime no. are = "+d); } }