BdMO National 2021 Junior Problem 10

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Anindya Biswas
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BdMO National 2021 Junior Problem 10

Unread post by Anindya Biswas » Mon Apr 12, 2021 12:22 pm

দুটো ধনাত্মক পূর্ণসংখ্যা \(a\) আর \(b\)-এর জন্য
\[0<\left\lvert\dfrac{a}{b}-\dfrac{3}{5}\right\rvert\leq\dfrac{1}{150}\]
\(b\)-এর সর্বনিম্ন সম্ভাব্য মান কত?

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 $b$?
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Naeem588
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Re: BdMO National 2021 Junior Problem 10

Unread post by Naeem588 » Mon Jul 26, 2021 2:26 pm

Bbbbbbbbbbbb

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Anindya Biswas
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Re: BdMO National 2021 Junior Problem 10

Unread post by Anindya Biswas » Tue Jul 27, 2021 12:15 am

Anindya Biswas wrote:
Mon Apr 12, 2021 12:22 pm
দুটো ধনাত্মক পূর্ণসংখ্যা \(a\) আর \(b\)-এর জন্য
\[0<\left\lvert\dfrac{a}{b}-\dfrac{3}{5}\right\rvert\leq\dfrac{1}{150}\]
\(b\)-এর সর্বনিম্ন সম্ভাব্য মান কত?

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 $b$?
The given inequality is equivalent to,
$\frac{89b}{150}\leq a<\frac{3b}{5}$ or $\frac{3b}{5}<a\leq\frac{91b}{150}$

So, it's sufficient to find the smallest $b$ such that there exist an integer $a$ between $\frac{89b}{150}$ and $\frac{91b}{150}$ and $a\neq\frac{3b}{5}$.

Now what? Checking and brute force?
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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F Nishat
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Re: BdMO National 2021 Junior Problem 10

Unread post by F Nishat » Tue Jul 27, 2021 2:43 pm

Anindya Biswas wrote:
Mon Apr 12, 2021 12:22 pm
দুটো ধনাত্মক পূর্ণসংখ্যা \(a\) আর \(b\)-এর জন্য
\[0<\left\lvert\dfrac{a}{b}-\dfrac{3}{5}\right\rvert\leq\dfrac{1}{150}\]
\(b\)-এর সর্বনিম্ন সম্ভাব্য মান কত?

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 $b$?
We can rewrite the inequality as $0<\left\lvert\dfrac{5a-3b}{5b}\right\rvert\leq\dfrac{1}{150}$. Since $a$ and $b$ are integers, we have $\left\lvert 5a-3b\right\rvert\geq 1$ is an integer. Of course, this implies $5b\geq 150\Rightarrow b\geq 30$.
Now, brute forcing is easy as we get the desired result when $\boxed{b=32}$ which is close to $30$.
"But in my opinion, all things in nature occur mathematically."-Rene Descartes

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