Solution For Problem 47 of Project Euler(Issue #15)

euler47.py

@@ -0,0 +1,53 @@
+def prime_no_collector(prime, n):
+    if n == 1:                       #checking whether n == 1, not appending if it is equal as 1 is not a prime number 
+        return prime
+    for i in range(2, int(n**0.5)+1): #Running loop from 2 to sq_root(n) to check whether the number is prime or not
+        if n%i == 0:                 #checking whether the numbe ris divisble or not (if divisible then it isn't a prime)
+            return prime             #if not prime then returning the array without appending
+    return prime.append(n)
+
+
+'''
+
+#Testing the prime_no_collector function
+prime = []
+for i in range(1,20):
+    prime_no_collector(prime,i)
+print(prime)
+#Excpected Output: 2, 3, 5, 7, 11, 13, 17, 19
+#Function Output: [2, 3, 5, 7, 11, 13, 17, 19]
+
+'''
+
+
+def prime_factors(prime, n):
+    factors = []
+    for i in prime:
+        if n%i == 0:
+            factors.append(i)
+    return factors
+
+
+'''
+
+#Testing function prime_factors:
+print(prime_factors(prime, 15))
+print(prime_factors(prime, 17))
+print(prime_factors(prime, 14))
+
+#Expected Output: [3, 5]    [17]    [2, 7]
+#Function Output: [3, 5]
+#                 [17]
+#                 [2, 7]
+
+'''
+
+
+def main():
+    prime = []
+    answer = 0
+    for i in range(1,1000000):              #looping till 1,000,000
+        prime_no_collector(prime,i)         #collecting all prime numbers as we iterate
+        if len(prime_factors(prime, i)) == 4 and len(prime_factors(prime, i+1)) == 4 and len(prime_factors(prime, i+2)) == 4 and len(prime_factors(prime, i+3)) == 4:
+            return i
+print(main())