def ans():
# all hexagon numbers are also triangle numbers
i: int = 144 # we should find the next after 143
while True:
if is_int(hex_and_pent_number(i)):
return polygonal_number(i, 6)
i += 1
def apply_step_backwards(number: int, step: str) -> Optional[int]:
"""
:param number: a result number
:param step: the step that was taken
:return: the origin number
"""
if step == "D":
return 3 * number
if step == "U":
res = (3 * number - 2) / 4
if is_int(res) and int(res) % 3 == 1:
return int(res)
if step == "d":
res = (3 * number + 1) / 2
if is_int(res) and int(res) % 3 == 2:
return int(res)
return None
def ans():
count_list: Dict[int, int] = defaultdict(int)
# go over all possible lengths
for a in range(1, 500):
ap2 = a ** 2
for b in range(1, a):
c = sqrt(ap2 + b ** 2)
if is_int(c):
# we found a triple
c = int(c)
p = a + b + c
if p <= 1000:
count_list[p] += 1
return max(count_list, key=lambda k: count_list[k])
def is_polygonal_number(num: int, shape: int) -> bool:
# n=(sqrt(8x(s-2)+(s-4)^2)+s-4)/2
n = (math.sqrt(8 * num * (shape - 2) +
(shape - 4)**2) + shape - 4) / (2 * (shape - 2))
return is_int(n)
def ans():
for i in range(1, 100000):
for j in range(1, i):
if is_int(pent_difference(i, j)) and is_int(pent_sum(i, j)):
return polygonal_number(int(pent_difference(i, j)), 5)