diff --git a/50/main.py b/50/main.py
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+'''
+The prime 41, can be written as the sum of six consecutive primes:
+
+41 = 2 + 3 + 5 + 7 + 11 + 13
+This is the longest sum of consecutive primes that adds to a prime below one-hundred.
+
+The longest sum of consecutive primes below one-thousand that adds to a prime, contains 21 terms, and is equal to 953.
+
+Which prime, below one-million, can be written as the sum of the most consecutive primes?
+'''
+
+import math
+import numpy as np
+
+def sieve(n):
+    assert n > 1
+
+    ns = [True] * n
+
+    for i in range(2, math.ceil(np.sqrt(n))):
+        if ns[i]:
+            j = pow(i, 2)
+
+            while j < n:
+                ns[j] = False
+                j = j + i
+
+    return [i for i,val in enumerate(ns) if val][2:]
+
+# Note: this subset sum implementation only backtracks the sum with the smallest values as possible
+def subsetSum(elements, target):
+    dyn = []
+
+    for _ in range(len(elements) + 1):
+        dyn.append([True])
+    
+    for _ in range(target):
+        dyn[0].append(False)
+
+    for i in range(1, len(elements) + 1):
+        for j in range(1, target + 1):
+            if elements[i-1] > j:
+                dyn[i].append(dyn[i-1][j])
+            else:
+                dyn[i].append(dyn[i-1][j] or dyn[i-1][j - elements[i-1]])
+
+    if not dyn[-1][-1]:
+        return []
+
+    j = target
+    i = len(elements)
+    res = []
+
+    while j > 0:
+        while dyn[i - 1][j]:
+            i -= 1
+        curEl = elements[i - 1]
+        j -= curEl
+        res.append(curEl)
+
+    return res
+
+def main():
+    print("Hello this is Patrick")
+
+    target = 1000000
+
+    primeList = sieve(target)
+    primeSet = set(primeList)
+    numberofprimes = len(primeList)
+
+    # s = 0
+    # n = 0
+    # while s < target:
+    #     s += primeList[n]
+    #     n += 1
+
+    # print(numberofprimes)
+    # print(n - 1, s - primeList[n-1])
+
+    # So the number of options is very, very, very very large, we can't brute force this
+    # print(math.factorial(numberofprimes) // (math.factorial(n-1) * math.factorial(numberofprimes - (n-1))))
+
+    # Additionally, this is exactly subset sum for all primes from big to small, since we probably want an as high as possible prime
+
+    # longest = 0
+    # prime = 0
+    # for p in reversed(primeList):
+    #     l = subsetSum(primeList, p)
+    #     if len(l) > longest:
+    #         longest = len(l)
+    #         prime = p
+    
+    # print(prime)
+
+    # Well it seems I completely misinterpreted the question and did something way more difficult than was intended
+
+    prime = 0
+    longest = 0
+
+    for i in range(len(primeList)):
+        s = 0
+        primes = []
+        for p in primeList[i:]:
+            s += p
+            primes.append(p)
+        
+            if s >= target:
+                break
+            if s in primeSet and len(primes) > longest:
+                prime = s
+                longest = len(primes)
+
+    print(prime)
+
+if __name__ == "__main__":
+    main()
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