বৃষ্টি একটা বিশেষ সেট \(A\) বানাতে চায়। সে \(A=\{0, 42\}\) দিয়ে শুরু করে। যেকোনো ধাপে সে একটা পূর্ণসংখ্যা \(x\)-কে \(A\)-তে ঢুকাতে পারবে যদি \(x\), \(A\)-তে ইতোমধ্যে থাকা সংখ্যাগুলোকে সহগ হিসেবে ব্যবহার করে বানানো কোনো বহুপদীর মূল হয়। এভাবে সে \(A\)-এ নতুন নতুন পূর্ণসংখ্যা ঢুকাতেই থাকে। যখন সে \(A\)-এ আর ঢুকানোর মতো নতুন সংখ্যা খুঁজে পাবে না, তখন \(A\)-তে কয়টা সংখ্যা থাকবে?
Bristy wants to build a special set \(A\). She starts with \(A=\{0, 42\}\). At any step, she can add an integer \(x\) to the set \(A\) if it is a root of a polynomial that uses the already existing integers in \(A\) as coefficients. She keeps doing this, adding more and more numbers to \(A\). After she eventually runs out of numbers to add to \(A\), how many numbers will be in \(A\)?
Re: BdMO National Higher Secondary 2020 P10
Posted: Thu Apr 01, 2021 11:27 pm
by Mehrab4226
Picking this up, as nationals are near, if there is any right time to post a solution to this problem, it must be now!
Re: BdMO National Higher Secondary 2020 P10laim
Posted: Fri Apr 02, 2021 11:08 am
by nimon
যেকোনো ধাপে সে একটা পূর্ণসংখ্যা x-কে A-তে ঢুকাতে পারবে যদি x, A-তে ইতোমধ্যে থাকা সংখ্যাগুলোকে সহগ হিসেবে ব্যবহার করে বানানো কোনো বহুপদীর মূল হয়।
Can you give more explaination to this line?
যেকোনো ধাপে সে একটা পূর্ণসংখ্যা x-কে A-তে ঢুকাতে পারবে যদি x, A-তে ইতোমধ্যে থাকা সংখ্যাগুলোকে সহগ হিসেবে ব্যবহার করে বানানো কোনো বহুপদীর মূল হয়।
Can you give more explaination to this line?
There are some integers in A. If you make a polynomial with coefficients are integers from A, and one of the roots of the polynomial is an integer $k \notin A$. Then you can add k in A.
For example,
$42x+42$is a polynomial with coefficients from A and root $-1$, but $-1$ is not in A, so we can add him in A. This process continues now $A=\{-1,0,42\}$. And other polynomials can also have degree more than 1.
Re: BdMO National Higher Secondary 2020 P10
Posted: Fri Apr 02, 2021 9:13 pm
by Anindya Biswas
Does $\pm14$ and $\pm21$ belongs to $A$? That's where I am stuck!
So far, we've got $M=\{0,\pm1,\pm2,\pm3,\pm6,\pm7,\pm14,\pm21,\pm42\}\subseteq S$
Here, observe that $x\in M$ if and only if $x=0$ or $x|42$.
Now we will show that $M=S$.
$\text{Proof :}$
Let's assume $a\in S$. Let $P$ be a polynomial such that $P(x)=x^k\cdot Q(x)$ where $k$ is a nonnegative integer and $Q$ is a polynomial with nonzero constant term and each coefficient of $P(x)$ is an element of $M$ and $P(a)=0$.
Now, if $a=0$ then $a\in M$
Now, let's assume $a\neq0$.
$\therefore Q(a)=0$.
Let $q$ be the constant term of $Q$.
By rational root theorem, we know that $a|q$. But since $q\in M$ and $q\neq0$, that means $q|42$. So, we conclude that $a|42$.
$\therefore a\in M\Longrightarrow S\subseteq M$
$\therefore S=M$
So, our answer should be $\boxed{|S|=17}$
So far, we've got $M=\{0,\pm1,\pm2,\pm3,\pm6,\pm7,\pm14,\pm21,\pm42\}\subseteq S$
Here, observe that $x\in M$ if and only if $x=0$ or $x|42$.
Now we will show that $M=S$.
$\text{Proof :}$
Let's assume $a\in S$. Let $P$ be a polynomial such that $P(x)=x^k\cdot Q(x)$ where $k$ is a nonnegative integer and $Q$ is polynomial with nonzero constant term and each coefficient of $P(x)$ is an element of $M$ and $P(a)=0$.
Now, if $a=0$ then $a\in M$
Now, let's assume $a\neq0$.
$\therefore Q(a)=0$.
Let $q$ be the constant term of $Q$.
By rational root theorem, we know that $a|q$. But since $q\in M$ and $q\neq0$, that means $q|42$. So, we conclude that $a|42$.
$\therefore a\in M\Longrightarrow S\subseteq M$
$\therefore S=M$
So, our answer should be $\boxed{|S|=17}$
Thank you, I was looking for a solution for a long time.