Let P ( x ) = a 0 + a 1 x + ⋯ + a n x n , where a 0 , … , a n are non-negative integers.
Given that P ( 1 ) = 4 and P ( 5 ) = 1 5 2 , find P ( 6 ) .
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nice problem
When x = 1
a 0 + a 1 + a 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . + a n = 4
Either except four of these or less, a 0 + a 1 + a 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a n all equal to 0
P ( 5 ) = 1 5 2
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 = 1 5 2 − 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 ( 6 ) = 2 5 4
The coefficients are non-negative: thus positive or 0. Use the method of undetermined coefficients.
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We have a 0 + a 1 + … + a n = 4 , implying max a 0 , a 1 , … , a n < 5 .
P ( 5 ) = a 0 + 5 ⋅ a 1 + 5 2 ⋅ a 2 + … + 5 n ⋅ a n
thus the unique base 5 representation of 1 5 2 . Since 1 5 2 = 1 1 0 2 5 ,
P ( x ) = x 3 + x 2 + 2 .
P ( 6 ) = 2 1 6 + 3 6 + 2 = 2 5 4 .