<s>[INST] Smallest value of $b$ when $0<\left\lvert \frac{a}{b}-\frac{3}{5}\right\rvert\leq\frac{1}{150}$ Problem
For positive integers $a$ and $b$, $$0<\left\lvert \dfrac{a}{b}-\dfrac{3}{5}\right\rvert\leq\dfrac{1}{150}$$
What is the smallest possible value of $b$? (BdMO 2021 Junior P10)
My approach
If $\dfrac{a}{b}>\dfrac{3}{5}$, then $\dfrac{a}{b} \leq \dfrac{91}{150}$ and if $\dfrac{a}{b}<\dfrac{3}{5}$, then $\dfrac{a}{b} \geq \dfrac{89}{150}$.
Hence the maximum value of $\dfrac{a}{b}$ is $\dfrac{91}{150}$. And $\dfrac{a}{b}$ is maximum when $b$ is the smallest.
So, I thought $150$ would be the smallest value of $b$.
But my friend said that the smallest value of $b$ would be $32$ as $\left\lvert \dfrac{19}{32}- \dfrac{3}{5}\right\rvert= \dfrac{1}{160}$ works actually. Though he said that he found this intuitively.
So, how to solve the problem with proper procedure?
 [/INST] Clearing denominators, we are effectively asked to find minimal $b$ s.t. $0 < |150a-90b| \leqslant b$, for $a, b \in \mathbb N$.
Note as the gcd $(150,90)=30$, the middle term will at least be $30$ (or a larger multiple), so $b\geqslant 30$.
It remains to to check if $150 \mid (30\pm 90b)$ or equivalently if $b \equiv \pm2 \pmod 5$ for $b\geqslant 30$ to conclude $b_{min}=32$.

The following constructive approach avoids brute force.
From Bézout's identity  $150\cdot(-1)+90\cdot2=30\tag{1}$ and the obvious $150\cdot(-3)+90\cdot5=0\tag{2}$ we can use the latter equation to scale up $b$ above $30$ without adding to the RHS error term.  Further, as we noted $b \equiv \pm2 \pmod 5$, all solutions will be of form equation $(2)\times n \,\pm$ equation $(1)$. If there is any solution which is not of this form, subtract $(2)$ multiple times from it to get a contradiction with $b \equiv \pm 2\pmod 5$.
Now as $5n\pm2 \geqslant 30$ has minimal solution $n=6$, equation $(2)\times 6\,+$ equation $(1)$ gives $150\cdot(-19)+90\cdot(32)=30 < 32 \implies b_{min}=32$.
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