How do you deal with the dark (negative) side of functions?

Let f ( x ) f(x) be a quartic polynomial in the form a x 4 + b x 3 + c x 2 + d x + e ax^4+bx^3+cx^2+dx+e where

f ( 7 ) = 4636 , f ( 6 ) = 2504 , f ( 5 ) = 1212. f(-7) = 4636 , f(-6) = 2504 , f(-5) = 1212.

Given that a , b , c , d a, b,c , d and e e are all positive integers less than 5. Evaluate the addition of the two 3-digit integers, a b c + c d e \overline{abc}+\overline{cde} .


The answer is 646.

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3 solutions

Eamon Gupta
Dec 27, 2015

This was my inspiration: Negative Integer Number Base

We can use negative integers bases of -5 and below since we are told that a , b , c . d . e < 5 a, b, c. d. e<5 and are all positive.

Hence we can take f ( 5 ) = 1212 f(-5)=1212 :

1212 1212 in base 5 \boxed{ -5} can be calculated as:

1212 ÷ 5 = 242 1212 \div\boxed{ -5} = \color{#D61F06}{-242} , remainder 2 \color{#EC7300}{2}

242 ÷ 5 = 49 \color{#D61F06}{-242} \div\boxed{ -5} =\color{#20A900}{ 49} , remainder 3 \color{#EC7300}{3}

49 ÷ 5 = 9 \color{#20A900}{49} \div\boxed{ -5} = \color{#3D99F6}{-9} , remainder 4 \color{#EC7300}{4}

9 ÷ 5 = 2 \color{#3D99F6}{-9} \div\boxed{ -5} = \color{#69047E}{2} , remainder 1 \color{#EC7300}{1}

2 ÷ 5 = 0 \color{#69047E}{2} \div\boxed{ -5} = 0 , remainder 2 \color{#EC7300}{2}

Reading the remainders from bottom to top we get 1212 = 2143 2 5 1212 = 21432_{-5}

2 ( 5 ) 4 + 1 ( 5 ) 3 + 4 ( 5 ) 2 + 3 ( 5 ) 4 + 2 ( 5 ) 0 = 1212 \color{#D61F06}{2} \cdot (-5)^4+\color{#3D99F6}{1} \cdot (-5)^3+\color{#20A900}{4} \cdot (-5)^2+\color{#EC7300}{3} \cdot (-5)^4+\color{#69047E}{2} \cdot (-5)^0 = 1212

Therefore the polynomial is 2 x 4 + 1 x 3 + 4 x 2 + 3 x + 2 \color{#D61F06}{2}x^4+\color{#3D99F6}{1}x^3+\color{#20A900}{4}x^2+\color{#EC7300}{3}x+\color{#69047E}{2}

a b c + c d e = 2 1 4 + 4 3 2 = 646 \overline{abc} + \overline{cde} = \color{#D61F06}{2}\color{#3D99F6}{1}\color{#20A900}{4}+\color{#20A900}{4}\color{#EC7300}{3}\color{#69047E}{2} = \boxed{646}

Moderator note:

Great solution using the idea of negative integer bases.

Mark Hennings
Jan 7, 2016

Lagrange interpolation can be used as well. We must have (the last three terms define the unique quadratic polynomial taking the desired values at 7 , 6 , 5 -7,-6,-5 , and the first term makes f ( x ) f(x) as general a quartic as possible): f ( x ) = ( α x + β ) ( x + 5 ) ( x + 6 ) ( x + 7 ) + 4636 2 ( x + 5 ) ( x + 6 ) 2504 ( x + 5 ) ( x + 7 ) + 1212 2 ( x + 6 ) ( x + 7 ) f(x) \;= \; (\alpha x + \beta)(x+5)(x+6)(x+7) + \tfrac{4636}{2}(x+5)(x+6) - 2504(x+5)(x+7) + \tfrac{1212}{2}(x+6)(x+7) for some values of α \alpha and β \beta . The coefficients of x 4 x^4 , x 3 x^3 and 1 1 in f ( x ) f(x) are α \alpha , 18 α + β 18\alpha+\beta and 7352 + 210 β 7352 + 210\beta respectively. Since each of these coefficients must be integers between 1 1 and 4 4 , inclusive, it follows that α \alpha and β \beta must both be integers, and indeed that α = 2 \alpha=2 and β = 35 \beta=-35 . Thus f ( x ) = 2 x 4 + x 3 + 4 x 2 + 3 x + 2 f(x) \; = \; 2x^4 + x^3 + 4x^2 + 3x + 2 and so the answer is 214 + 432 = 646 214 + 432 = 646 .

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