# Check if a number can be written as a sum of ‘k’ prime numbers

• Difficulty Level : Hard
• Last Updated : 31 Aug, 2021

Given two numbers N and K. We need to find out if ‘N’ can be written as sum of ‘K’ prime numbers.
Given N <= 10^9

Examples :

```Input  : N = 10 K = 2
Output : Yes
10 can be written as 5 + 5

Input  : N = 2 K = 2
Output : No```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

The idea is to use Goldbach’s conjecture which says that every even integer (greater than 2) can be expressed as sum of two primes.
If the N >= 2K and K = 1: the answer will be Yes iff N is a prime number
If N >= 2K and K = 2: If N is an even number answer will be Yes(Goldbach’s conjecture) and if N is odd answer will be No if N-2 is not a prime number and Yes if N-2 is a prime number. This is because we know odd + odd = even and even + odd = odd. So when N is odd, and K = 2 one number must be 2 as it is the only even prime number so now the answer depends on whether N-2 is odd or not.
If N >= 2K and K >= 3: Answer will always be Yes. When N is even N – 2*(K-2) is also even so N – 2*(K – 2) can be written as sum of two prime numbers (Goldbach’s conjecture) p, q and N can be written as 2, 2 …..K – 2 times, p, q. When N is odd N – 3 -2*(K – 3) is even so it can be written as sum of two prime numbers p, q and N can be written as 2, 2 …..K-3 times, 3, p, q

## C++

```// C++ implementation to check if N can be
// written as sum of k primes
#include<bits/stdc++.h>
using namespace std;

// Checking if a number is prime or not
bool isprime(int x)
{

// check for numbers from 2 to sqrt(x)
// if it is divisible return false
for (int i = 2; i * i <= x; i++)
if (x % i == 0)
return false;
return true;
}

// Returns true if N can be written as sum
// of K primes
bool isSumOfKprimes(int N, int K)
{
// N < 2K directly return false
if (N < 2*K)
return false;

// If K = 1 return value depends on primality of N
if (K == 1)
return isprime(N);

if (K == 2)
{
// if N is even directly return true;
if (N % 2 == 0)
return true;

// If N is odd, then one prime must
// be 2. All other primes are odd
// and cannot have a pair sum as even.
return isprime(N - 2);
}

// If K >= 3 return true;
return true;
}

// Driver function
int main()
{
int n = 10, k = 2;
if (isSumOfKprimes (n, k))
cout << "Yes" << endl;
else
cout << "No" << endl;
return 0;
}
```

## Java

```// Java implementation to check if N can be
// written as sum of k primes
public class Prime
{
// Checking if a number is prime or not
static boolean isprime(int x)
{
// check for numbers from 2 to sqrt(x)
// if it is divisible return false
for (int i=2; i*i<=x; i++)
if (x%i == 0)

return false;
return true;
}

// Returns true if N can be written as sum
// of K primes
static boolean isSumOfKprimes(int N, int K)
{
// N < 2K directly return false
if (N < 2*K)
return false;

// If K = 1 return value depends on primality of N
if (K == 1)
return isprime(N);

if (K == 2)
{
// if N is even directly return true;
if (N%2 == 0)
return true;

// If N is odd, then one prime must
// be 2. All other primes are odd
// and cannot have a pair sum as even.
return isprime(N - 2);
}

// If K >= 3 return true;
return true;
}

public static void main (String[] args)
{
int n = 10, k = 2;
if (isSumOfKprimes (n, k))
System.out.print("Yes");
else
System.out.print("No");
}
}
// Contributed by Saket Kumar
```

## Python3

```# Python implementation to check
# if N can be written as sum of
# k primes

# Checking if a number is prime
# or not

def isprime(x):

# check for numbers from 2
# to sqrt(x) if it is divisible
# return false
i = 2
while(i * i <= x):
if (x % i == 0):
return 0
i += 1
return 1

# Returns true if N can be written
# as sum of K primes

def isSumOfKprimes(N, K):

# N < 2K directly return false
if (N < 2 * K):
return 0

# If K = 1 return value depends
# on primality of N
if (K == 1):
return isprime(N)

if (K == 2):

# if N is even directly
# return true;
if (N % 2 == 0):
return 1

# If N is odd, then one
# prime must be 2. All
# other primes are odd
# and cannot have a pair
# sum as even.
return isprime(N - 2)

# If K >= 3 return true;
return 1

# Driver function
n = 15
k = 2
if (isSumOfKprimes(n, k)):
print("Yes")
else:
print("No")

# This code is Contributed by Sam007.
```

## C#

```// C# implementation to check if N can be
// written as sum of k primes
using System;

class GFG {

// Checking if a number is prime or not
static bool isprime(int x)
{
// check for numbers from 2 to sqrt(x)
// if it is divisible return false
for (int i = 2; i * i <= x; i++)
if (x % i == 0)

return false;
return true;
}

// Returns true if N can be written as sum
// of K primes
static bool isSumOfKprimes(int N, int K)
{
// N < 2K directly return false
if (N < 2 * K)
return false;

// If K = 1 return value depends on primality of N
if (K == 1)
return isprime(N);

if (K == 2)
{
// if N is even directly return true;
if (N % 2 == 0)
return true;

// If N is odd, then one prime must
// be 2. All other primes are odd
// and cannot have a pair sum as even.
return isprime(N - 2);
}

// If K >= 3 return true;
return true;
}

// Driver function
public static void Main ()
{
int n = 10, k = 2;
if (isSumOfKprimes (n, k))
Console.Write("Yes");
else
Console.Write("No");
}
}

// This code is contributed by Sam007
```

## PHP

```<?php
// PHP implementation to check
// if N can be written as sum
// of k primes

// Checking if a number
// is prime or not
function isprime(\$x)
{
// check for numbers from 2
// to sqrt(x) if it is
// divisible return false
for (\$i = 2; \$i * \$i <= \$x; \$i++)
if (\$x % \$i == 0)
return false;
return true;
}

// Returns true if N can be
// written as sum of K primes
function isSumOfKprimes(\$N, \$K)
{
// N < 2K directly return false
if (\$N < 2 * \$K)
return false;

// If K = 1 return value
// depends on primality of N
if (\$K == 1)
return isprime(\$N);

if (\$K == 2)
{
// if N is even directly
// return true;
if (\$N % 2 == 0)
return true;

// If N is odd, then one prime
// must be 2. All other primes
// are odd and cannot have a
// pair sum as even.
return isprime(\$N - 2);
}

// If K >= 3 return true;
return true;
}

// Driver Code
\$n = 10; \$k = 2;
if (isSumOfKprimes (\$n, \$k))
echo "Yes";
else
echo"No" ;

// This code is contributed by vt
?>

```

## Javascript

```<script>
// javascript implementation to check if N can be
// written as sum of k primes

// Checking if a number is prime or not
function isprime(x)
{

// check for numbers from 2 to sqrt(x)
// if it is divisible return false
for (i = 2; i * i <= x; i++)
if (x % i == 0)

return false;
return true;
}

// Returns true if N can be written as sum
// of K primes
function isSumOfKprimes(N, K)
{

// N < 2K directly return false
if (N < 2 * K)
return false;

// If K = 1 return value depends on primality of N
if (K == 1)
return isprime(N);

if (K == 2)
{

// if N is even directly return true;
if (N % 2 == 0)
return true;

// If N is odd, then one prime must
// be 2. All other primes are odd
// and cannot have a pair sum as even.
return isprime(N - 2);
}

// If K >= 3 return true;
return true;
}

// Driver code
var n = 10, k = 2;
if (isSumOfKprimes(n, k))
document.write("Yes");
else
document.write("No");

// This code is contributed by gauravrajput1
</script>
```

Output :

`Yes`

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