BDMO higher secondary P6
Forum rules
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
Please don't post problems (by starting a topic) in the "X: Solved" forums. Those forums are only for showcasing the problems for the convenience of the users. You can always post the problems in the main Divisional Math Olympiad forum. Later we shall move that topic with proper formatting, and post in the resource section.
-
- Posts:194
- Joined:Sat Jan 02, 2021 9:28 pm
Let $S= {1,2,...,15}$. How many sets $x \subseteq S$ are there such that if $x \in X$ and $3x \in S$ then $3x \in X$
Hmm..Hammer...Treat everything as nail
-
- Posts:194
- Joined:Sat Jan 02, 2021 9:28 pm
Re: BDMO higher secondary P6
We first establish a bijection.
Notice there are $5$ multiples of $3$ in $S$ which we write in pairs like $(1,3),(2,6),(3,9),(4,12),(5,15)$
now notice the desired sets are just subsets of this multiset except $\phi$ so the desired number is $2^5 -1=31$
Is it right?
Notice there are $5$ multiples of $3$ in $S$ which we write in pairs like $(1,3),(2,6),(3,9),(4,12),(5,15)$
now notice the desired sets are just subsets of this multiset except $\phi$ so the desired number is $2^5 -1=31$
Is it right?
Hmm..Hammer...Treat everything as nail
- Mehrab4226
- Posts:230
- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: BDMO higher secondary P6
Let me take one of your subsets,Asif Hossain wrote: ↑Sat Mar 27, 2021 3:48 pmWe first establish a bijection.
Notice there are $5$ multiples of $3$ in $S$ which we write in pairs like $(1,3),(2,6),(3,9),(4,12),(5,15)$
now notice the desired sets are just subsets of this multiset except $\phi$ so the desired number is $2^5 -1=31$
Is it right?
$X=\{(1,3),(2,6),(4,12),(5,15)\}$
$3\in X$
But $3\times 3 = 9 \notin X$
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
-Henri Poincaré
- Mehrab4226
- Posts:230
- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: BDMO higher secondary P6
My solution is like this,
Last edited by Mehrab4226 on Sat Mar 27, 2021 9:37 pm, edited 1 time in total.
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
-Henri Poincaré
-
- Posts:194
- Joined:Sat Jan 02, 2021 9:28 pm
Re: BDMO higher secondary P6
Does the question say that for "every" $x \in X$ the property holds? (sorry i didn't notice that i missed some countings..)Mehrab4226 wrote: ↑Sat Mar 27, 2021 7:42 pmLet me take one of your subsets,Asif Hossain wrote: ↑Sat Mar 27, 2021 3:48 pmWe first establish a bijection.
Notice there are $5$ multiples of $3$ in $S$ which we write in pairs like $(1,3),(2,6),(3,9),(4,12),(5,15)$
now notice the desired sets are just subsets of this multiset except $\phi$ so the desired number is $2^5 -1=31$
Is it right?
$X=\{(1,3),(2,6),(4,12),(5,15)\}$
$3\in X$
But $3\times 3 = 9 \notin X$
Hmm..Hammer...Treat everything as nail
- Mehrab4226
- Posts:230
- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: BDMO higher secondary P6
Yes. The question said if $x$ is in $X$, then $3x$ is also in $X$. That is enough to let us know that $x$ can be any number in $X$Asif Hossain wrote: ↑Sat Mar 27, 2021 9:21 pmDoes the question say that for "every" $x \in X$ the property holds?Mehrab4226 wrote: ↑Sat Mar 27, 2021 7:42 pmLet me take one of your subsets,Asif Hossain wrote: ↑Sat Mar 27, 2021 3:48 pmWe first establish a bijection.
Notice there are $5$ multiples of $3$ in $S$ which we write in pairs like $(1,3),(2,6),(3,9),(4,12),(5,15)$
now notice the desired sets are just subsets of this multiset except $\phi$ so the desired number is $2^5 -1=31$
Is it right?
$X=\{(1,3),(2,6),(4,12),(5,15)\}$
$3\in X$
But $3\times 3 = 9 \notin X$
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
-Henri Poincaré
-
- Posts:194
- Joined:Sat Jan 02, 2021 9:28 pm
Re: BDMO higher secondary P6
Didn't it said if "$x \in X$ and $3x \in S$" then $3x \in X$Mehrab4226 wrote: ↑Sat Mar 27, 2021 9:32 pmYes. The question said if $x$ is in $X$, then $3x$ is also in $X$. That is enough to let us know that $x$ can be any number in $X$Asif Hossain wrote: ↑Sat Mar 27, 2021 9:21 pmDoes the question say that for "every" $x \in X$ the property holds?Mehrab4226 wrote: ↑Sat Mar 27, 2021 7:42 pm
Let me take one of your subsets,
$X=\{(1,3),(2,6),(4,12),(5,15)\}$
$3\in X$
But $3\times 3 = 9 \notin X$
Hmm..Hammer...Treat everything as nail
- Mehrab4226
- Posts:230
- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: BDMO higher secondary P6
Yes, it said if $x$ is in $X$ and $3x$ in $S$. Then $3x$ must be in $X$. It implies that we cannot find any $x$ inside $X$ whose $3x$ if in $S$ not in $X$.Asif Hossain wrote: ↑Sat Mar 27, 2021 9:36 pmDidn't it said if "$x \in X$ and $3x \in S$" then $3x \in X$Mehrab4226 wrote: ↑Sat Mar 27, 2021 9:32 pmYes. The question said if $x$ is in $X$, then $3x$ is also in $X$. That is enough to let us know that $x$ can be any number in $X$Asif Hossain wrote: ↑Sat Mar 27, 2021 9:21 pm
Does the question say that for "every" $x \in X$ the property holds?
Last edited by Mehrab4226 on Sat Mar 27, 2021 9:49 pm, edited 1 time in total.
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
-Henri Poincaré
-
- Posts:194
- Joined:Sat Jan 02, 2021 9:28 pm
Re: BDMO higher secondary P6
well even if my assumption is right is the ans $31$ right..Mehrab4226 wrote: ↑Sat Mar 27, 2021 9:45 pmYou maybe right. But in let's say set builders method we write a set,Asif Hossain wrote: ↑Sat Mar 27, 2021 9:36 pmDidn't it said if "$x \in X$ and $3x \in S$" then $3x \in X$Mehrab4226 wrote: ↑Sat Mar 27, 2021 9:32 pm
Yes. The question said if $x$ is in $X$, then $3x$ is also in $X$. That is enough to let us know that $x$ can be any number in $X$
$K=\{x:x\in \mathbb{N}\}$
Then $K=\{1,2,3,4,\cdots\}
You know what, I don't know which one is right.
Hmm..Hammer...Treat everything as nail
- Mehrab4226
- Posts:230
- Joined:Sat Jan 11, 2020 1:38 pm
- Location:Dhaka, Bangladesh
Re: BDMO higher secondary P6
Depends on your assumption,Asif Hossain wrote: ↑Sat Mar 27, 2021 9:47 pmwell even if my assumption is right is the ans $31$ right..Mehrab4226 wrote: ↑Sat Mar 27, 2021 9:45 pmYou maybe right. But in let's say set builders method we write a set,Asif Hossain wrote: ↑Sat Mar 27, 2021 9:36 pm
Didn't it said if "$x \in X$ and $3x \in S$" then $3x \in X$
$K=\{x:x\in \mathbb{N}\}$
Then $K=\{1,2,3,4,\cdots\}
You know what, I don't know which one is right.
If you assume that if any single element of a subset of $S$ follows that criteria,then that subset is a valid $X$, then your answer is probably not correct.
If you assumed that $X$ can only contain the numbers which have $3x \in S$. Then you are maybe correct.
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
-Henri Poincaré