<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] HINT:
If $\frac{a}{b}$ satisfies the condition then
$$|5 a - 3 b| = 1$$
the explanation being that in the parallelogram on the vectors $(3,5)$ and $(a,b)$ there should be no other integer points excepts the vertices. Now, the solutions of
$$5 a - 3 b = 1$$
are $$a = 2 + 3 t \\ b = 3 + 5 t$$
while those of
$$5 a - 3 b = -1$$
are
$$a = 1 + 3 t \\ b = 2 + 5 t $$
Now if $|5 a - 3 b| = 1$, then
$$\left|\frac{a}{b} - \frac{3}{5}\right| = \frac{1}{5 b}$$
Therefore, we need $$5 b \ge 150$$, or $$b \ge 30$$
Now, the smallest $b \equiv 2 \text{ or } 3 \mod 5$ and $\ge 30$ is $32$ ( and not $33$).
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