project-euler/solutions/python/045.py at master · ThomasJackDalby/project-euler · GitHub

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# COMPLETED
"""
    Problem 45
    ==========

    Triangle, pentagonal, and hexagonal numbers are generated by the following
    formulae:

    Triangle     T[n]=n(n+1)/2    1, 3, 6, 10, 15, ...
    Pentagonal   P[n]=n(3n−1)/2   1, 5, 12, 22, 35, ...
    Hexagonal    H[n]=n(2n−1)     1, 6, 15, 28, 45, ...

    It can be verified that T[285] = P[165] = H[143] = 40755.

    Find the next triangle number that is also pentagonal and hexagonal.

    Answer: 30dfe3e3b286add9d12e493ca7be63fc

"""
from common import check

PROBLEM_NUMBER = 45
ANSWER_HASH = "30dfe3e3b286add9d12e493ca7be63fc"

tn = 285
pn = 165
hn = 143

triangle = 0
pentagonal = 0
hexagonal = 0
while True:
    min_index = min(enumerate([triangle, pentagonal, hexagonal]), key=lambda a: a[1])[0]
    if min_index == 0:
        tn += 1
        triangle = int(tn * (tn + 1) / 2)
    elif min_index == 1:
        pn += 1
        pentagonal = int(pn * (3 * pn - 1) / 2)
    elif min_index == 2:
        hn += 1
        hexagonal = int(hn * (2 * hn - 1))
    if triangle == pentagonal == hexagonal:
        print(f"T[{tn}] = P[{pn}] = H[{hn}] = {triangle}")
        check(triangle, PROBLEM_NUMBER, ANSWER_HASH)
        exit()