This is an interesting question.  As far as your example goes, I'd say
that using a 6k resister and a 2k would yield 1/4 the voltage at the
junction.  The 12k and 2k would be 1/7 the voltage.

To answer the question though, there's little point to trying to get
integer multiples of resistors to build nice bridges.  The reason the
values jump oddly from one value to the next is simply because, in the
case of 5% resistors, 5% of 1.3 and 1.5 are large enough that having a
1.4 would really be redundant.

In the 5% resistor world there are really only 24 values:
10, 11, 12, 13, 15, 16, 18, 20, 22, 24, 27, 30,
33, 36, 39, 43, 47, 51, 56, 62, 68, 75, 82, 91
Which are close to 10% away from each other.  You create all the
standard 5% resistor values by multiplying by some magnitude of 10.

I don't have my digikey catalog handy, otherwise I'd look in its list of
1% resistors to see what those standard values are.  However, there are
good reasons for having the list we have.

Currently the above 24 values, when factored, contain 11 of the 25 prime
numbers from 2 to 100:
2, 3, 5, 7, 11, 13, 17, 31, 41, 43, 47
Missing the following:
19, 23, 29, 37, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

This means that it's possible to have very few resistors to represent a
divider of any combination of those primes.  You want a divide by four?
Use the 10 and 30.  Divide by 8? 56 and 91 are divisible by 7.  91 is
easier to use since its factors are 7 and 13, so parallel the 91 with 13
and you'll have a 7 ohm resister.  Use a 1 ohm to get your 1/8th.

3/8?  Put two 30s and a 20 in series and use one of the junctions.

In the end it's all about factors, whether you're using the 1% or 5%.
Since there are more options among the 1% you probably have a wider
range of choices, and it becomes more complex - I don't think there'll
be an 'easy' in your head calculation that will give you what you want,
but if you practise I'm sure you can learn the factors of the resistors
and do the calculation in your head.

Interesting trivia:
13 resistors are divisible by 2 (have factors of 2)
11 have factors of 3
5 have factors of 5
2 have factors of 7
3 have factors of 11
3 have factors of 13
2 have factors of 17
1 each have factors of 31, 41, 43, 47

So you can represent any resistance which only has factors of those
primes with a 'small' number of resistors.  In theory you could
represent any resistance with an infinite number of resistors.

Probably doesn't answer your question, but I'm pretty bored right this
minute, and I've always found use for factorization.  It would be
interesting to generate a table of fractions to 1/8, but once you start
using more than a few resistors the tolerance just gets to large.  It
probably wouldn't be useful to do much finer than 1/8, even with 1%
resistors.

-Adam

Ken Pergola wrote:

>Sorry guys -- I made a mistake in the text -- I meant 12.0 K ohms for the
>upper leg instead of 6.0 K ohms. Sorry for the confusion and brain lapse.
>Corrected text below.
>
>Regards,
>
>Ken Pergola
>
>
>
>
>
>
>
>
>Hi,
>
>Kind of a strange question but very basic:
>
>Without pulling out the calculator or spreadsheet and doing it manually,
>what is the quickest way to come up with resistors that are exact *integer*
>multiples of each other (from the 1% standard value list)?
>
>Let me elaborate: In other words, say I want to make a 'divide by 4' voltage
>divider using 1% resistors and *only* two resistors (no building of new
>resistance values allowed).
>
>For example, let me pick 2.00 K ohms as the bottom leg of the voltage
>divider.  That means I need 12.00 K ohms as the upper leg of the voltage
>divider. Well, 12.00 K ohms is not a standard value, so I need to try again:
>So I'll try the next standard value for the bottom leg -- 2.05 K ohms
>(standard value). This would require 12.3 K ohms as the top resistor which
>is not a standard value. Try again...ad nauseam...
>
>Some values are exact integer multiples of each other -- what's the quickest
>and easiest way to identify them?
>Does anyone have something in their 'toolbox' they'd like to share?
>
>Does anyone know of any existing chart, software, or Excel spreadsheet that
>already exists to do this. I realize this is simple math and I could go off
>and do this myself, but it would be nice having something that already
>exists. I'm sure a lot of people have had a need for this at one time or
>another.
>
>Thanks very much.
>
>Best regards,
>
>Ken Pergola
>
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>
>
>
>

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