Rounded percentages

Inspired by Jeremy Galvagni and this problem

Every integer percentage 0%p100% 0\% \leq p \leq 100\% can be written as a100,a=p \frac a {100}, a=p , but of course for many of them there is a value b<100 b < 100 that also makes ab=p \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) q(p) be the smallest possible value of b b such that there exists some integer a a with ab=aq(p)p \frac a b = \frac a {q(p)} \approx p when rounded to the nearest whole percent.

If we sort all values of q(p) q(p) by size, what is the sum of the values of p p that give the 10 largest q(p) 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 ab \frac a b is exactly 0.5 we will round towards 50%, so 12.5%13% 12.5\% \approx 13\% , but 87.5%87% 87.5\% \approx 87\% . This preserves the symmetry 1=ab+1ab 1 = \frac a b + \frac {1-a} b .
  • As an example, q(62%)=8 q(62\%)=8 because 58=0.6250.62 \frac 58 = 0.625 \approx 0.62 is the fraction with the smallest denominator (8) that, when rounded with the specified rules, gives 62% 62\% .

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