But 2016 isn't here yet!

Algebra Level 3

Evaluate the expression 201 6 4 + 201 6 2 + 1 201 6 3 + 1 \large {\frac{2016^{4}+2016^{2}+1}{2016^{3}+1}} .

If the answer can be expressed as a mixed fraction in simplest form a b c a \frac{b}{c} , find a + b + c . a+b+c.


Note: Solve this problem without using a calculator.


The answer is 4034.

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1 solution

Kay Xspre
Nov 16, 2015

Let x = 2016 x = 2016 . We can rewrote this expression into x 4 + x 2 + 1 x 3 + 1 \frac{x^4+x^2+1}{x^3+1} . This equals to x 4 + x 2 + 1 x 3 + 1 = x 4 + x 2 + x 2 + 1 x 2 x 3 + 1 = ( x 4 + 2 x 2 + 1 ) x 2 x 3 + 1 = ( x 2 + 1 ) 2 x 2 x 3 + 1 \frac{x^4+x^2+1}{x^3+1} = \frac{x^4+x^2+x^2+1-x^2}{x^3+1} = \frac{(x^4+2x^2+1)-x^2}{x^3+1} = \frac{(x^2+1)^2-x^2}{x^3+1} Which may be simplified to ( x 2 + x + 1 ) ( x 2 x + 1 ) ( x + 1 ) ( x 2 x + 1 ) = x 2 + x + 1 x + 1 = x + 1 x + 1 \frac{(x^2+x+1)(x^2-x+1)}{(x+1)(x^2-x+1)} = \frac{x^2+x+1}{x+1} = x+\frac{1}{x+1} Therefore a + b + c = x + 1 + ( x + 1 ) = 2 x + 2 = 2 ( 2016 ) + 2 = 4034 a+b+c = x+1+(x+1) = 2x+2 = 2(2016)+2 = 4034

Correct! That's how I did it. Just being picky, you could show how you manipulated the numerator in order to get a difference of two squares by adding and subtracting x 2 x^{2} .

Hobart Pao - 5 years, 7 months ago

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I do not want the solution to be lengthy than necessary so I decided not to wrote that part, but for the sake of completeness, I edit the solution. Please see whether it suits you.

Kay Xspre - 5 years, 7 months ago

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Looks great now! Thanks!

Hobart Pao - 5 years, 6 months ago

I am in class 8th I can't understand properly.Can you try another method to explain.

Ritik Devnani - 3 years, 11 months ago

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what don't you understand

Hobart Pao - 3 years, 11 months ago

This = ( x^4 + 2x^2 + 1 ) -x^2 / x^3 + 1 = ( x^2 + 1 )^2 - x^2 / x^3 + 1

Ritik Devnani - 3 years, 11 months ago

Brilliant!!!

Praveena Nookala - 1 year, 11 months ago

Brilliant!

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