I know that the top three candidates received 14.14%, 13.13% and 12.12% of a vote. I also know that the total number of votes cannot be more that 300, and the numbers of votes cast and the total must be integers, I also believe that the percentages are rounded to two decimal places.
How can I calculate how many votes were cast in total?
My workings so far: I have managed to create the equations:
$\frac{a}{x*100} = 14.14$
$\frac{b}{x*100} = 13.13$
$\frac{c}{x*100} = 12.12$
which gives:
$\frac{a}{x*100} +\frac{b}{x*100} + \frac{c}{x*100} = 14.14+13.13+12.12$
that is:
$\frac{a+b+c}{3x} = 39.39$
and I don't know where to go from there..
Another route is from intuitively:
$A+B+C+D=100$
in this case $D = 60.61$ so we can say $\frac{d}{x*100}=60.61$ although I don't know how this helps...
Now from looking at simultaneous equations on Wikipedia I realise we should solve each equation for one variable which means we can get to: $a = x1414$, $b = x1313$ and $c = x1212$ or better put as: $\frac{a}{1414} =x$, $\frac{b}{1313} =x$ and $\frac{c}{1212} =x$
so: $\frac{a}{1414} =\frac{b}{1313} =\frac{c}{1212}$
and: $\frac{a}{1414} =\frac{b}{1313}$ means that: $a=b1.076923$
and $\frac{b}{1313} =\frac{c}{1212}$ means that: $b=c1.08333$
and $\frac{a}{1414} =\frac{c}{1212}$ means that: $a=c.166666$
therefore: $c1.166666 = (c1.08333)1.076923$ and this breaks when we expand the brackets i.e.: $c1.166666 = (c1.076923)(1.166666)$ hence my question.
As a general approach to find integers $m,n$ whose ratio closely approximates a decimal fraction, we can use the method of continued fraction expansion. I will illustrate it for the number $13.13\%$ or $0.1313$ appearing among the vote percentages.
First we take the absolute value, in case we were given a negative number. Here the number is positive and nothing needs to be done. We will construct a sequence of positive integer pairs $m_i,n_i$ such for their size give the best possible rational approximation.
Pull off the integer (whole number) part to get started, setting $m_0=0$ in this case and put that over a denominator $n_0=1$. That leaves a remainder $0.1313 - 0/1$ as our approximation "error".
At each successive step we have a nonegative error of approximation which is less than one, e.g. $0.1313$ in this instance. If zero error, quit. Otherwise taking the reciprocal of the error gives a new value greater than one:
$$ \frac{1}{0.1313} = 7 + 0.61614623\ldots $$
Considered in terms of using the single digit $7$ as an approximation to the reciprocal error, we have:
$$ 0.1313 \approx 0 + 1/7 $$
and if we continue in this fashion we will get a succession of improved approximations until we hit exactly:
$$ 0.1313 = 1313/10000 $$
For your problem you would carry the approxidation procedure out as long as the denominator remained under $300$ until a value rounding to two places in the percentage (four places without such a percentage sign). The procedure would fail if the denominator went over $300$, but in the immediate case we succeed with:
$$ 0.1313 \approx 13/99 $$