What is the integer part of the following fraction: $\frac{2012^{2013}+2013^{2014}}{2012^{2012}+2013^{2013}}$ 
What is the integer part of the following fraction: $\dfrac{2012^{2013}+2013^{2014}}{2012^{2012}+2013^{2013}}$.

This is a competition problem for 7th grade students.
The answer to this question is $2012$.
Is there any way to simplify/evaluate it so we can see the integer part clearly? Or can we just estimate?
 A: $2012=\frac{2012^{2012}\times2012+2013^{2013}\times2012}{2012^{2012}+2013^{2013}}<\frac{2012^{2013}+2013^{2014}}{2012^{2012}+2013^{2013}}<\frac{2012^{2012}\times2013+2013^{2013}\times2013}{2012^{2012}+2013^{2013}}=2013$
A: Let $y=2012^{2013}+2013^{2014}$
Now $2012^{2013}+2013^{2014}=2012^{2012}(2012)+2013^{2013}(2013)=2012\Big(2012^{2012}+2013^{2013}\Big)+2013^{2013}$
Let $x=2012^{2012}+2013^{2013}$
Thus $y=2012x+2013^{2013}\Rightarrow \dfrac{y}{x}=2012+\dfrac{2013^{2013}}{x}$
Since $x>2013^{2013}$, $\dfrac{2013^{2013}}{x}<1$
$\therefore $ Integral part of $\dfrac{y}{x}$ is $2012$.
A: Some manipulation of the numerator gives us
$$
2012^{2013}+2013^{2014}=2012\cdot2012^{2012}+(2012+1)2013^{2013}\\
=2012\left(2012^{2012}+2013^{2013}\right)+2013^{2013}
$$
Now it is much easier to see the result of dividing by $2012^{2012}+2013^{2013}$.
A: Observe that for any $a,b > 0,$ we have $\dfrac{a^{a+1}+b^{b+1}}{a^a + b^b} = \dfrac{a(a^a + b^b)+(b-a)b^{b}}{a^a + b^b} = a + (b-a)\dfrac{b^b}{a^a+b^b}.$ Clearly, $b^b < a^a + b^b.$ So in order for $\dfrac{b^b}{a^a+b^b}$ to be a fraction, we need $\dfrac{b^b}{a^a+b^b} < \frac{1}{b-a}.$ Here we have $a = 2012, b = 2013.$ Hence $(b-a)\dfrac{b^b}{a^a+b^b} = \dfrac{b^b}{a^a+b^b} < 1,$ so the integer part is $a = 2012.$
