Compute $\dfrac{2017!+2014!}{2016!+2015!}$ to the nearest integer.

My solution to the problem is 2016. Just wanted to check if it's correct.



(We can ignore the 1, since it's value, when divided by the denominator, will be negligible)

Thus we have $\dfrac{(2015)(2016)(2017)}{(2015)(2016)+2015}$=$\dfrac{(2016)(2017)}{(2016)+1}$=$\dfrac{(2016)(2017)}{2017}$=2016

  • $\begingroup$ How did you arrive at $2016$? Consider adding that to your post. $\endgroup$ – Jordan Glen Apr 6 '15 at 16:11
  • 3
    $\begingroup$ To verify your result, just enter the expression into wolframalpha.com. $\endgroup$ – Martin R Apr 6 '15 at 16:12
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    $\begingroup$ Since $2015\cdot 2016+2015=2015(2016+1)=2015\cdot 2017$, you can have $2016+\frac{1}{2015\cdot 2017}$. $\endgroup$ – mathlove Apr 6 '15 at 16:30



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