clementi

products = set()

for pair in [(a, b) for a in range(2, 101) for b in range(2, 101)]:
    products.add(pair[0] ** pair[1])

print len(products) 

clementi

from itertools import *

pairs = product(range(1, 100), range(1, 100))
exponentials = imap(lambda pair: pair[0] ** pair[1], pairs)
strings = imap(lambda exponential: str(exponential), exponentials)
sums = imap(lambda string: sum(map(lambda digit: int(digit), string)), strings)

print max(sums) 

clementi

current = 0
step = 2

sum = 0
limit = 1001 ** 2

spiral = range(1, limit + 1)
while current < limit:
    for i in range(4):
        if current > limit:

clementi

import operator

def factorial(n):
    return 1 if n == 0 else reduce(operator.mul, range(1, n + 1))

def combo(n, r):
    return factorial(n) / (factorial(r) * factorial(n - r))

count = 0

clementi

digits = ""

for n in range(1, 1000001):
    digits += str(n)

product = 1
for i in range(7):
    product *= int(digits[10 ** i - 1])

print product

clementi

(defn prime-factorization-aux [n candidate]
  (if (= n 1)
	'()
	(if (= (mod n candidate) 0)
	  (cons candidate (prime-factorization-aux (/ n candidate) candidate))
	  (prime-factorization-aux n (+ candidate 1)))))

(defn prime-factorization [n]
  (prime-factorization-aux n 2))

clementi

def prime_factorization(n):
    candidate = 2
    while n > 1:
        factors = []
        while n % candidate == 0:
            factors.append(candidate)
            n /= candidate
        if factors:
            yield factors
        candidate += 1

clementi

def triangle_numbers():
    n = 1
    while True:
        yield n * (n + 1) / 2
        n += 1

def prime_factorization(n):
    candidate = 2
    while n > 1:
        factors = []

clementi

def is_even(n):
    return n % 2 == 0

def collatz_length(n):
    length = 0
    while n > 1:
        length += 1
        if is_even(n):
            n /= 2
        else:

clementi

from itertools import permutations

MILLIONTH = 10 ** 6 - 1

print reduce(lambda x, y: str(x) + str(y), list(permutations(range(10), 10))[MILLIONTH])