Answer / makthar basha

Write a program that takes a 3 digit number n and finds out
whether
the number 2^n + 1 is prime, or if it is not prime find out
its
factors.

Answer / sreenu

please send source code for for a program that takes a 3
digit number n and finds out
whether
the number 2^n + 1 is prime, or if it is not prime find out
its
factors in c.

Answer / salma

i also face sthe same question please send me the full code

Answer / sreejesh1987

//Sorry,this code is wrong.
//I'm not removing this,because it may serve some other purpose.

#include<stdio.h>
#include<conio.h>
void main()
{
int i,j,n,x,flag=0,d;
clrscr();
printf("\tPRIME CHECK\n");
printf("Enter a 3digit no:");
scanf("%d",&n);

if((n/100>0)&&(n/100<10))
printf("\nNumber is of 3 Digits");
else
printf("\nNot a 3 digit number");

for(j=2;j<n,n%j!=0;j++)
if(n-1==j)
{
printf("\n\t%d is a prime",n);

flag=1;

d=n-1;
while(d%2==0)
d=d/2;

if(d==1)
printf("\nNumber%d is in 2^n+1 format",n);
else
printf("\nNumber%d is not in 2^n+1 format",n);
}

if(!flag)
{
printf("\nNumber %d is not a prime",n);
printf("\nIts factors are:\n\t");

x=2;
while(n!=1)
{
while(n%x==0)
{
n=n/x;
printf("%d ",x);
}
x++;
}
}
getch();
}

Answer / amit

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
int n, i, m=0, flag=0;
cout << "Enter the power of 2 ";
cin >> n;
m=pow(2,n)+1;
for(i = 2; i <= m; i++)
{
if(m % i == 0)
{
cout<<"Number is not Prime."<<endl;
flag=1;
break;
}
}
if (flag==0)
cout << "Number is Prime."<<endl;
return 0;
}

main(){int a=5,b 10,c=2, d;a=b c;d=++a=(--c)*2; printf("%d%d%d%d,a,b,c,d; return o;}

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