Minimum number of elements needed for a given average Let's say something can be given a rating from 1 to 10 (only integer numbers). When we see a rating like 7.00 it usually looks like it was the result of only one rating (it could be the result of any number of ratings but it could be only one rating). Similarly, 4.50 could be calculated from just two ratings but not from just one.
Is there a way to get that minimum number for any given average? (i.e. 9.43)
 A: If there is rounding, then you need to use continued fractions to find rationals with small denominator near the given decimal.
To use your example:  $9.43=\{9;2,3,14\}$ and the convergents are $\{9,\frac {19}2,\frac {66}7,\frac {943}{100}\}$.  The first two are out, but $\frac {66}7=9.428571429\cdots $ which would round to $9.43$ so the answer is $7$. (Indeed, we confirm that ratings of $(10,10,10,10,10,10,6)$ have an average of $\frac {66}7$).
As a general algorithm, write out the convergents and test each one to see if it rounds to the desired decimal under whatever rounding scheme you are using.  Go in order, as you can stop the first time the rounding works.
Side note: this comes up all the time in baseball.  If you see a batter with a $750$ average, you can be quite sure that he's had very few at bats.  A lot tougher if the average is $281$, say.  For that example, we use the algorithm to see that $.281\approx \frac 9{32}$ so the batter might have had as few as $32$ at bats, but not fewer.
