# Check if the sum of primes is divisible by any prime from the array

Given an array arr[], the task is to check whether the sum of primes in the array is divisible by any of the primes in the array. If it is then print YES, otherwise print NO.

Examples:

Input: arr[] = {2, 3}
Output: NO
Primes: 2, 3
Sum = 2 + 3 = 5 which is neither divisible by 2 nor 3

Input: arr[] = {1, 2, 3, 4, 5}
Output: YES
2 + 3 + 5 = 10 is divisible by 2 as well as 5

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The idea is to generate all primes upto maximum element from the array using Sieve of Eratosthenes.

• Traverse the array and check if the current element is prime or not. If it is prime then update sum = sum + arr[i].
• Traverse the array again and check if sum % arr[i] = 0 where arr[i] is prime. If it is then print YES. Otherwise print NO in the end.

Below is the implementation of the above approach:

## C++

 // C++ program to check if sum of primes from an array // is divisible by any of the primes from the same array #include using namespace std;    // Function to print "YES" if sum of primes from an array // is divisible by any of the primes from the same array void SumDivPrime(int A[], int n) {     int max_val = *(std::max_element(A, A + n)) + 1;        // USE SIEVE TO FIND ALL PRIME NUMBERS LESS     // THAN OR EQUAL TO max_val     // Create a boolean array "prime[0..n]". A     // value in prime[i] will finally be false     // if i is Not a prime, else true.     vector prime(max_val + 1, true);        // Remaining part of SIEVE     prime[0] = false;     prime[1] = false;     for (int p = 2; p * p <= max_val; p++) {            // If prime[p] is not changed, then         // it is a prime         if (prime[p] == true) {                // Update all multiples of p             for (int i = p * 2; i <= max_val; i += p)                 prime[i] = false;         }     }        int sum = 0;        // Traverse through the array     for (int i = 0; i < n; ++i) {         if (prime[A[i]])             sum += A[i];     }        for (int i = 0; i < n; ++i) {         if (prime[A[i]] && sum % A[i] == 0) {             cout << "YES";             return;         }     }        cout << "NO"; }    // Driver program int main() {     int A[] = { 1, 2, 3, 4, 5 };     int n = sizeof(A) / sizeof(A[0]);        SumDivPrime(A, n);        return 0; }

## Java

 // Java program to check if sum of primes from an array  // is divisible by any of the primes from the same array  class Solution {     //returns the maximum value static int max_element(int A[]) {     int max=Integer.MIN_VALUE;            for(int i=0;i

## Python3

 # Python3 program to check if sum of  # primes from an array is divisible # by any of the primes from the same array import math    # Function to print "YES" if sum of primes  # from an array is divisible by any of the  # primes from the same array def SumDivPrime(A, n):        max_val = max(A) + 1        # USE SIEVE TO FIND ALL PRIME NUMBERS      # LESS THAN OR EQUAL TO max_val     # Create a boolean array "prime[0..n]".      # A value in prime[i] will finally be      # false if i is Not a prime, else true.     prime = [True] * (max_val + 1)        # Remaining part of SIEVE     prime[0] = False     prime[1] = False     for p in range(2, int(math.sqrt(max_val)) + 1):            # If prime[p] is not changed,          # then it is a prime         if prime[p] == True :                # Update all multiples of p             for i in range(2 * p, max_val + 1, p):                 prime[i] = False        sum = 0        # Traverse through the array     for i in range(0, n):         if prime[A[i]]:             sum += A[i]            for i in range(0, n):         if prime[A[i]] and sum % A[i] == 0:             print("YES")             return                print("NO")    # Driver Code A = [ 1, 2, 3, 4, 5 ] n = len(A)    SumDivPrime(A, n)    # This code is contributed  # by saurabh_shukla

## C#

 // C# program to check if sum of primes // from an array is divisible by any of  // the primes from the same array  class GFG  {     //returns the maximum value  static int max_element(int[] A)  {      int max = System.Int32.MinValue;             for(int i = 0; i < A.Length; i++)          if(max < A[i])              max = A[i];             return max;  }        // Function to print "YES" if sum of  // primes from an array  is divisible  // by any of the primes from the same array  static void SumDivPrime(int[] A, int n)  {      int max_val = (max_element(A)) + 1;         // USE SIEVE TO FIND ALL PRIME NUMBERS LESS      // THAN OR EQUAL TO max_val      // Create a boolean array "prime[0..n]". A      // value in prime[i] will finally be false      // if i is Not a prime, else true.      bool[] prime=new bool[max_val+1];             //initilize the array      for(int i = 0; i <= max_val; i++)          prime[i] = true;         // Remaining part of SIEVE      prime[0] = false;      prime[1] = false;      for (int p = 2; p * p <= max_val; p++)      {             // If prime[p] is not changed, then          // it is a prime          if (prime[p] == true)          {                 // Update all multiples of p              for (int i = p * 2; i <= max_val; i += p)                  prime[i] = false;          }      }      int sum = 0;         // Traverse through the array      for (int i = 0; i < n; ++i)     {          if (prime[A[i]])              sum += A[i];      }         for (int i = 0; i < n; ++i)      {          if (prime[A[i]] && sum % A[i] == 0)          {              System.Console.WriteLine( "YES");              return;          }      }      System.Console.WriteLine("NO");  }     // Driver code public static void Main()  {      int []A = { 1, 2, 3, 4, 5 };      int n = A.Length;      SumDivPrime(A, n);  }  }     // This code is contributed by mits

## PHP



Output:

YES

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