Reverse-engineering the number of cases used in an experiment I Sometimes a published article will report experimental results in terms of percentages, but not state how many subjects or cases they used. I have seen ways to estimate these numbers based on common divisors etc., in the reported percentages, but I cannot find any such algorithm right now. Anyone know what I am talking about?
For example, I'm reading a paper where all the accuracy results (correct / $N$) are whole numbers or end in either .6667 or .3333. Clearly the number of cases is a multiple of 3, but what is the total? The values include 34, 84, 60.6667, and 69.3333. Based also on other clues, it seems that $N = 150$ would account for these numbers. But what is the general algorithm?
(PS. I am not asking this on statistics.so because as I remember it is simple combinatorics.)
 A: Just taking your example of $34$, $84$, $60.6667$, and $69.3333$, these are actually percentages, so divide by $100$ to give $0.34$, $0.84$, $0.606667$, and $0.693333$
Now write these as rationals in lowest terms: $\frac{17}{50}$, $\frac{21}{25}$, $\frac{91}{150}$,  $\frac{52}{75}$
The lowest common multiple of the denominators $50,25,150,75$ is $150$, so you are looking for an answer of $150$ or a multiple of $150$.
The problem comes when the sample is reasonably large and the percentages are rounded: for example could $31.4\%$ be $\frac{11}{35}$ or $\frac{16}{51}$ or something else? A possibility is to look at plausible rationals and hope that common factors with those from other percentages or other clues help: for example, seeing $14.3\%$ (perhaps $\frac{1}{7}$) and $30.0\%$ (perhaps $\frac{3}{10})$ as well, might push you towards the $\frac{11}{35}$ option.
One useful trick is to take the difference between two close percentages. Although you might double any rounding error in this process, it could still work.  For example the difference between  $31.4\%$ and  $30.0\%$ is $1.4\%$.  Assuming this represents some number between $0.013$ and $0.015$, then these are the reciprocals of something between $66.66$ and $76.93$ and the answer will be a multiple of some number in this range  or a multiple of this range. 
