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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?
Does 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 $3\times 3 = 9 \notin X$
(sorry i didn't notice that i missed some countings..)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 $3\times 3 = 9 \notin X$
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 $3\times 3 = 9 \notin X$
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?
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.
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.