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Try the following code, which will actually returns all the factors, efficiently and very quickly of a number "n".
from functools import reducedef factors(n): return set(reduce(list.__add__, ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
from functools import reduce
def factors(n):
return set(reduce(list.__add__,
([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
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