Polynomial Problem

Algebra Level 4

Let P ( x ) P(x) = a 0 + a 1 x + + a n x n a_0+a_1x+ \cdots +a_nx^n , where a 0 , , a n a_0,\ldots ,a_n are non-negative integers.

Given that P ( 1 ) = 4 P(1)=4 and P ( 5 ) = 152 P(5)=152 , find P ( 6 ) P(6) .


The answer is 254.

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3 solutions

James Christian
Nov 5, 2016

We have a 0 + a 1 + + a n = 4 , a_0+a_1+\ldots + a_n = 4, implying max a 0 , a 1 , , a n < 5. \max{a_0, a_1, \ldots, a_n} < 5.

P ( 5 ) = a 0 + 5 a 1 + 5 2 a 2 + + 5 n a n P(5) = a_0 + 5 \cdot a_1 + 5^2 \cdot a_2 + \ldots + 5^n \cdot a_n

thus the unique base 5 5 representation of 152 152 . Since 152 = 110 2 5 , 152 = 1102_5,

P ( x ) = x 3 + x 2 + 2. P(x)=x^3+x^2+2.

P ( 6 ) = 216 + 36 + 2 = 254 . P(6)=216+36 + 2 = \boxed{254}.

nice problem

Kushal Bose - 4 years, 7 months ago
Anshu Garg
Nov 19, 2016

When x = 1 x=1

a 0 + a 1 + a 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . + a n = 4 a_0+a_1+a_2................................................+a_n=4

Either except four of these or less, a 0 + a 1 + a 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a n a_0+a_1+a_2.......................................................a_n all equal to 0 0

P ( 5 ) = 152 P(5)=152

Hence the only possible value of constant term is 2.

maximum three terms are possible and all other are 0.

Let 5 x + 5 y = 152 2 5^x + 5^y=152-2 where x and y are natural no.s

The only possible pair is 2,3.

Hence the P ( x ) = x 3 + x 2 + 2 P(x)=x^3+x^2+2

P ( 6 ) = 254 P(6)=254

William Isoroku
Nov 14, 2016

The coefficients are non-negative: thus positive or 0. Use the method of undetermined coefficients.

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