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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.

$\dfrac{2017!+2014!}{2016!+2015!}$=$\dfrac{2014!(2015)(2016)(2017)+2014!}{2014!(2015)(2016)+2014!(2015)}$

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

(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

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  • $\begingroup$ How did you arrive at $2016$? Consider adding that to your post. $\endgroup$ – Jordan Glen Apr 6 '15 at 16:11
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    $\begingroup$ 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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$$\frac{2017!+2014!}{2016!+2015!}=\frac{2014!(1+2017\times2016\times2015)}{2014!(2015+2016\times2015)}=\frac{1+2017\times2015\times2016}{2015+2016\times2015}=\frac{1+2017\times2015\times2016}{2015(1+2016)}=\frac{1+2017\times2015\times2016}{2015\times2017}\simeq\frac{2017\times2015\times2016}{2015\times2017}=2016$$

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