Exemplo n.º 1
0
def is_stupid(n, d):
    ndigits = get_digits(n)
    ddigits = get_digits(d)
    if ndigits[1] == ddigits[0]:
        if ddigits[1] != 0:
            if Fraction(ndigits[0], ddigits[1]) == Fraction(n, d):
                return True
    return False
Exemplo n.º 2
0
def is_stupid(n, d):
    ndigits = get_digits(n)
    ddigits = get_digits(d)
    if ndigits[1] == ddigits[0]:
        if ddigits[1] != 0:
            if Fraction(ndigits[0], ddigits[1]) == Fraction(n, d):
                return True
    return False
Exemplo n.º 3
0
def rotations(n):
    "Return the set of rotations of the given n. e.g. 37 outputs {37, 73}."
    digits = get_digits(n)
    result = set()
    for i in range(0, len(digits)):
        result.add(combine_digits(digits[i:] + digits[0:i]))
    return result
Exemplo n.º 4
0
def rotations(n):
    "Return the set of rotations of the given n. e.g. 37 outputs {37, 73}."
    digits = get_digits(n)
    result = set()
    for i in range(0, len(digits)):
        result.add(combine_digits(digits[i:] + digits[0:i]))
    return result
Exemplo n.º 5
0
def find_curious_numbers():
    # Precompute factorials
    facts = [factorial(x) for x in range(0, 10)]

    # Numbers obeying this rule cannot be larger than 7 digits, because
    # 9! * 8 is 2903040, a 7-digit number.  (An 8 digit number can never
    # produce itself in this way.)
    limit = facts[9] * 7 + 1
    is_curious = lambda n: sum([facts[x] for x in get_digits(n)]) == n
    return filter(is_curious, xrange(10, limit))
Exemplo n.º 6
0
def get_permutations(x):
    "Return a list of all permutations of digits of x (as integers)."
    digits = get_digits(x)
    if len(digits) == 1:
        return digits
    result = []
    for i in range(0, len(digits)):
        combine = lambda d, e: int("".join(map(str, d)) + "".join(map(str, e)))
        result += map(lambda x: x*10 + digits[i],
                get_permutations(combine(digits[:i], digits[i+1:])))
    return list(set(result)) # Hacky duplicate filter
Exemplo n.º 7
0
def get_permutations(x):
    "Return a list of all permutations of digits of x (as integers)."
    digits = get_digits(x)
    if len(digits) == 1:
        return digits
    result = []
    for i in range(0, len(digits)):
        combine = lambda d, e: int("".join(map(str, d)) + "".join(map(str, e)))
        result += map(lambda x: x * 10 + digits[i],
                      get_permutations(combine(digits[:i], digits[i + 1:])))
    return list(set(result))  # Hacky duplicate filter
Exemplo n.º 8
0
Arquivo: 4.py Projeto: Anks/euler
def is_palindrome(number):
    """ Return true if the number is a palindrome.
    
    Start dividing the number by 10, 100, 1000, etc to get the digits
    from the right-hand side of the number.
    """

    digits = get_digits(number)

    palindrome = True
    for (k, v) in zip(digits, reversed(digits)):
        if k != v:
            palindrome = False
            break

    return palindrome
Exemplo n.º 9
0
def same_digits(numbers):
    expect = sorted(get_digits(numbers[0]))
    for i in range(1, len(numbers)):
        if sorted(get_digits(numbers[i])) != expect:
            return False
    return True
Exemplo n.º 10
0
def is_narcissistic(n, exp):
    value = 0
    digits = get_digits(n)
    for digit in digits:
        value += digit**exp
    return value == n
Exemplo n.º 11
0
#!/usr/bin/env python

"""
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 digits import get_digits

if __name__ == "__main__":
    maxsum = 0
    for a in range(0, 100):
        for b in range(0, 100):
            s = sum(get_digits(a ** b))
            if s > maxsum:
                maxsum = s
                maxa = a
                maxb = b
    print "%d^%d = %d" % (maxa, maxb, maxa ** maxb)
    print "Sum: %d" % maxsum
Exemplo n.º 12
0
#!/usr/bin/env python

"""
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 digits import get_digits

if __name__ == '__main__':
    maxsum = 0
    for a in range(0, 100):
        for b in range(0, 100):
            s = sum(get_digits(a**b))
            if s > maxsum:
                maxsum = s
                maxa = a
                maxb = b
    print "%d^%d = %d" % (maxa, maxb, maxa**maxb)
    print "Sum: %d" % maxsum
Exemplo n.º 13
0
Arquivo: 16.py Projeto: Anks/euler
#!/usr/bin/env python
# -*- coding: utf-8 -*-

import sys
sys.path.append('../lib')
from digits import get_digits

print sum(reversed(get_digits(2 ** 1000)))
Exemplo n.º 14
0
from digits import get_digits
from fractions import Fraction
from itertools import count


def compute_continued_fraction(seq, limit):
    "Expand the given continued fration sequence to the nth convergent."
    a = seq.next()
    if limit == 1:
        result = Fraction(a)
    else:
        result = a + Fraction(1, compute_continued_fraction(seq, limit - 1))
    return result


def e_seq():
    "The sequence of values given in the problem to compute e."
    yield 2
    for n in count(2):
        if n % 3 == 0:
            yield n / 3 * 2
        else:
            yield 1


if __name__ == '__main__':
    result = compute_continued_fraction(e_seq(), 100)
    print '100th convergent:', result
    print 'Numerator digit sum:', sum(get_digits(result.numerator))
Exemplo n.º 15
0
Find the sum of digits in the numerator of the 100th convergent of the
continued fraction for e.
"""

from digits import get_digits
from fractions import Fraction
from itertools import count

def compute_continued_fraction(seq, limit):
    "Expand the given continued fration sequence to the nth convergent."
    a = seq.next()
    if limit == 1:
        result = Fraction(a)
    else:
        result = a + Fraction(1, compute_continued_fraction(seq, limit-1))
    return result

def e_seq():
    "The sequence of values given in the problem to compute e."
    yield 2
    for n in count(2):
        if n % 3 == 0:
            yield n/3 * 2
        else:
            yield 1

if __name__ == '__main__':
    result = compute_continued_fraction(e_seq(), 100)
    print '100th convergent:', result
    print 'Numerator digit sum:', sum(get_digits(result.numerator))