I did some research on what three colors a 6-bit monitor can't do. Here's what I found, partly based on some research about what these bits actually mean, and a dash of guesswork added for flavor:
Each RGB color is represented by a byte, which is 8 bits of information. Thus if the byte's value is 00000000 in binary (00 in hex), then it's black. If the byte's value is 11111111, (FF in hex), then it's white. This is how the color is stored in the computer.
So if we think of this in terms of fractions, we want 256 gradations, evenly spaced, going from 0 to 1. This actually means 256 shades that are spaced 1/255 apart, the 255 in the denominator (not 256) because you're counting from 1 to 255, along with 0, to get the 256 colors. Thus the shades are:
0/255
1/255
2/255
3/255
and so on of the maximum brightness (255/255).
Now, say you want to use 6 bits to represent this. You could try doing the same thing again, which means representing all shades as such:
0/63
1/63
2/63
3/63
and so forth. Once again, it's out of 63 (not 64) because you're counting from 1 to 63, then add on the 0, for the 64 different shades.
The problem with doing this is that 63 does not divide into 255 evenly (they have a 21:85 ratio, not a 1:4 ratio, even though it's 64 shades vs 256 shades) -- in order to truly represent say 148/255 brightness, you'd take the two closest values (36/63 brightness and 37/63 brightness), and make sure that they're distributed throughout the picture such that the average of all of them becomes 148/255 brightness. But this means that 37/85 of all pixels would be at 36/63 brightness, and 48/85 of all pixels would be set to 37/63 brightness. Much too unwieldy, considering you're basically trying to dither on an 85-pixel basis.
No, a much simpler method is to space out the color gradations as 0/255, 4/255, 8/255, 12/255, etc. This way, you only need a 4-pixel dithering. If you want to represent say 193/255, then you just take the the values closest to it (192/255 and 196/255). Then, for every 4 pixels, make sure that three of them are at 192/255 and one of them is at 196/255, and thus they average out to 193/255.
The problem then is that this gradation system only goes up to 252/255 (which is the highest value). In other words, the missing three shades are 253/255, 254/255, and 255/255. Thus, 6-bit color can only produce 253 shades this way, but it's a fair tradeoff, since you basically get 6-bit (faster) performance, and you get the vast majority of the possible colors, without complicated dithering schemes.
Now, if you notice, many computers nowadays support 32-bit color (true color). What this actually is, is simply the three 8-bit values for red green blue, plus a dummy 8-bit. The advantage of doing this is that each color takes up 4 bytes, which is convenient in terms of memory addressing. So while 24-bit (3 bytes) would be more efficient (take up less memory), 32-bit (4 bytes) is a lot quicker, at the expense of more memory needed. However, assuming you want to do some fancy dithering, you should be able to use that dummy byte to produce at least 10-bit color using 8-bit plus dithering. While this means 1.06 billion possible colors (1021 x 1021 x 1021) instead of the current 16.7 million, I'm pretty sure most people have figured that we already can't tell the difference between adjacent shades as it is, so I don't think people have done much research in this area. But, theoretically, it is possible.
Chuck Hsiao
Amptron
If I would have shot you when I had the chance, I would be out by now 
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