Question

I am trying to write a program that takes a number with a single leading integer and an arbitrary number of trailing zeros and then prints all possible combinations of two factors.

ie.

100

factors are 2^2, 5^2

so the program would print:

(2,50),(4,25),(5,20)

or

600

factors are 2^3,3,5^2

(2,300),(4,150),(8,75),(3,200),(5,120),(25,24),(6,100),(12,50),(15,40),(30,20),(60,10)

…I think that’s all of them? Is it? I could use a program to check…

import itertools

facts=[[2,2,2],[3],[5,5]]
for n in itertools.product(*facts)
    print(n)

I see that I am using this incorrectly but that was my first stab at it.

This just gives (2,3,5) ten times.

I want something like (2) * (2,3,5,5) and (2,2) * (3,5.5) and the like…

Answer 1

To generate all factors of a number given its prime factors:

#!/usr/bin/env python
import itertools, operator

def all_factors(prime_dict):
    series = [[p**e for e in range(maxe+1)] for p, maxe in prime_dict.items()]
    for multipliers in itertools.product(*series):
        yield reduce(operator.mul, multipliers)

Example

prime_dict = {2:3, 3:1, 5:2}
L = sorted(all_factors(prime_dict))
number_of_divisors = reduce(lambda prod, e: prod*(e+1), prime_dict.values(),1)
assert len(L) == number_of_divisors
# -> [1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24,
#     25, 30, 40, 50, 60, 75, 100, 120, 150, 200, 300, 600]

To produce pairs:

n, isodd = divmod(len(L), 2)
print(zip(L[:n], reversed(L[n + isodd:])))
if isodd: # number is perfect square
   print((L[n], L[n]))

Output

[(1, 600), (2, 300), (3, 200), (4, 150), (5, 120), (6, 100),
 (8, 75), (10, 60), (12, 50), (15, 40), (20, 30), (24, 25)]

It works for small numbers. You could use it to test your solution that could take into account the special form of your numbers: x00000...