Yes, things are clearing up.

Now, given that eeprom is only storing a BYTE and not a WORD I can understand that generating a RANDOM WORD to try and stuff into the eeprom is pointless, but how is adding 23 any less pointless? If the eeprom is storing a BYTE of 0 and 1's isn't 23 a WORD? I know that 23 is well within the 256 limit, but is there some internal math going on that is saying, "okay, I'm going to convert that 23 to binary to make up my byte then add it to what is there already" that it CAN'T do with 65536 which is beyond the 256 limit?

So, I also assume that to stuff a WORD into eeprom there is a way to separate the low and high bytes and put into two eeprom addresses?

This is getting past the random issue now. I am looking at the inner workings of word and byte.

Further. Is this written correct? You wrote:

Read 0,B0
W0=W0+23
Write 0,B0

As I see this we are just putting the same information back into address 0 without really adding anything to it since the +23 is going on a word variable. Is line two possibly b0=w0+23 then only the lowbyte is being written back?

Bart