From 8d23cd41eb9f4e0cf06715abe27c3999aa80aee9 Mon Sep 17 00:00:00 2001
From: Charles <sircharlesaze@gmail.com>
Date: Mon, 18 Nov 2019 23:22:33 +0100
Subject: problem 50 python

---
 python/038-pandigital_multiples.py  | 18 ++++++++++++++++++
 python/050-consecutive_prime_sum.py | 31 +++++++++++++++++++++++++++++++
 python/wip/026-reciprocal_cycles.py |  5 -----
 3 files changed, 49 insertions(+), 5 deletions(-)
 create mode 100644 python/038-pandigital_multiples.py
 create mode 100644 python/050-consecutive_prime_sum.py
 delete mode 100644 python/wip/026-reciprocal_cycles.py

(limited to 'python')

diff --git a/python/038-pandigital_multiples.py b/python/038-pandigital_multiples.py
new file mode 100644
index 0000000..be8fedc
--- /dev/null
+++ b/python/038-pandigital_multiples.py
@@ -0,0 +1,18 @@
+# ###
+# Pandigital multiples
+# Problem 38
+# 
+# Take the number 192 and multiply it by each of 1, 2, and 3:
+# 192 × 1 = 192
+# 192 × 2 = 384
+# 192 × 3 = 576
+# By concatenating each product we get the 1 to 9 pandigital, 192384576. We will call 192384576 the concatenated product of 192 and (1,2,3)
+# The same can be achieved by starting with 9 and multiplying by 1, 2, 3, 4, and 5, giving the pandigital, 918273645, which is the concatenated product of 9 and (1,2,3,4,5).
+# What is the largest 1 to 9 pandigital 9-digit number that can be formed as the concatenated product of an integer with (1,2, ... , n) where n > 1?
+# ###
+
+
+def concatenated_prod(n):
+    cprod = ""
+    
+
diff --git a/python/050-consecutive_prime_sum.py b/python/050-consecutive_prime_sum.py
new file mode 100644
index 0000000..0fde061
--- /dev/null
+++ b/python/050-consecutive_prime_sum.py
@@ -0,0 +1,31 @@
+# ###
+# Consecutive prime sum
+# Problem 50
+# 
+# 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?
+# ###
+
+
+from helper.prime import primes_loop, is_prime
+
+
+sumed = []
+ps = []
+max_len = 0
+for p in primes_loop():
+    if p > 20000:
+        break
+    ps.append(p)
+    for i in range(len(ps)):
+        s = sum(ps[i:])
+        if s > 1000000:
+            break
+        if is_prime(s) and len(ps) > max_len:
+            sumed.append(s)
+            max_len = len(ps)
+
+print(sumed[-1])
diff --git a/python/wip/026-reciprocal_cycles.py b/python/wip/026-reciprocal_cycles.py
deleted file mode 100644
index b502c9a..0000000
--- a/python/wip/026-reciprocal_cycles.py
+++ /dev/null
@@ -1,5 +0,0 @@
-from decimal import Decimal
-
-for i in range(1, 11):
-    # decimals = int(str(1/i)[2:])
-    print(d)
-- 
cgit