This seems to be a popular topic as of late. The definition of a practical number:

wiki said:

For the definition, the wiki is a good resource, but for practical/implementation guidance, it's poor. The initial way to approach this is to iterate through 1 to number-1 and check to see if each can be expressed by the sum of the divisors. I haven't written a solution that follows this thought process, although I may at some point. This approach has large time complexity and can be tricky. Another way to approach this problem is to get the prime factors and use a formula based on the summation of these factors. I have yet to find a decent source that I can reference for this and as of this writing did not write a solution following this approach either.

So you're probably asking yourself, what do you have Knowles? Using the "properties" of practical numbers:

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bool isPractical(int num){
	//if the number is less then zero or odd, return false immediately
	//Get the divisors of num
	//----Go through the divisors 
	//	  if the sum is greater then or equal to the most recent divisor-1 add it to the sum
	//	  otherwise return false 

	if(num < 0) return false;
	else if (num == 1) return true; //1 is a divisor of itself 
	else if (num&1) return false; //odd check 

	//divisors
	vector<int> divisors;
	divisors.push_back(1); //save ourselves an iteration
	for(int i = 2; i < num; i++){
		if((num % i) == 0)
			divisors.push_back(i);
	}
	int sum = 0;

	for(int i = 0; i < divisors.size(); i++){
		if (sum >= divisors.at(i)-1) sum += divisors.at(i);
		else return false;
	}

	return true;
}
int main(){

	//253 to show the practical numbers up to and including 252
	for(int i = 1; i < 253; i++){
		if(isPractical(i)){
			cout << i << " ";
		}
	}
	return 0;
}

The result from the above program:

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A list of the first few practical numbers, from here

Quote

Neat stuff. As I mentioned above there are quite a few ways to approach this problem (aren't numbers awesome?) so this one method certainly isn't the be all end all.