diff --git a/CSClasses/Algorithms.cs b/CSClasses/Algorithms.cs
index 982c654..1d05be1 100644
--- a/CSClasses/Algorithms.cs
+++ b/CSClasses/Algorithms.cs
@@ -318,6 +318,143 @@ namespace mee{
 					foundFactor = false;
 				}
 			}
+
+			//Sort the list before returning it
+			primes.Sort();
+			return primes;
+		}
+		//This function gets a certain number of primes
+		public static List<int> getNumPrimes(int numberOfPrimes){
+			List<int> primes = new List<int>();	//Holds the prime numbers
+			bool foundFactor = false;	//A flag for whether a factor of the current number has been found
+
+			//If the number is 0 or negative return an empty list
+			if(numberOfPrimes <= 1){
+				return primes;
+			}
+			//Otherwise the number is at least 2, so 2 should be added to the list
+			else{
+				primes.Add(2);
+			}
+
+			//We can now start at 3 and skip all even numbers, because they cannot be prime
+			for(int possiblePrime = 3;primes.Count < numberOfPrimes;possiblePrime += 2){
+				//Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor
+				int topPossibleFactor = (int)Math.Ceiling(Math.Sqrt(possiblePrime));
+				//We can safely assume that there will be at least 1 element in the primes list because of 2 being added before this
+				for(int primesCnt = 0;primes[primesCnt] <= topPossibleFactor;){
+					if((possiblePrime % primes[primesCnt]) == 0){
+						foundFactor = true;
+						break;
+					}
+					else{
+						++primesCnt;
+					}
+					//Check if the index has gone out of bounds
+					if(primesCnt >= primes.Count){
+						break;
+					}
+				}
+
+				//If you didn't find a factor then the current number must be prime
+				if(!foundFactor){
+					primes.Add(possiblePrime);
+				}
+				else{
+					foundFactor = false;
+				}
+			}
+
+			//Sort the list before returning it
+			primes.Sort();
+			return primes;
+		}
+		public static List<long> getNumPrimes(long numberOfPrimes){
+			List<long> primes = new List<long>();	//Holds the prime numbers
+			bool foundFactor = false;	//A flag for whether a factor of the current number has been found
+
+			//If the number is 0 or negative return an empty list
+			if(numberOfPrimes <= 1){
+				return primes;
+			}
+			//Otherwise the number is at least 2, so 2 should be added to the list
+			else{
+				primes.Add(2);
+			}
+
+			//We can now start at 3 and skip all even numbers, because they cannot be prime
+			for(long possiblePrime = 3;primes.Count < numberOfPrimes;possiblePrime += 2){
+				//Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor
+				long topPossibleFactor = (long)Math.Ceiling(Math.Sqrt(possiblePrime));
+				//We can safely assume that there will be at least 1 element in the primes list because of 2 being added before this
+				for(int primesCnt = 0;primes[primesCnt] <= topPossibleFactor;){
+					if((possiblePrime % primes[primesCnt]) == 0){
+						foundFactor = true;
+						break;
+					}
+					else{
+						++primesCnt;
+					}
+					//Check if the index has gone out of bounds
+					if(primesCnt >= primes.Count){
+						break;
+					}
+				}
+
+				//If you didn't find a factor then the current number must be prime
+				if(!foundFactor){
+					primes.Add(possiblePrime);
+				}
+				else{
+					foundFactor = false;
+				}
+			}
+
+			//Sort the list before returning it
+			primes.Sort();
+			return primes;
+		}
+		public static List<BigInteger> getNumPrimes(BigInteger numberOfPrimes){
+			List<BigInteger> primes = new List<BigInteger>();	//Holds the prime numbers
+			bool foundFactor = false;	//A flag for whether a factor of the current number has been found
+
+			//If the number is 0 or negative return an empty list
+			if(numberOfPrimes <= 1){
+				return primes;
+			}
+			//Otherwise the number is at least 2, so 2 should be added to the list
+			else{
+				primes.Add(2);
+			}
+
+			//We can now start at 3 and skip all even numbers, because they cannot be prime
+			for(BigInteger possiblePrime = 3;primes.Count < numberOfPrimes;possiblePrime += 2){
+				//Check all current primes, up to sqrt(possiblePrime), to see if there is a divisor
+				BigInteger topPossibleFactor = (BigInteger)Math.Exp(BigInteger.Log(possiblePrime) / 2);
+				//We can safely assume that there will be at least 1 element in the primes list because of 2 being added before this
+				for(int primesCnt = 0;primes[primesCnt] <= topPossibleFactor;){
+					if((possiblePrime % primes[primesCnt]) == 0){
+						foundFactor = true;
+						break;
+					}
+					else{
+						++primesCnt;
+					}
+					//Check if the index has gone out of bounds
+					if(primesCnt >= primes.Count){
+						break;
+					}
+				}
+
+				//If you didn't find a factor then the current number must be prime
+				if(!foundFactor){
+					primes.Add(possiblePrime);
+				}
+				else{
+					foundFactor = false;
+				}
+			}
+
 			//Sort the list before returning it
 			primes.Sort();
 			return primes;