Confused about function of resistor in this circuit....

The circuit below is the digital input section of a commercial I/O board. Manufacturer says inputs will work from 400V down down to 24V (guaranteed) perhaps as low as 12V. I am perplexed as to how one can obtain that wide of a range from a fixed set of resistances. Could someone explain the function of the resistor in parallel with the opto diode (R15 for instance) My thought is that once the diode is forward biased, the 12K resistor ceases to have much impact on the circuit.
Am I missing something obvious here? it seems too simple.
PS, i believe the PIC is just a input stretcher to allow for AC signals .
Thanks.

The datasheet for the PC815 opto says that its minimum current transfer ratio is 6 times and when you calculate the input current then the circuit will work with an input of 12V to 400V.

The 12k resistors might protect the IR LED from reverse leakage current in the rectifiers.

The 12 k resistors will make sure that the LEDs won't turn on until the voltage is above about 5 V. Also the 12 k resistors mean that small leakage currents won't turn on the LEDs.

It's also the bottom half of a potential divider (who's ratio is about 4 to 1), so increases the low input voltage.