I could write the program quickly, but I want to leave you part of the fun.
Just write a loop for the numbers from 2-100 (1 is not considered a Prime number). Write a second loop to test divisibility of each number, "n", with all factors lower than that number (from 2 to n-1). If such a factor exists, the number is NOT a prime number.
To test for divisibility, use the remainder of a division.
if (n % factor == 0)
// If true, it is divisible.
import java.io.*;
class Fun
{
public void even(int m,int n) //prints even numbers
{
for(int i=(m+1);i<n;i++)
{
if(i%2==0)
System.out.println(i);
System.out.println("\n");
}
}
public void prime(int x,int y ) //prints prime numbers
{
for(int j=(x+1);j<y;j++)
{
int f=0;
for(int k=2;k<j/2;k++)
{
if(j%k==0)
{
f=1;
break;
}
}
if(f==0)
{ System.out.println(j);
System.out.println("\n");
}
class exe
{
public static void main (String[] args)
{
Fun h=new Fun();
h.even(100,200);
h.prime(100,200);
}
}
public class PerfectNumber{
public static void main(String[]args){
int sum=0, x=0;
for(int num=1;num<500;num++){
for(int i=1;i x=num%i; if(x==0) sum=sum+i; } if(sum==num){ System.out.println("Perfect Number is: "+num); System.out.println("Factors are: "); for(int i=1;i x=num%i; if(x==0) System.out.println(i); } } sum=0; } } }
package javaapplication1;
public class JavaApplication1 {
public static boolean is_prime(int val){
if (val<0) val*=-1;
if (val<2) return false;
if (val%2==0) return val==2;
int max = (int)Math.sqrt(val)+1;
for (int div=3; div<max; ++div)
if (val%div==0)
return false;
return true;
}
public static void main(String[] args) {
System.out.print ("List of primes from 0 to 100:\n");
for (int i=0; i<=100; ++i){
if (is_prime(i)) {
System.out.print (i);
System.out.print (" is prime\n");
}
}
}
}
package javaapplication1;
public class JavaApplication1 {
public static boolean is_prime(int val){
if (val<0) val*=-1; // make val positive
if (val<2) return false;
if (val%2==0) return val==2;
int max = (int)Math.sqrt(val)+1;
for (int div=3; div<max; ++div)
if (val%div==0)
return false;
return true;
}
public static void main(String[] args) {
System.out.print ("List of primes from 0 to 100:\n");
for (int i=0; i<=100; ++i){
if (is_prime(i)) {
System.out.print (i);
System.out.print (" is prime\n");
}
}
}
}
look man that would take alot >>> Ill give you the way =============================================================================== this answer i write it now on net ! i read about that befor in it uni where i study i will chick out the answer and re write it best wishes 2024
PRINT 2,3,5,7,11,13,17,19,23,29,31,37
Here is a simple program to generate prime numbers upto a given specific number /*Prime No. from 1 to 50*/ /*By-Himanshu Rathee*/ #include<stdio.h> #include<conio.h> void main() { int i,j,n; clrscr(); printf(" Enter the number upto which we have to find the prime number: "); scanf("%d",&n); printf("\n"); for(i=2;i<=n;i++) { for(j=2;j<=i-1;j++) if(i%j==0) break; /*Number is divisble by some other number. So break out*/ if(i==j) printf("\t%d",i); /*Number was divisible by itself (that is, i was same as j)*/ } /*Continue loop upto nth number*/ getch(); }
(defun prime (num) (if (< 2 num) (do ((dividend 2 (1 + dividend)) (chk-to (sqrt num))) ((equal (rem num dividend) 0)) (when (<= chk-to dividend) (return t))) t))
Loop through some numbers - for example, 2 through 100 - and check each one whether it is a prime number (write a second loop to test whether it is divisible by any number between 2 and the number minus 1). If, in this second loop, you find a factor that is greater than 1 and less than the number, it is not a prime, and you can print it out.
VBnet program to find the prime numbers between 100 to 200?
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Prime numbers are numbers that are only divisible by themselves and the number 1. You can write a program to print all prime numbers from 1 to 100 in FoxPro.
Since there is an infinite set of prime numbers the answer would be infinity.
By learning how to program on C+.
Write a function that implements an algorithm that checks to see if a particular integer is prime (returning a boolean). Write a program that uses that function on each number from 1 to 100, and if true, displays that number.
This would require some computer knowledge. It can make it easier to find out the prime numbers without figuring it out in your head.
To get all tutorials of "c programming" Reference:cprogramming-bd.com/c_page2.aspx# prime number
A number as a product of prime numbers would be "x".
Numbers divisible by 1 & number itself are called prime numbers. These numbers also have the property to be odd numbers.
You can't write that as the sum of two prime numbers. Note: Goldbach's Conjecture (for expressing numbers as the sum of two prime numbers) applies to EVEN numbers.
Prime factorization of 27 = 3x3x3