def pe12(n=500): '''What is the value of the first triangle number to have over five hundred divisors?''' m = 1 while True: if utils.n_div(utils.triangle(m)) > n: return utils.triangle(m) break m += 1
def part2() -> int:
m = mean(DATA)
pos_low = int(m)
pos_high = int(m + 1)
return min(
sum(utils.triangle(abs(i - pos)) for i in DATA)
for pos in (pos_high, pos_low))
"""
The nth term of the sequence of triangle numbers is given by, tn = (n * (n + 1)) / 2; so the first ten triangle numbers are:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
By converting each letter in a word to a number corresponding to its alphabetical position and adding these values
we form a word value. For example, the word value for SKY is 19 + 11 + 25 = 55 = t10. If the word value is a
triangle number then we shall call the word a triangle word.
Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common
English words, how many are triangle words?
"""
from utils import triangle
triangles = [triangle(n) for n in range(1,20)]
f = open("Data/words.txt", "r")
words = f.readlines()[0].replace("\"", "").split(",")
print len([word for word in words if sum(ord(c) - 64 for c in word) in triangles])
def problem_twelve():
for tri in triangle():
num_divisors = len(factors(tri))
print tri, num_divisors
if num_divisors > 500:
return tri
while x % prime == 0:
if not prime in factors:
factors[prime] = 0
factors[prime] += 1
x //= prime
if x == 1:
break
return [(prime,factors[prime]) for prime in factors if factors[prime] > 0]
def factor_count(x):
factors = prime_factors(x)
if len(factors) == 0:
return 2
total = 1
for prime,pow in factors:
total *= pow + 1
return total
if __name__ == "__main__":
for i in xrange(1,200000):
num = triangle(i)
#print "Num", num
count = factor_count(num)
#print num, ":", count
if count > 500:
print num
break
def problem_twelve():
for tri in triangle():
num_divisors = len(factors(tri))
print tri, num_divisors
if num_divisors > 500:
return tri