Skeptic Percentage

Note that whatever is the value of $a$ , we could find a value of $b=3a$ which makes the ratio rendered as 33% (although b=3a violate the coprime rule, it is a good starting point.)

Because of $\gcd(a,3a\pm1)=1$ , we want to find as minimum as possible that $3a\pm1$ satisfy the condition, now let me declare this first inequality: $0.325\leq\frac ab<0.335$

If $b=3a-1$ , then $\dfrac{a}{b}=\dfrac{a}{3a-1}>\dfrac{a}{3a}>0.325$ , so it is only necessary to check the upper bound: $\begin{aligned}\dfrac{a}{3a-1}&<0.335\\&=\frac{67}{200}\\200a&<201a-67\\a&>67\end{aligned}$ So at this case the minimum value of $a$ will be $68$ , then $b=203$ , $\dfrac ab =0.334975...\approx0.33$ .

Else if $b=3a+1$ , then $\dfrac{a}{b}=\dfrac{a}{3a+1}<\dfrac{a}{3a}<0.335$ , so we only need to check bottom bound: $\begin{aligned}\dfrac a{3a+1}&\geq0.325\\&=\frac{13}{40}\\40a&\geq39a+13\\a&\geq13\end{aligned}$ at this point, the minimum of $a$ will be $13$ and $b$ is $40$ , which is far lesser then previous value $203$ . we have $\dfrac{13}{40}=0.325\approx0.33$ which fullfil our purpose.

• This is making sense because the gap of $\dfrac13$ between bottom bound is larger than upper bound, so the answer will be found in the bottom bound section.

We consider two cases by looking at the continued fractions of $x = \frac{a}{b}$ :

• $\frac{1}{3} < x < 0.335 = \frac{67}{200}$
• $0.325 = \frac{13}{40} \leq x < \frac{1}{3}$

In the first case: $\frac{200}{67} < x^{-1} < 3 \\ \frac{66}{67} < x^{-1} - 2 < 1 \\ 1 < (x^{-1} - 2)^{-1} < \frac{67}{66} \\ 0 < (x^{-1} - 2)^{-1}-1 < \frac{1}{66} \\ 66 < ((x^{-1}-2)^{-1}-1)^{-1}$ This last step of the algorithm is the first one that admits an integer solution, and the smallest such solution is $67$ , so we solve $((x^{-1}-2)^{-1}-1)^{-1} = 67 \implies x = \frac{68}{203}$

In the second case: $3 < x^{-1} \leq \frac{40}{13} \\ 0 < x^{-1} - 3 \leq \frac{1}{13} \\ 13 \leq (x^{-1}-3)^{-1}$ Again, this last step admits an integer solution, and the smallest such solution is $13$ , so we solve $(x^{-1}-3)^{-1} = 13 \implies x = \frac{13}{40}.$

Between the two of these, the smallest denominator is $\boxed{40}$ .