## BdMO National Higher Secondary 2020 P6

\(f\) একটা এক-এক ফাংশন যার ডোমেইন আর কোডোমেইন উভয়ই ধনাত্মক পূর্ণসংখ্যার সেট এবং \(f(xy)=f(x)f(y)\)। \(f(2020)\)-এর সম্ভাব্য সর্বনিম্ন মান বের করো।

\(f\) is a one-to-one function from the set of positive integers to itself such that \(f(xy) = f(x)f(y)\). Find the minimum possible value of \(f(2020)\).

\(f\) is a one-to-one function from the set of positive integers to itself such that \(f(xy) = f(x)f(y)\). Find the minimum possible value of \(f(2020)\).

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- Mehrab4226
**Posts:**230**Joined:**Sat Jan 11, 2020 1:38 pm**Location:**Dhaka, Bangladesh

### Re: BdMO National Higher Secondary 2020 P6

I am not sure this solution is correct or not but meh,

*The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.*

-Henri Poincaré

### Re: BdMO National Higher Secondary 2020 P6

How did you get $f(2)=2$,$f(p)=p$

- Mehrab4226
**Posts:**230**Joined:**Sat Jan 11, 2020 1:38 pm**Location:**Dhaka, Bangladesh

### Re: BdMO National Higher Secondary 2020 P6

We want the minimum value of $f(2020) = f(2)f(2)f(5)f(101)$,

So it is natural to assume that factors should have the minimum value possible.

$f(2) = 2$ since 1 is already been taken by $f(1)$ and $f$ is a one-one function. That's it.

My solution is not complete. If BDMO 2020 was a written exam then the marks I would get is 0. But it was not a written exam , so....

And keeping $f(p)=p$ for the rest of the primes gives us a valid function $f$ fulfilling the given criteria.

*The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.*

-Henri Poincaré

- Anindya Biswas
**Posts:**263**Joined:**Fri Oct 02, 2020 8:51 pm**Location:**Magura, Bangladesh-
**Contact:**

### Re: BdMO National Higher Secondary 2020 P6

Almost done, now showing that for all primes $p,q, r$,

$f(p)^2f(q)f(r)\geq 2^2\cdot3\cdot5$ should do the work, isn't it?

"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."

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**John von Neumann**- Mehrab4226
**Posts:**230**Joined:**Sat Jan 11, 2020 1:38 pm**Location:**Dhaka, Bangladesh

### Re: BdMO National Higher Secondary 2020 P6

Hmm, but that should be obvious.Anindya Biswas wrote: ↑Wed Mar 03, 2021 8:25 pmAlmost done, now showing that for all primes $p,q, r$,

$f(p)^2f(q)f(r)\geq 2^2\cdot3\cdot5$ should do the work, isn't it?

*The Mathematician does not study math because it is useful; he studies it because he delights in it, and he delights in it because it is beautiful.*

-Henri Poincaré

- Anindya Biswas
**Posts:**263**Joined:**Fri Oct 02, 2020 8:51 pm**Location:**Magura, Bangladesh-
**Contact:**

### Re: BdMO National Higher Secondary 2020 P6

Yeah then we are done, you've constructed a function $f$ for which $f(2020)=60$ and also it can be shown that $60$ is the minimum. So, your solution is valid.Mehrab4226 wrote: ↑Wed Mar 03, 2021 11:29 pmHmm, but that should be obvious.Anindya Biswas wrote: ↑Wed Mar 03, 2021 8:25 pmAlmost done, now showing that for all primes $p,q, r$,

$f(p)^2f(q)f(r)\geq 2^2\cdot3\cdot5$ should do the work, isn't it?

"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."

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**John von Neumann**### Re: BdMO National Higher Secondary 2020 P6

What if we assume that f(n) = 1

For all positive integers.

??

For all positive integers.

??

- Anindya Biswas
**Posts:**263**Joined:**Fri Oct 02, 2020 8:51 pm**Location:**Magura, Bangladesh-
**Contact:**

### Re: BdMO National Higher Secondary 2020 P6

$f$ is one-to-one function.

"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."

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**John von Neumann**### Re: BdMO National Higher Secondary 2020 P6

So that's why we made f(101)=5

So that is the lowest possible value.

I got it now.

So that is the lowest possible value.

I got it now.