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;

}

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AnandSir ,

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

jahangir kumars=1x2x3x4+2x3x4x5+3x4x5x6+4x5x6x7+5x6x7x8

jahangir kumars=(1+2)/(1×2)+(1+2+3)/(1x2x3+)+…….+(1+2+3+……n terms)/(1x2x3x……….n terms)

please help in this series

jahangir kumarproblem s=1+12+123+1234+…………+123….n terms