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| Author | SHA1 | Date | |
|---|---|---|---|
| a1fa273624 | |||
| cf417eaca2 | |||
| 605c9c7f7f | |||
| e6ed590b73 | |||
| 233fc85708 | |||
| 2fb728a4ba | |||
| e900cb117e | |||
| a036c30fc4 | |||
| 8c3da33225 | |||
| 24fb140672 | |||
| 2df02a8204 | |||
| 8580160d67 | |||
| 5cee1f7998 |
@@ -22,7 +22,7 @@ def create_problem():
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https://projecteuler.net/problem={args["problem"]}
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"""
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from utils import timeit
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from project_euler_python.utils import timeit
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@timeit("Problem {(args["problem"]):0>3}")
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@@ -9,8 +9,9 @@ Solution for problem 008 of Project Euler
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https://projecteuler.net/problem=8
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"""
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import collections
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from collections import deque
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from itertools import islice
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from math import prod
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from project_euler_python.utils import timeit
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@@ -19,7 +20,7 @@ from project_euler_python.utils import timeit
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def sliding_window(iterable, n):
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# sliding_window('ABCDEFG', 4) -> ABCD BCDE CDEF DEFG
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it = iter(iterable)
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window = collections.deque(islice(it, n), maxlen=n)
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window = deque(islice(it, n), maxlen=n)
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if len(window) == n:
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yield tuple(window)
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for x in it:
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@@ -65,12 +66,10 @@ def compute():
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num = NUM.replace("\n", "").replace(" ", "")
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adjacent_digits = 13
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ans = 0
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for nums in sliding_window(num, adjacent_digits):
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prod = 1
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for num in nums:
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prod *= int(num)
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if prod > ans:
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ans = prod
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for digits in sliding_window(num, adjacent_digits):
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window_product = prod(int(digit) for digit in digits)
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ans = max(ans, window_product)
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return ans
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@@ -24,9 +24,9 @@ def compute():
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Find the product abc.
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"""
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upper_limit = 1000
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for a in range(1, upper_limit + 1):
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for b in range(a + 1, upper_limit + 1):
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upper_limit = 1000 + 1
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for a in range(1, upper_limit):
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for b in range(a + 1, upper_limit):
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c = upper_limit - a - b
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if a * a + b * b == c * c:
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# It is now implied that b < c, because we have a > 0
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@@ -9,7 +9,7 @@ Solution for problem 012 of Project Euler
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https://projecteuler.net/problem=12
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"""
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from itertools import count
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from itertools import accumulate, count
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from math import floor, sqrt
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from project_euler_python.utils import timeit
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@@ -53,12 +53,10 @@ def compute():
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divisors?
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"""
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triangle = 0
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for i in count(1):
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# This is the ith triangle number, i.e. num = 1 + 2 + ... + i =
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# = i*(i+1)/2
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triangle += i
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if num_divisors(triangle) > 500:
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limit = 500
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for triangle in accumulate(count(1)):
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# This is the ith triangle number, i.e. num = 1 + 2 + ... + i = i*(i+1)/2
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if num_divisors(triangle) > limit:
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return str(triangle)
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@@ -12,14 +12,14 @@ https://projecteuler.net/problem=14
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from project_euler_python.utils import timeit
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def chain_length(n, terms):
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def chain_length(n: int, terms: dict[int, int]) -> int:
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length = 0
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while n != 1:
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if n in terms:
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length += terms[n]
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break
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if n % 2 == 0:
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n = n / 2
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n //= 2
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else:
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n = 3 * n + 1
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length += 1
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@@ -27,7 +27,7 @@ def chain_length(n, terms):
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@timeit("Problem 014")
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def compute():
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def compute() -> int:
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"""
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The following iterative sequence is defined for the set of positive
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integers:
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@@ -49,15 +49,17 @@ def compute():
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NOTE: Once the chain starts the terms are allowed to go above one million.
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"""
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ans = 0
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limit = 1_000_000
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score = 0
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terms = dict()
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for i in range(1, limit):
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terms[i] = chain_length(i, terms)
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if terms[i] > score:
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score = terms[i]
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ans = i
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ans = 0
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longest_length = 0
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chain_lengths = {1: 0}
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for start in range(2, limit):
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current_length = chain_length(start, chain_lengths)
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chain_lengths[start] = current_length
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if current_length > longest_length:
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ans = start
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longest_length = current_length
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return ans
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@@ -34,7 +34,7 @@ def compute():
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names = sorted(f.read().replace('"', "").split(","))
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ans = 0
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for idx, name in enumerate(names, 1):
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for idx, name in enumerate(names, start=1):
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ans += sum(ord(char) - 64 for char in name) * idx
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return ans
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@@ -9,9 +9,21 @@ Solution for problem 023 of Project Euler
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https://projecteuler.net/problem=23
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"""
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from itertools import combinations_with_replacement
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from project_euler_python.utils import timeit
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def _get_abundant_numbers(limit):
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divisor_sums = [0 for _ in range(limit)]
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for divisor in range(1, limit):
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for multiple in range(divisor * 2, limit, divisor):
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divisor_sums[multiple] += divisor
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return [number for number in range(1, limit) if divisor_sums[number] > number]
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@timeit("Problem 023")
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def compute():
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"""
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@@ -35,25 +47,19 @@ def compute():
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sum of two abundant numbers.
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"""
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limit = 28124
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divisor_sum = [0] * limit
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for i in range(1, limit):
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for j in range(i * 2, limit, i):
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divisor_sum[j] += i
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limit = 28123 + 1
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abundant_numbers = _get_abundant_numbers(limit)
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abundant_nums = [i for (i, x) in enumerate(divisor_sum) if x > i]
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expressible_sums = {
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first + second
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for first, second in combinations_with_replacement(
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abundant_numbers,
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2,
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)
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if first + second < limit
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}
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expressible = [False] * limit
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for i in abundant_nums:
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for j in abundant_nums:
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if i + j < limit:
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expressible[i + j] = True
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else:
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break
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ans = sum(i for (i, x) in enumerate(expressible) if not x)
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return ans
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return sum(number for number in range(1, limit) if number not in expressible_sums)
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if __name__ == "__main__":
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@@ -9,7 +9,7 @@ Solution for problem 024 of Project Euler
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https://projecteuler.net/problem=24
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"""
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from itertools import permutations
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from itertools import islice, permutations
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from project_euler_python.utils import timeit
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@@ -28,9 +28,10 @@ def compute():
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0, 1, 2, 3, 4, 5, 6, 7, 8 and 9?
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"""
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limit = 1_000_000
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digits = list(range(10))
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_permutations = list(permutations(digits))
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ans = "".join(str(digit) for digit in _permutations[999_999])
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millionth_permutation = next(islice(permutations(digits), limit - 1, limit))
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ans = "".join(str(digit) for digit in millionth_permutation)
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return ans
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@@ -37,11 +37,17 @@ def compute():
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cycle_length = 0
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ans = 0
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# Because denominators with factor 2 contribute only to the terminating
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# part of a decimal. The repeating behavior comes from the part of the
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# denominator that is coprime with 10, so evens are skipped
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for number in range(3, 1000, 2):
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if number % 5 == 0:
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continue
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p = 1
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while 10**p % number != 1:
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remainder = 10 % number
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while remainder != 1:
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# avoid huge exponentiation
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remainder = (remainder * 10) % number
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p += 1
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if p > cycle_length:
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cycle_length, ans = p, number
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@@ -29,11 +29,10 @@ def compute():
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"""
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ans = set()
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pandigital = ["1", "2", "3", "4", "5", "6", "7", "8", "9"]
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pandigital = [str(number) for number in range(1, 10)]
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for x in range(1, 100):
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for y in range(100, 10_000):
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# product = x * y
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if sorted(str(x) + str(y) + str(x * y)) == pandigital:
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ans.add(x * y)
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@@ -9,11 +9,7 @@ Solution for problem 036 of Project Euler
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https://projecteuler.net/problem=36
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"""
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from project_euler_python.utils import timeit
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def is_palidrome(num):
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return str(num) == str(num)[::-1]
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from project_euler_python.utils import is_palindrome, timeit
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@timeit("Problem 036")
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@@ -28,7 +24,7 @@ def compute():
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ans = 0
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for i in range(1, 1_000_001, 2):
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if is_palidrome(i) and is_palidrome(bin(i)[2:]):
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if is_palindrome(i) and is_palindrome(bin(i)[2:]):
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ans += i
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return ans
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@@ -12,14 +12,6 @@ https://projecteuler.net/problem=41
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from project_euler_python.utils import is_prime, timeit
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def is_pandigital(number):
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number = sorted(str(number))
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check = [str(i) for i in range(1, len(number) + 1)]
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if number == check:
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return True
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return False
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@timeit("Problem 041")
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def compute():
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"""
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@@ -30,8 +22,10 @@ def compute():
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What is the largest n-digit pandigital prime that exists?
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"""
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pandigital = [str(number) for number in range(1, 10)]
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for ans in range(7654321, 1, -1):
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if is_pandigital(ans):
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if sorted(str(ans)) == pandigital[: len(str(ans))]:
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if is_prime(ans):
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return ans
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@@ -36,7 +36,7 @@ def compute():
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"""
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ans = []
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pandigital = ["0", "1", "2", "3", "4", "5", "6", "7", "8", "9"]
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pandigital = [str(number) for number in range(1, 10)]
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for n in permutations(pandigital):
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n_ = "".join(n)
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