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313 def factors(number): 314 """ 315 316 Find all of the factors of a number and return it as a list 317 318 number: 319 The number to find the factors for 320 321 """ 322 323 factors = [] 324 for i in range(1, number + 1): 325 if number % i == 0: 326 factors.append(i) 327 return factors

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30 def get_factors(n): 31 """[summary] 32 33 Arguments: 34 n {[int]} -- [to analysed number] 35 36 Returns: 37 [list of lists] -- [all factors of the number n] 38 """ 39 40 def factor(n, i, combi, res): 41 """[summary] 42 helper function 43 44 Arguments: 45 n {[int]} -- [number] 46 i {[int]} -- [to tested divisor] 47 combi {[list]} -- [catch divisors] 48 res {[list]} -- [all factors of the number n] 49 50 Returns: 51 [list] -- [res] 52 """ 53 54 while i * i <= n: 55 if n % i == 0: 56 res += combi + [i, int(n/i)], 57 factor(n/i, i, combi+[i], res) 58 i += 1 59 return res 60 return factor(n, 2, [], [])

34 def inverse_factorial(number, round_up=True): 35 ''' 36 Get the integer that the factorial of would be `number`. 37 38 If `number` isn't a factorial of an integer, the result will be rounded. By 39 default it'll be rounded up, but you can specify `round_up=False` to have 40 it be rounded down. 41 42 Examples: 43 44 >>> inverse_factorial(100) 45 5 46 >>> inverse_factorial(100, round_up=False) 47 4 48 49 ''' 50 assert number >= 0 51 if number == 0: 52 return 0 53 elif number < 1: 54 return int(round_up) # Heh. 55 elif number == 1: 56 return 1 57 else: 58 current_number = 1 59 for multiplier in itertools.count(2): 60 current_number *= multiplier 61 if current_number == number: 62 return multiplier 63 elif current_number > number: 64 return multiplier if round_up else (multiplier - 1)

241 def closest_square_factors(integer, larger_first=True): 242 243 factors = [] 244 root = ceil(sqrt(integer)) 245 for i in range(root, integer+1): 246 if integer % i == 0: 247 return (i, int(integer / i)) if larger_first else (int(integer / i), i) 248 249 return (1, integer)

17 def factor_modulus(n, d, e): 18 """ 19 Efficiently recover non-trivial factors of n 20 See: Handbook of Applied Cryptography 21 8.2.2 Security of RSA -> (i) Relation to factoring (p.287) 22 http://www.cacr.math.uwaterloo.ca/hac/ 23 """ 24 t = (e * d - 1) 25 s = 0 26 27 while True: 28 quotient, remainder = divmod(t, 2) 29 30 if remainder != 0: 31 break 32 33 s += 1 34 t = quotient 35 36 found = False 37 38 while not found: 39 i = 1 40 a = random.randint(1,n-1) 41 42 while i <= s and not found: 43 c1 = pow(a, pow(2, i-1, n) * t, n) 44 c2 = pow(a, pow(2, i, n) * t, n) 45 46 found = c1 != 1 and c1 != (-1 % n) and c2 == 1 47 48 i += 1 49 50 p = fractions.gcd(c1-1, n) 51 q = n // p 52 53 return p, q

746 def prime_number_factorisation(n): 747 if n < 2: 748 return [n] 749 i = 2 750 factors = [] 751 while i * i <= n: 752 if n % i: 753 i += 1 754 else: 755 n //= i 756 factors.append(i) 757 if n > 1: 758 factors.append(n) 759 return factors