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\(f\) 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 \(f(2020)\).

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

Mehrab4226 wrote: ↑

Wed Feb 10, 2021 12:58 am

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

Clearly,
$f(1)=1$
If we can find the values for prime numbers we can get the values of composite numbers very easily.

Let,
f(2)=2,f(5)=3,f(101)=5, f(3)=101,
And then all other prime p, $f(p)=p$
Since all numbers have a different prime power factorization all numbers will have different values in the function so it is still one-one and also follows the given law.
So minimum of $f(2020) = f(2)f(2)f(5)f(101) = 2 \times 2\times 3 \times 5 = 60$

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

Dustan wrote: ↑

Fri Feb 12, 2021 4:08 pm

Mehrab4226 wrote: ↑

Wed Feb 10, 2021 12:58 am

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

Clearly,
$f(1)=1$
If we can find the values for prime numbers we can get the values of composite numbers very easily.

Let,
f(2)=2,f(5)=3,f(101)=5, f(3)=101,
And then all other prime p, $f(p)=p$
Since all numbers have a different prime power factorization all numbers will have different values in the function so it is still one-one and also follows the given law.
So minimum of $f(2020) = f(2)f(2)f(5)f(101) = 2 \times 2\times 3 \times 5 = 60$

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

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.

Mehrab4226 wrote: ↑

Wed Feb 10, 2021 12:58 am

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

Clearly,
$f(1)=1$
If we can find the values for prime numbers we can get the values of composite numbers very easily.

Let,
f(2)=2,f(5)=3,f(101)=5, f(3)=101,
And then all other prime p, $f(p)=p$
Since all numbers have a different prime power factorization all numbers will have different values in the function so it is still one-one and also follows the given law.
So minimum of $f(2020) = f(2)f(2)f(5)f(101) = 2 \times 2\times 3 \times 5 = 60$

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?

Anindya Biswas wrote: ↑

Wed Mar 03, 2021 8:25 pm

Mehrab4226 wrote: ↑

Wed Feb 10, 2021 12:58 am

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

Clearly,
$f(1)=1$
If we can find the values for prime numbers we can get the values of composite numbers very easily.

Let,
f(2)=2,f(5)=3,f(101)=5, f(3)=101,
And then all other prime p, $f(p)=p$
Since all numbers have a different prime power factorization all numbers will have different values in the function so it is still one-one and also follows the given law.
So minimum of $f(2020) = f(2)f(2)f(5)f(101) = 2 \times 2\times 3 \times 5 = 60$

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?

Hmm, but that should be obvious.

Mehrab4226 wrote: ↑

Wed Mar 03, 2021 11:29 pm

Anindya Biswas wrote: ↑

Wed Mar 03, 2021 8:25 pm

Mehrab4226 wrote: ↑

Wed Feb 10, 2021 12:58 am

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

Clearly,
$f(1)=1$
If we can find the values for prime numbers we can get the values of composite numbers very easily.

Let,
f(2)=2,f(5)=3,f(101)=5, f(3)=101,
And then all other prime p, $f(p)=p$
Since all numbers have a different prime power factorization all numbers will have different values in the function so it is still one-one and also follows the given law.
So minimum of $f(2020) = f(2)f(2)f(5)f(101) = 2 \times 2\times 3 \times 5 = 60$

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?

Hmm, but that should be obvious.

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.

What if we assume that f(n) = 1
For all positive integers.
??

Naeem588 wrote: ↑

Tue Apr 06, 2021 12:48 pm

What if we assume that f(n) = 1
For all positive integers.
??

$f$ is one-to-one function.

So that's why we made f(101)=5
So that is the lowest possible value.
I got it now.