For which $n \in \mathbb{Z^+}$ is $(4n+9)/(2n^2+7n+6)$ a terminating decimal? I saw this problem here.
My approach:
Let $A = (4n+9)/(2n^2+7n+6) = \frac{6}{2n + 3} - \frac{1}{n + 2}$
If $\frac{6}{2n + 3}$ and $\frac{1}{n + 2}$ are terminating decimals, then so is $A$.
A number terminates if and only if it can be written as a ratio of the form $\frac{k}{2^m\cdot5^n}$
We know that $2\nmid2n+3$
So $2n+3=3^x\cdot5^y$ where $0\le x\le1$, $1\le y$ and $n+2=(3^x\cdot5^y+1)/2$
So we have $A = \frac{6}{3^x\cdot5^y} - \frac{2}{3^x\cdot5^y+1}$
But I don't even know if this approach makes any sense.

The only prime that can simultaneously divide numerator and denominator is $3$. For if an odd prime $p$ divides $4n+9$ and $2n+3$, it must divide $(4n+9)-2(2n+3)=3$. And it is easy to see that $4n+9$ and $n+2$ are relatively prime.
We first examine the case where $3$ does not simultaneously divide numerator and denominator. Then $2n+3$ must be a power of $5$. What about $n+2$? In principle it could be divisible by a power of $2$ greater than $1$. However, that would make $2n+3$ of the shape $4k+1$, which no power of $5$ can be. 
So $2n+3$ is a power of $5$ and $n+2$ is a power of $2$. This is obviously impossible if $n$ is positive (it only happens if $n=-1$). 
It remains to deal with the case where $3$ divides both $4n+9$ and $2n+3$. Then $3$ must divide $n$. Let $n=3m$. Then our expression becomes
$$\frac{4m+3}{(2m+1)(3m+2)}.$$
It is easy to see that $3$ cannot divide both numerator and denominator. So $2m+1$ is a power of $5$. Since $2m+1$ and $3m+2$ are relatively prime, that forces $3m+2$ to be a power of $2$. There are the obvious solutions $m=0$ (which the wording of the problem does not allow) and $m=2$. That gives the solution $n=4$. 
Are there other solutions? Note that $2(3m+2)-3(2m+1)=1$. So we need to consider solutions of the exponential Diophantine equation $2^x-3\cdot 5^y=1$.  We do not know whether this equation has solutions other than the ones mentioned. 
Added: Thanks to Gerry Myerson for pointing out that the exponential Diophantine equation has no other solutions. 
Added 2: Zafer Cesur has pointed out that the analysis of the equation $2^x-3\cdot 5^y=1$ does not require fancy machinery.  By working modulo $3$, we can see that $x$ is even, say $x=2n$. Then the equation can be rewritten as $(2^n-1)(2^n+1)=3\cdot 5^y$. Because $2^n-1$ and $2^n+1$ are relatively prime, the only possibilities are $2^n-1=1$ and $2^n-1=3$. 
