BdMO National 2021 Higher Secondary Problem 4

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Anindya Biswas
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BdMO National 2021 Higher Secondary Problem 4

Unread post by Anindya Biswas » Sun Apr 11, 2021 9:24 pm

\(P(x)\) হলো অঋণাত্মক পূর্ণসাংখ্যিক সহগবিশিষ্ট \(x\)-এর একটা বহুপদী। যদি \(P(1)=5\) আর \(P(P(1))=177\) হয়, তাহলে \(P(10)\)-এর সম্ভাব্য সব মানের যোগফল কত?

$P(x)$ is a polynomial in $x$ with non-negative integer coefficients. If $P(1)=5$ and $P(P(1))=177$, what is the sum of all possible values of $P(10)$?
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Pro_GRMR
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Re: BdMO National 2021 Higher Secondary Problem 4

Unread post by Pro_GRMR » Sun Apr 11, 2021 10:48 pm

Anindya Biswas wrote:
Sun Apr 11, 2021 9:24 pm
$P(x)$ is a polynomial in $x$ with non-negative integer coefficients. If $P(1)=5$ and $P(P(1))=177$, what is the sum of all possible values of $P(10)$?
Let $P(x)= a_0+a_1x+a_2x^2+ \dots+a_nx^n$ where $a_n \geq 1$.
Notice, $P(1)= a_0+a_1\cdot1+a_2\cdot1^2+ \dots+a_n\cdot1^n = a_0+a_1+a_2+ \dots+a_n= 5$
Also, $P(P(1))=P(5)=a_0+a_1\cdot5+a_2\cdot5^2+ \dots +a_n\cdot5^n=177$

But if $n \geq 4$, $P(5)\geq 1\cdot5^4=625$, because $a_n \geq 1$
Again if $n \leq 2$, $P(5)\leq5\cdot5^2=125$, because $a_0+a_1+a_2+ \dots+a_n= 5$ and we must maximize value of $a_n$ to maximize $P(n)$.

So n must be $3$ and $P(x)$ of the form $a_0+a_1x+a_2x^2+ \dots+a_3x^3$ where $a_n \geq 1$.
Now if $a_3 \geq 2, P(5) \geq 250$ So $a_1$ must be $1$

We get that $P(x)= x^3 + 2x^2 + 2$ by using similar logic.

Now we plug in 10 and get the only value of $P(10)=10^3+2\cdot 10^2+2=\boxed{1202}$
"When you change the way you look at things, the things you look at change." - Max Planck

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