Hell, I don’t even think all of E12 is necessary. I’ll stick to E6 most of the time.
Sense resistor? 0.1 ohm.
Resistor for an LED: 100 ohm
Pull up resistor: 10k
Bias resistor for some mosfet gate: 10M
Voltage divider to measure the battery voltage with an ADC: two 100k resistors.
It's super rare I need anything else. I hate fiddling about with switching the reels on the pick'n'place anyway.
100 Ω sounds like way too much current for modern LEDs. I often end up using 100 kΩ especially for green LEDs. They are very visible under indoor lighting even with 1 MΩ and 3.3 V supply.
For pulling down FETs, you want something in the range of 10 kΩ. 10 MΩ sounds way too high, which makes your circuit sensitive to being touched or affected by moisture, especially if there are near by components connected to the power rail.
My digital electronics grab bag consist of 22 mΩ for sensing, 100 kΩ for battery voltage divider, 22 kΩ for one of the 3.3 V buck converter feedback dividers, 10 kΩ for everything else like I2C pulling.
In fact, just hold an LED between your fingers in a dark enough room and you'll sometimes see them glow from stray magnetic fields inducing enough current in your body to light them.
Of course as stated by another comment, our eyes are also incredible, and can pick up very faint amount of light.
Yeah, that's why I can read a book by the blue LEDs on my alarm clock...
Depending on the colour bleed though. It may wipe out all visibility of the clock numbers.
edit: Actually, I'm not so sure anymore that the tolerances would add up in series... I should probably just look this stuff up, since I'm not awake enough to intuit correctly, I think.
Suppose you had a pair of 105 Ohm resistors that are nominally 100 Ohm. In parallel you get:
1/(1/105 + 1/105) = 105/2 = 52.5 Ohm (5% over expected 50 Ohm)
If one is over nominal and the other is under, they'll cancel out for the most part:
1/(1/105 + 1/95) = 49.875 Ohm (0.25% under expected 50 Ohm)
22 - 5% = 20.9
47 - 5% = 44.65
Actual resistance in series: 65.55
Nominal resistance in series: 69
69 - 5% = 65.55
So the combination of the components still appears to maintain the 5% tolerance.
In either case though, the tolerance divides.
The combined tolerance becomes more accurate the more resistors there are in total, whether parallel or serial. The highs and lows, and the chances of high or low, cancel each other out and you get a final actual value that is closer to the nominal statistical center of the bell curve the more individual parts there are. (same goes for other components, just resistors are simpler to talk about because their behavior is simple.)
In series, a single 10K might really be 9K or 11K, but if you chain 10 10Ks in series, you don't get a "maybe 90K maybe 110K". That is technically possible but statistics means that what what you actually get is if there was N% chance that a given 10K is 9K or 11K, the there is 1/10th of N% chance (or less, I bet the actual equation is more complicated) that the chain of 10 is 90K or 110K. If the individual 10Ks were 10%, then you get 100K with something like 1% tolerance.
(except also in reality, there is such a thing as batches, where all the parts in a given batch are all high or low the same way, because the process was drifting a little high or low while it was cranking out thousands of them that hour. So Ideally your 10 individuals need to come from 10 different batches or even 10 different manufacturers if that were practical or in a pure math world.)
In parallel, the statistical division is the same though the value centers on the value/N rather than value*N. 10 10% 10Ks in parallel = 1 1% 1k
https://en.wikipedia.org/wiki/Central_limit_theorem
I’m a bit tired otherwise I’d write something more rigorous. There are different ways the central limit theorem is expressed and proved here—there are more powerful ways to state it that require more complicated proofs, and there are simpler versions that are simple to prove.
A simple version will suffice here. Treat the resistors as iid variables with finite variance σ². When you average them, the variance of the average is σ²/n. More or less… this means that if your resistors are ±10%, and you have 16 of them, you get something with (fuzzy math) ±10%/√16 = ±2.5%.
There’s a lot of unstated assumptions in what I just wrote. But you’ll see the “grows proportional to √n” a lot in stats.