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

Correct answer: 34 aligned & a_0+a_1+a_2+ +a_n=4 & a_i ≤ 4 & a_0+5 a_1+5^2 a_2+ .+a 5^n a_n=136 & a_0=1+5 λ a_0=1 & Hence 5 a_1+5^2 a_2+ .+5^n a_n=135 & a_1+5 a_2+ .5^ n-1 a_ n-1 =27 & a_1=5 λ+2 a_1=2 & 5 a_2+ 5^ n-1 a_ n-1 =25 & a_2+5 a_3+ . .5^ n-2 a_ n-2 =5 & a_2=5 λ a_2=0 & a_3+5 a_4+ .+5^ n-3 a_ n-3 =1 & a_3=1 & a_4+5 a_5+ . .+5^ n-4 a_ n-3 =0 & a_4=a_5= . a_n=0 & Hence P(n)=x^3+2 x+1 & P(3)=34 aligned

Olympiad workbook

Exam: Olympiad workbook
Subject: IOQM
Chapter: 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)=a_0+a_1 x+a_2 x^2+\ldots .+a_n x^n

be a polynomial in which

a_i

is non-negative integer for each

i \in\{0,1,2,3, \ldots, n\}

. If

P(1)=4

and

P(5)=136

, what is the value of

P(3)

Correct answer

Step-by-step solution

\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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