Normally a PIC will only handle a maximum voltage of 5 volts. When you said
# Torch voltage is scaled from 120vdc to 1.2vdc.
I thought you were doing this with a voltage divider to bring the 120 down to 1.2, stating that the divider you planned on using was 100 to 1.
So...

I was thinking that being the expected voltages from the torch are 60 to 185 a voltage divider that would bring the 60 volts down to 1 volt for the low end and the 185 high end volts would then be ~3 volts. That is where the divide (scale) by 60 came in.

Many PICs have a 10 bit ADC. Even with a 10bit if you want you only have to use the first 8.
The difference is the resolution. There is also a voltage reference available when using the ADC. VDD can be used or some other voltage less than VDD. Some have a low end VREF but I figure on this one zero volts will be fine.

This all means that if an 8 bit (256 steps)ADC setup is used with VREF set for 4 volts, each step represents ~0.015 volts. In real life with the torch voltage being divided by 60, 0.015 would equal 0.94 volts. The torch voltage would have to change 0.94 volts before the PIC sees the change.

A 10 bit setup works the same but with 1024 steps. Now the PIC can see a 0.23 volt change on the torch.

This could be tightened up by lowering the VREF and/or scaling the 60 to maybe 0.5 volts, but I figure a little cushion would be good on both ends. I may have more than is needed...

Now using all of this. Say you want the torch voltage to be 120. With the 4 volt VREF and the voltage dividing, 4 volts at the ADC is equal to 240 at the torch, so we want to see 2 at the ADC to be good. The ADC will give a value of 512. If the ADC is + or - 512 the torch height will need adjusted.

Let me know how much I confused things