def find_chain(n):
ans = 1
while n not in [169, 871, 872, 145] and sum_digits(n, lambda x: factorial(x)) != n:
n = sum_digits(n, lambda x: factorial(x))
if n < limit:
visited[n] = 1
ans += 1
if n == 169:
return ans + 2
elif n == 871 or n == 872:
return ans + 1
else:
return ans
def problem_92(r, c89=0):
for x in xrange(1, r):
while x not in [1, 89]:
x = sum_digits(x, 2)
if x == 89:
c89 += 1
return c89
def compute():
limit = 10000000
m = [0] * limit
m[0], m[1], m[89] = 2, 2, 1
for i in range(len(m)):
if m[i] == 0:
sequence = set()
while m[i] == 0:
sequence.add(i)
i = sum_digits(i, lambda x: x**2)
for s in sequence:
m[s] = m[i]
return len([s for s in range(limit) if m[s] == 1])
# A googol (10^100) is a massive number: one followed by one-hundred zeros; 100^100 is almost unimaginably
# large: one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is
# only 1.
#
# Considering natural numbers of the form, a^b, where a, b < 100, what is the maximum digital sum?
from euler import sum_digits
max_sum = 0
for a in range(100):
for b in range(100):
max_sum = max(max_sum, sum_digits(a**b))
print(max_sum)
# the whole thing as a one liner... ugh, readability!
print(
max([
sum([int(d) for d in str(a**b)]) for a in range(100)
for b in range(100)
]))
# getting crazy now
import itertools
print(max( map(lambda x: sum([int(d) for d in str(x)]), \
map(lambda n: n[0] ** n[1], itertools.product(range(100), repeat=2)))))
# Euler 65
# convergents for e are [2; 1,2,1, 1,4,1, 1,6,1, 1,8,1, ... 1,2k,1]
# So the first 10 terms are 2, 3, 8/3, 11/4, 19/7, 82/32 ... 1457/536.
# Find the sum of digits in the numerator of the 100th convergent
# of the continued fraction for e.
from fractions import Fraction
from math import ceil
from euler import sum_digits, convert_convergents_to_fraction
def generate_convergents_for_e(n):
result = [2]
groups = ceil((n - 1) / 3)
for k in range(1, groups + 1):
result += [1, 2 * k, 1]
return result[0:n]
if __name__ == '__main__':
for i in [100]: #range(1,11):
convergents = generate_convergents_for_e(i)
convergent_fraction = convert_convergents_to_fraction(convergents)
sum_digits_numerator = sum_digits(convergent_fraction.numerator)
print("{}: {} (sum digits numerator: {})".format(
i, convergent_fraction, sum_digits_numerator))
def compute():
limit = factorial[9] * 7
return sum([n for n in range(3, limit) if sum_digits(n, lambda x: factorial[x]) == n])
'''
Created on Jan 10, 2017
The number 512 is interesting because it is equal to the sum of its digits raised to some power: 5 + 1 + 2 = 8, and 8^3 = 512.
Another example of a number with this property is 614656 = 28^4.
We shall define an to be the nth term of this sequence and insist that a number must contain at least two digits to have a sum.
You are given that a2 = 512 and a10 = 614656.
Find a30.
@author: mstackpo
'''
from datetime import datetime
start = datetime.now()
from euler import sum_digits
pows = set()
for base in range(2, 100):
for exp in range(2, 20):
x = base**exp
s = sum_digits(x)
if x > 10 and s == base:
pows.add(x)
print(sorted(pows)[29])
end = datetime.now()
print("runtime = %s" % (end - start))
def compute():
return max([sum_digits(a**b) for a in range(1,100) for b in range(1,100)])
def compute():
return sum_digits(2 ** 1000)
'''
Created on Jan 2, 2017
A googol (10^100) is a massive number: one followed by one-hundred zeros; 100^100 is almost unimaginably large:
one followed by two-hundred zeros. Despite their size, the sum of the digits in each number is only 1.
Considering natural numbers of the form, ab, where a, p < 100, what is the maximum digital sum?
@author: mstackpo
'''
from datetime import datetime
start = datetime.now()
from euler import sum_digits
max_sum = 0
for a in range(100):
for p in range(100):
max_sum = max(sum_digits(a**p), max_sum)
print(max_sum)
end = datetime.now()
print("runtime = %s" % (end - start))