Prashant | Sat, 13 Jun, 2020 | 930

Here we will see a program to return two prime numbers such that their sum is equal to a given number, and in each case that is possible.

So how can we solve it ?

Well there are lot of solutions to this questions, but we will do it using the best optimal solution.

Given an even number (greater than 2), return two prime numbers whose sum will be equal to given number. There are several combinations possible. Print only first such pair.

**NOTE:** A solution will always exist, read Goldbach’s conjecture. Also, solve the problem in linear time complexity, i.e., O(n).

**Input:**

The first line contains T, the number of test cases. The following T lines consist of a number each, for which we'll find two prime numbers.

**Note**: The number would always be an even number.

**Output:**

For every test case print two prime numbers space separated, such that the smaller number appears first. Answer for each test case must be in a new line.

**Constraints:**

1 ≤ T ≤ 70

2 < N ≤ 10000

**Example:****Input:**

5

74

1024

66

8

9990

**Output:**

3 71

3 1021

5 61

3 5

17 9973

```
#include<bits/stdc++.h>
using namespace std;
bool isprime(int n){
if(n<2)return 0;
if(n%2 == 0)return 0;
if(n == 2)return 1;
for(int i=3;i*i<=n;i+=2) {
if(n%i == 0) {
return 0;
}
}
return 1;
}
int main() {
int n,num;
cin>>n;
while(n--) {
cin>>num;
for(int i=2;i<=num/2;i++) {
if(isprime(i) && isprime(num-i)) {
cout<<i<<" "<<num-i<<endl;
break;
}
}
}
return 0;
}
```

Correct Answer Execution Time:0.01.