Simple Simplification

Algebra Level 3

Compute

⌊ 201 6 3 2014 ⋅ 2015 − 201 4 3 2015 ⋅ 2016 ⌋ \left\lfloor \dfrac {2016^3}{2014 \cdot 2015} - \dfrac {2014^3}{2015 \cdot 2016} \right\rfloor


The answer is 8.

Danish Ahmed by Danish Ahmed , 24, India

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1 solution

Danish Ahmed
Feb 26, 2015

Let n = 2015 n=2015 .

( n + 1 ) 3 n ( n − 1 ) − ( n − 1 ) 3 n ( n + 1 ) \dfrac{(n+1)^3}{n(n-1)}-\dfrac{(n-1)^3}{n(n+1)}

= ( n + 1 ) 4 − ( n − 1 ) 4 n ( n − 1 ) ( n + 1 ) =\dfrac{(n+1)^4-(n-1)^4}{n(n-1)(n+1)}

= [ ( n + 1 ) 2 − ( n − 1 ) 2 ] [ ( n + 1 ) 2 + ( n − 1 ) 2 ] n ( n − 1 ) ( n + 1 ) =\dfrac{[(n+1)^2-(n-1)^2][(n+1)^2+(n-1)^2]}{n(n-1)(n+1)}

= ( 4 n ) ( 2 n 2 + 2 ) n ( n − 1 ) ( n + 1 ) =\dfrac{(4n)(2n^2+2)}{n(n-1)(n+1)}

= 8 ⋅ n 2 + 1 n 2 − 1 =8\cdot\dfrac{n^2+1}{n^2-1}

The fraction here is obviously only slightly more than one, so the greatest integer less than or equal to this is 8 \boxed{8}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Nice solution :) I used a similar method

Curtis Clement - 6 years, 3 months ago

But it should be mentioned in the question that the answer is approx. I got my answer as 8.00000034 and kept thinking for 5 minutes if it was correct ir not, because the answer had to be given in an integer . But still got the answer right !!!!!!!!

Parth Bhardwaj - 6 years, 3 months ago

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