# Rounded percentages

Inspired by Jeremy Galvagni and this problem

Every integer percentage $0\% \leq p \leq 100\%$ can be written as $\frac a {100}, a=p$, but of course for many of them there is a value $b < 100$ that also makes $\frac a b = p$ true, because many fractions can be reduced.

If we now allow rounding to the nearest whole percent, this again gives more possibilities and therefore smaller denominators.

Let $q(p)$ be the smallest possible value of $b$ such that there exists some integer $a$ with $\frac a b = \frac a {q(p)} \approx p$ when rounded to the nearest whole percent.

If we sort all values of $q(p)$ by size, what is the sum of the values of $p$ that give the 10 largest $q(p)$?

Submit your answer just without the %. So, if you get 100%, give 100 as the answer.

Details and assumptions

• If the fractional part of $\frac a b$ is exactly 0.5 we will round towards 50%, so $12.5\% \approx 13\%$, but $87.5\% \approx 87\%$. This preserves the symmetry $1 = \frac a b + \frac {1-a} b$.
• As an example, $q(62\%)=8$ because $\frac 58 = 0.625 \approx 0.62$ is the fraction with the smallest denominator (8) that, when rounded with the specified rules, gives $62\%$.
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