$t=\frac{30^{65}-29^{65}}{30^{64}-29^{64}}$, find the closest pair of integers, a and b, such that, $a \lt t \lt b$. $t=\frac{30^{65}-29^{65}}{30^{64}-29^{64}}$
find the closest pair of integers, a and b, such that, $a \lt t \lt b$.
$30=1+29$
$(1+29)^{65}=(1+29)(1+29)^{64}$

I actually found this easier to do without algebra. The numerator looks like $30*$ denominator + error term, so you can write 
$$\frac{30^{65}-29^{65}}{30^{64}-29^{64}} = \frac{30^{65} - 30*29^{64} + 30*29^{64} - 29^{65}}{30^{64} - 29^{64}}$$
$$= 30 + \frac{30*29^{64} - 29^{65}}{30^{64} - 29^{64}}$$
$$= 30 + \frac{30*29^{64} - 29*29^{64}} {30^{64} - 29^{64}}$$
$$= 30 + \frac{29^{64}}{30^{64} - 29^{64}}$$
The idea is that the denominator here should be much larger than the numerator, so the overall fraction is between $30$ and $31$. There are a few ways to show this. One is to expand $30^{64} = (29 + 1)^{64}$ via the binomial theorem, obtaining
$$30^{64} = 29^{64} + 64*29^{63} + ...$$
Thus the denominator $30^{64} - 29^{64}$ is greater than $64*29^{63}$, which in turn is greater than the numerator $29^{64}$ since $64 > 29$. Thus indeed the second term is less than $1$, so that the overall fraction is between $30$ and $31$.
