From c99d4e2e180ac519c405e7bb7f799eb479aa0cea Mon Sep 17 00:00:00 2001 From: mddicicco Date: Wed, 15 Jun 2022 14:27:54 -0500 Subject: java and python solutions --- challenge-169/michael-dicicco/python/ch-1.py | 24 ++++++++++++++++++ challenge-169/michael-dicicco/python/ch-2.py | 38 ++++++++++++++++++++++++++++ 2 files changed, 62 insertions(+) create mode 100644 challenge-169/michael-dicicco/python/ch-1.py create mode 100644 challenge-169/michael-dicicco/python/ch-2.py (limited to 'challenge-169/michael-dicicco/python') diff --git a/challenge-169/michael-dicicco/python/ch-1.py b/challenge-169/michael-dicicco/python/ch-1.py new file mode 100644 index 0000000000..102d3eaeb5 --- /dev/null +++ b/challenge-169/michael-dicicco/python/ch-1.py @@ -0,0 +1,24 @@ +def is_brilliant(factors): + """a brilliant number has only two prime factors of the same length""" + return len(factors) == 2 and len(str(int(factors[0]))) == len(str(int(factors[1]))) + + +def prime_factors(n): + output = [] + + while n % 2 == 0: + output.append(2) + n /= 2 + + for i in range(3, int(n ** 1 / 2 + 1), 2): + while n % i == 0: + output.append(i) + n /= i + + if n > 2: + output.append(int(n)) + return output + + +if __name__ == '__main__': + print(", ".join([str(i) for i in range(4, 300) if is_brilliant(prime_factors(i))])) diff --git a/challenge-169/michael-dicicco/python/ch-2.py b/challenge-169/michael-dicicco/python/ch-2.py new file mode 100644 index 0000000000..8052d94e24 --- /dev/null +++ b/challenge-169/michael-dicicco/python/ch-2.py @@ -0,0 +1,38 @@ +def is_achilles(factors, given): + """Achilles was powerful but imperfect""" + for factor in factors: + """a powerful number is divisible by the squares of its prime factors""" + if not given % factor ** 2 == 0: + return False + for i in range(2, 10): + """a number is a perfect power if it has any integer roots""" + if nth_root(given, i) % 1 == 0: + return False + """the number is powerful and imperfect, like Achilles""" + return True + + +def nth_root(some_number, n): + """raising a number to a fractional power of 1/n is the same as taking the nth root""" + return some_number ** (1 / n) + + +def prime_factors(n): + output = [] + + while n % 2 == 0: + output.append(2) + n /= 2 + + for i in range(3, int(n ** 1 / 2 + 1), 2): + while n % i == 0: + output.append(i) + n /= i + + if n > 2: + output.append(int(n)) + return output + + +if __name__ == '__main__': + print(", ".join([str(i) for i in range(72, 1801) if is_achilles(prime_factors(i), i)])) -- cgit