# Blog Archives

## Project Euler Problem#7 solution in C++ (Brute Force)

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.

What is the 10 001st prime number?

Brute Force Solution:

#include <iostream>

using namespace std;

int find_prime (long *numArray, int maxNum)
{
long factor = 2; // we will make 2 as the starting point.
numArray = 0; // rule out 1 from our logic to avoid incorrect results.

// loop condition will check, if we are in our maximum number limit.
//maxNum is the number till there we can find the prime numbers.
while ((factor * factor) < maxNum)
{
for (int i = 2; i < maxNum; i++) // we start our itration from number 2 not from 0 or 1 to get correct results.
{
if (numArray[i] != 0) //if a number on current array index is not zero, it is a prime or we havne’t yet checked it.
{
// we are putting zeros on all multiples of prime numbers, one by one.
if (numArray[i] != factor && (numArray[i]% factor) == 0)
{
numArray[i] = 0;
}
}
}
++factor;
}// after the loop, array will have zeros on all non prime locations and prime numbers.
}

int main ()
{
int prime_count = 0;
int nth_prime = 0;
int maxNum = 0;
int i = 0;

cout << “enter max number for array: “; // you need to test some upper limit values.
cin >> maxNum;
cout << endl << “enter nth prime number you are searching for: “;
cin >> nth_prime;

long *myArray = new long [maxNum];

//we fill up the array till the number we want to find the smallest positive number that is evenly divisible.
for (int i = 0; i < maxNum; i++)
{
myArray[i]= i;
}

// we will get prime numbers till the maxNum by calling below funtions.
find_prime(myArray, maxNum);

for (; i < maxNum; i++)
{
if (myArray[i] != 0)
{
prime_count++;
}
if (prime_count == nth_prime){break;}
}

cout << “the nth prime number is: ” << “value of prime count is: “<< prime_count << ” – ” << myArray[i] << endl;

}