The answer is not 11!

Algebra Level 4

Suppose f ( x ) f(x) is polynomial of degree 5 5 with leading coefficient of 2001. With f ( 1 ) = 1 f(1)=1 , f ( 2 ) = 3 f(2)=3 , f ( 3 ) = 5 f(3)=5 , f ( 4 ) = 7 f(4)=7 , f ( 5 ) = 9 f(5)=9 . What is the value of f ( 6 ) f(6) ?


The answer is 240131.

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Sai Ram by Sai Ram , 19, India

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1 solution

Surya Prakash
Aug 14, 2015

Let g ( x ) = f ( x ) − 2 x + 1 g(x) = f(x) - 2x + 1 .

So, 1 1 , 2 2 , 3 3 , 4 4 and 5 5 are the roots of g ( x ) = 0 g(x)=0 .

So, g ( x ) = 2001 ( x − 1 ) ( x − 2 ) ( x − 3 ) ( x − 4 ) ( x − 5 ) g(x) = 2001(x-1)(x-2)(x-3)(x-4)(x-5)

f ( 6 ) = g ( 6 ) + 11 = 2001 × 5 ! + 11 = 240131 f(6) = g(6) +11 = 2001 \times 5! +11 = \boxed{240131}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Simple standard approach via the Remainder-Factor Theorem.

In response to Surya Prakash : How did u get the term (-2x + 1 ) is this a matter of observation or something else by any formula ?? I also think that f(x) should be placed in place of g(x) because (dividend=divisor X quotient + remainder )

Chirayu Bhardwaj - 5 years, 6 months ago

Highly Overrated.

Mehul Arora - 5 years, 10 months ago

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What do you mean by that?

Surya Prakash - 5 years, 10 months ago

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I mean that's the problem was not worth 110 points, the amount of points I solved it for .

Mehul Arora - 5 years, 10 months ago

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