# Counting prime number that is expressed as the sum of consecutive prime numbers

Question:

Some prime numbers can be expressed as a sum of other consecutive prime numbers.
For example,
5 = 2 + 3
17 = 2 + 35 + 7
41 = 2 + 3 + 5 + 7 + 11 + 13.

Now the task is to find out
how many prime numbers which satisfy this property are
present in the range 3 to N subject to a constraint that summation should always start with number 2.

Input Format:
First-line contains a number N
Output Format:
Print the total number of all such prime numbers which are
less than or equal to N.

Constraints:
2<N<=12,000,000,000

Sample Input 1 :
20
Sample Output 1 :
2
Explanation :
5 = 2+3
17 = 2+3+5+7
Sample Input 2 :
15
Sample Output 2 :
1
Explanation :
5 = 2+3

Solution:

C Language

```//code in C to understand the program
//codewindow.in

#include <stdio.h>

//prime function
int prime(int b)
{
int j,cnt;
cnt=1;

for(j=2;j<=b/2;j++)
{
if(b%j==0)
cnt=0;
}

if(cnt==0)
return 1;
else
return 0;
}

//main
int main() {
int i,j,n,cnt,a,c,sum=0,count=0,k=0;
scanf("%d",&n);

for(i=2;i<=n;i++)
{
cnt=1;
for(j=2;j<=n/2;j++)
{
if(i%j==0)
cnt=0;
}

if(cnt==1)
{
a[k]=i;
k++;
}
}

for(i=0;i<k;i++)
{
sum=sum+a[i];
c= prime(sum); //calling the function prime
if(c==1)
count++;
}

printf("%d",count);

return 0;
}
```

JAVA

```//code in JAVA to understand the program
//codewindow.in

import java.util.Scanner;

class Main
{
static int prime(int b)
{

int j,cnt;
cnt=1;

for (j = 2; j <= b/2; j++)
{
if(b%j==0)
cnt=0;
}

if(cnt==0)
return 1;
else
return 0;

}

public static void main(String[] args) {
Scanner sc = new Scanner(System.in);

int i,j,n=0,cnt,c=0,sum=0,count=0,k=0;

Main t = new Main();
int[] a = new int;

System.out.println("Enter no");
n = sc.nextInt();

for (i = 2; i <=n ; i++)
{
cnt=1;
for (j = 2; j <= n/2; j++)
{
if(i%j==0)
cnt=0;
}

if(cnt==1)
{
a[k]=i;
k++;
}
}

for (i = 0; i < k; i++)
{
sum=sum+a[i];
c=t.prime(sum);

if(c==1)
count++;
}

System.out.println(count);

}
}
```

PYTHON

```#code in Python to understand the program
#codewindow.in

num = int(input())
arr = []
sum = 0
count = 0
if num > 1:
for i in range(2, num + 2):
for j in range(2, i):
if i % j == 0:
break
else:
arr.append(i)
def is_prime(sum):
for i in range(2, (sum // 2) +2):
if sum % i == 0:
return False
else:
return True
for i in range(0, len(arr)):
sum = sum + arr[i]
if sum <= num:
if is_prime(sum):
count = count + 1
print(count)
```

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