write a progam to display the factors of a given number and disply how many pr

write a progam to display the factors of a given number and
disply how many prime numbers are there?

Question Posted / sowmya

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write a progam to display the factors of a given number and disply how many prime numbers are ther..

Answer / sreejesh1987

//This code displays prime factors only

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

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

flag=1;
}

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);
count++;
}
x++;
}
printf("\nNumber of factors:%d",count);
}
getch();
}

Is This Answer Correct ?

3 Yes

1 No

write a progam to display the factors of a given number and disply how many prime numbers are ther..

Answer / kracekumar

to check a number is prime or not ,it is enough to check till square root(floor) of the number is divisible or not.
#include<stdio.h>
#include<math.h>
/*
WHILE COMPILING IN LINUX USE gcc -lm -o factor factor.c and execute ./factor
*/
void isprime(int factor)
{
/*This function will check the given factor of the no is prime or not if so it will print the number factor as (prime)*/
int inner_loop,sqrt_of_factor;
sqrt_of_factor=(int)sqrt((double)factor);//in sqrt function will take float or double argument
if(factor==2)
printf("(prime)");
else
{/*To find out the no is prime or not you need to find out square root and from 2 to square root of the no if the no is not divisble by the no it is called as prime no*/
for(inner_loop=2;inner_loop<=sqrt_of_factor;++inner_loop)
{
if(factor%inner_loop==0)//if it divisble it exit the looop
break;
}
if(inner_loop>sqrt_of_factor)
/*best part is this checking ,if the no(factor) is prime it should be exited the loop with the loop variable value greater than condition varibale value */
printf("(prime)");
}
}
void factor(int no)
{
int start,no_by_2=no/2;
/*no_by_2 is used because factor of a no will be in the range 1 to half the number */
printf("\n factors of %d are %d",no,1);
for(start=2;start<no_by_2;++start)
{
if(no%start==0)
{
printf("\t %d",start);
isprime(start);//checking the factor is prime or not
}
}
printf("\t %d",no);
isprime(no);
}
int main()
{
system("clear");//work only in LINUX
int no;
printf("\n enter the integer no for which factor is to be found:");
scanf("%d",&no);
factor(no);
printf("\n \n=================================================\n");
return 0;
}

Is This Answer Correct ?

1 Yes

1 No

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