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Olympiad workbook
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IOQM
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Functions

Olympiad workbook · IOQM · Functions

Let P(x)=a_0+a_1 x+a_2 x^2+ .+a_n x^n be a polynomial in which a_i...

Let P(x)=a0+a1x+a2x2+….+anxnP(x)=a_0+a_1 x+a_2 x^2+\ldots .+a_n x^n be a polynomial in which aia_i is non-negative integer for each i∈{0,1,2,3,…,n}i \in\{0,1,2,3, \ldots, n\}. If P(1)=4P(1)=4 and P(5)=136P(5)=136, what is the value of P(3)P(3) ?

Correct answer

34

Step-by-step solution

a0+a1+a2+……+an=4⇒ai≤4a0+5a1+52a2+….+a5nan=136⇒a0=1+5λ⇒a0=1 Hence 5a1+52a2+….+5nan=135a1+5a2+….5n−1an−1=27⇒a1=5λ+2⇒a1=2⇒5a2+…5n−1an−1=25a2+5a3+…..5n−2an−2=5⇒a2=5λ⇒a2=0a3+5a4+……….+5n−3an−3=1a3=1⇒a4+5a5+…..+5n−4an−3=0a4=a5=….an=0 Hence P(n)=x3+2x+1P(3)=34\begin{aligned} & a_0+a_1+a_2+\ldots \ldots+a_n=4 \\ & \Rightarrow a_i \leq 4 \\ & a_0+5 a_1+5^2 a_2+\ldots .+a 5^n a_n=136 \\ & \Rightarrow a_0=1+5 \lambda \Rightarrow a_0=1 \\ & \text { Hence } 5 a_1+5^2 a_2+\ldots .+5^n a_n=135 \\ & a_1+5 a_2+\ldots .5^{n-1} a_{n-1}=27 \\ & \Rightarrow a_1=5 \lambda+2 \Rightarrow a_1=2 \\ & \Rightarrow 5 a_2+\ldots 5^{n-1} a_{n-1}=25 \\ & a_2+5 a_3+\ldots . .5^{n-2} a_{n-2}=5 \\ & \Rightarrow a_2=5 \lambda \Rightarrow a_2=0 \\ & a_3+5 a_4+\ldots \ldots \ldots .+5^{n-3} a_{n-3}=1 \\ & a_3=1 \\ & \Rightarrow a_4+5 a_5+\ldots . .+5^{n-4} a_{n-3}=0 \\ & a_4=a_5=\ldots . a_n=0 \\ & \text { Hence } P(n)=x^3+2 x+1 \\ & P(3)=34\end{aligned}
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