diff --git a/Algorithms.hpp b/Algorithms.hpp
index dca330a..3dd21b7 100644
--- a/Algorithms.hpp
+++ b/Algorithms.hpp
@@ -196,20 +196,34 @@ std::vector<T> getFactors(T goalNumber){
 template<class T>
 std::vector<T> getDivisors(T num){
 	std::vector<T> divisors;	//Holds the number of divisors
-	//You only need to go to sqrt(number). cnt * cnt is faster than sqrt()
-	for(uint64_t cnt = 1;cnt * cnt <= num;++cnt){
+	//Ensure the parameter is a valid number
+	if(num <= 0){
+		return divisors;
+	}
+	else if(num == 1){
+		divisors.push_back(1);
+	}
+	else{
+		divisors.push_back(1);
+		divisors.push_back(num);
+	}
+	//You only need to check up to sqrt(num)
+	T topPossibleDivisor = ceil(sqrt(num));
+	for(uint64_t possibleDivisor = 1;possibleDivisor <= topPossibleDivisor;++possibleDivisor){
 		//Check if the counter evenly divides the number
 		//If it does the counter and the other number are both divisors
-		if((num % cnt) == 0){
-			if(!isFound(cnt, divisors)){
-				divisors.push_back(cnt);
-			}
-			if(!isFound(num/cnt, divisors)){
-				divisors.push_back(num / cnt);
+		if((num % possibleDivisor) == 0){
+			//We don't need to check if the number already exists because we are only checking numbers <= sqrt(num), so there can be no duplicates
+			divisors.push_back(possibleDivisor);
+			//We still need to account for sqrt(num) being a divisor
+			if(possibleDivisor != topPossibleDivisor){
+				divisors.push_back(num / possibleDivisor)
 			}
 		}
 	}
+	//Sort the vector for neatness
 	std::sort(divisors.begin(), divisors.end());
+	//Return the vector of divisors
 	return divisors;
 }