BdMO National Higher Secondary 2020 P6

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Mursalin
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BdMO National Higher Secondary 2020 P6

Unread post by Mursalin » Mon Feb 08, 2021 4:06 pm

\(f\) 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 \(f(2020)\).
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Mehrab4226
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Re: BdMO National Higher Secondary 2020 P6

Unread post by Mehrab4226 » 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 $f(2020) = f(2)f(2)f(5)f(101) = 2 \times 2\times 3 \times 5 = 60$

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

Dustan
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Re: BdMO National Higher Secondary 2020 P6

Unread post by Dustan » 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 $f(2)=2$,$f(p)=p$

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Mehrab4226
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Re: BdMO National Higher Secondary 2020 P6

Unread post by Mehrab4226 » Fri Feb 12, 2021 8:10 pm

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.
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

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Anindya Biswas
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Re: BdMO National Higher Secondary 2020 P6

Unread post by Anindya Biswas » 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?
"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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Mehrab4226
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Re: BdMO National Higher Secondary 2020 P6

Unread post by Mehrab4226 » 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.
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

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Anindya Biswas
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Re: BdMO National Higher Secondary 2020 P6

Unread post by Anindya Biswas » Thu Mar 04, 2021 10:27 pm

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.
"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 Higher Secondary 2020 P6

Unread post by Naeem588 » Tue Apr 06, 2021 12:48 pm

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

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Anindya Biswas
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Re: BdMO National Higher Secondary 2020 P6

Unread post by Anindya Biswas » Tue Apr 06, 2021 12:52 pm

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.
"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 Higher Secondary 2020 P6

Unread post by Naeem588 » Tue Apr 06, 2021 1:01 pm

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

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