# PROBLEM LINK:

*Author:* Ashraf Khan

*Tester:* Ashraf Khan

*Editorialist:* Ashraf Khan

# DIFFICULTY:

EASY, EASY-MEDIUM

# PREREQUISITES:

Knowledge of Prime Numbers, Basic Mathematics, Basic Programming

# PROBLEM:

Chef has developed interest in prime numbers and is looking out for a challenge to showcase his skills.

He picks a random number N, then finds N prime numbers beginning from 2. For example, if N=5, then the list of prime numbers will be 2,3,5,7,11.

Given the value of N, you have to find the sum of those N prime numbers.

# QUICK EXPLANATION:

Find the first *N* prime numbers starting from *2* and then print the sum of those numbers.

# EXPLANATION:

The very first line of input contains the number of testcases, *T*. Then the following *T* lines contains the values of *N*.

- So, first take the input for
*T*. - Then start a loop which will run
*T*times. - Take the input for
*N*inside this loop. - Initialize a variable (say
*count = 1*) to keep the count of prime numbers found, . - Initialize a variable (say
*sum = 2*) to store the summed up prime numbers. - Initialize another variable (say
*i = 3*) to represent numbers. This variable will keep incrementing by*2*to get the numbers which will be checked for prime number. - Initialize a while loop as
*while(count < N)*. Inside this loop, we will check for the possible prime numbers. If prime number is found then it will be summed up in*sum*variable and*count*will be incremented by*1*.

# Solutions:

## Setter's Solution

```
#Solution code in python
from math import sqrt
T = int(input())
while T>0:
T -= 1
N = int(input())
count = 1
sum = 2
i = 3
while count<N:
found = 0
for j in range(3, int(sqrt(i))+1, 2):
if i % j == 0:
found = 1
break
if found == 0:
count += 1
sum += i
i += 2
print(sum)
```

## Tester's Solution

```
//Solution code in C++
#include <iostream>
#include<math.h>
using namespace std;
int main() {
int T;
cin >> T;
while(T--) {
int N;
cin >> N;
int count=1, sum=2, i=3;
while(count < N) {
int found = 0;
for(int j = 3; j < sqrt(i)+1; j+=2) {
if(i%j == 0) {
found = 1;
break;
}
}
if(found == 0) {
count++;
sum += i;
}
i+=2;
}
cout << sum << endl;
}
return 0;
}
```

## Editorialist's Solution

```
#Solution code in python
from math import sqrt
T = int(input())
while T>0:
T -= 1
N = int(input())
count = 1
sum = 2
i = 3
while count<N:
found = 0
for j in range(3, int(sqrt(i))+1, 2):
if i % j == 0:
found = 1
break
if found == 0:
count += 1
sum += i
i += 2
print(sum)
```