math.remainder performs *floating point* remainder. That means it has to convert the arguments to floats first, and there is no float capable of representing either of those two numbers exactly:

    py> '%f' % float(12345678901234567890)
    '12345678901234567168.000000'
    py> '%f' % float(12345678901234567891)
    '12345678901234567168.000000'

That's just the way floats work. They don't have infinite precision, so sometimes they have rounding error.

If you want exact integer remainder, use the `%` operator:

    py> 12345678901234567890 % 3
    0
    py> 12345678901234567891 % 3
    1

See the tutorial and the FAQs

https://docs.python.org/3/tutorial/floatingpoint.html

https://docs.python.org/3/faq/design.html#why-are-floating-point-calculations-so-inaccurate