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NS2 Installation in Ubuntu 22.04

C Program 4–To check whether a given number is prime or not

A prime number can be divided by 1 and itself, there are no other divisors,

Examples are : 2 3, 5, 7, 11, …..

To find out whether a given number is prime or not, here is the logic

1. Get the number

2. divide the given number from 2 to n-1 (Example if 6 is the number divided by 2,3,4,5 will get the remainder respectively 0,0,2,3)

3. increment a counter to 1 if the remainder is 0

4. if there counter variable is 0, then the given number is prime (because we didn’t get any remainder) else non prime

Here is the program

#include <stdio.h>
#include <conio.h>

int main()
{
    int a,i,count=0;
    printf("enter a"); //Let the given number is a
    scanf("%d",&a); //get the number
    for(i=2;i<a;i++) //divide the number a from 2 to a-1
    {
    if(a%i==0) 
    count++; //increment a counter if the divisibility is 0
    }
    if(count !=0) //if the counter is not zero, then prime
    printf("a is not a prime number");
    else
    printf("a is a prime number");
getch();
return 0;
}

Comments

  1. Sir ,
    for ( i=2 ; i < math.pow(a,0.5) ; i++ ) will satisfy . Its enough if we iterate till the square root of the number .

    ReplyDelete
  2. problem s=1+12+123+1234+............+123....n terms

    ReplyDelete
  3. s=(1+2)/(1x2)+(1+2+3)/(1x2x3+)+.......+(1+2+3+......n terms)/(1x2x3x..........n terms)

    please help in this series

    ReplyDelete
  4. s=1x2x3x4+2x3x4x5+3x4x5x6+4x5x6x7+5x6x7x8

    ReplyDelete

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