Can 720! be written as the difference of two positive integer powers of 3? Does the equation:
$$3^x-3^y=720!$$
have any positive integer solution?
 A: As alex.jordan writes, $3^x-3^y$ factors as $3^y(3^{x-y}-1)$, so $y$ must be the number of factors of $3$ in $720!$.
I don't actually need to count the number of $3$s in $720!$, so let's just define the notation $720!_3$ for "$720!$ with all of the powers of $3$ divided out". This must yield the other factor $3^{x-y}-1$, so we need to investigate whether $720!_3$ is one less than a power of $3$. To do this we will compute it modulo $3$.
In general we have that
$$ (3k)!_3 \equiv (1\cdot 2)^k \cdot k!_3 \equiv (-1)^k k!_3 \pmod 3 $$
And therefore
$$ \begin{align} 720!_3
\equiv 240!_3 
&\equiv 80!_3 \equiv 80\cdot 79 \cdot 78!_3 \equiv - 78!_3 \\
&\equiv -26!_3 \equiv -26\cdot 25 \cdot 24!_3 \equiv 24!_3 \\
&\equiv 8!_3 \equiv 8\cdot 7\cdot 6!_3 \equiv -6!_3 \\
&\equiv -2!_3 \equiv -2 \equiv 1 &\pmod 3
\end{align}$$
which is not one less than a multiple of 3, so certainly not one less than a power of $3$.
So the answer is no.
A: Let's write $n=x-y$.  If $720!=3^x-3^y=3^y(3^n-1)$, then $3^n-1$ is divisible by the primes $17$, $31$, $43$, and $79$ (among others, of course).  As it happens, $3$ is a primitive root for those primes.  This means that $n$ is divisible by $16$, $30$, $42$, and $78$, hence by the lcm of these, or
$$16\cdot15\cdot7\cdot39=65{,}520$$
But this is vastly larger than $\log_3(720!)\approx3660$.  Thus $720!$ is not the difference of two powers of $3$.
A: If $$720!=3^x-3^y=3^y\left(3^{x-y}-1\right)$$ and since the power of $3$ dividing $720!$ is $$\left\lfloor\frac{720}{3}\right\rfloor+\left\lfloor\frac{720}{9}\right\rfloor+\left\lfloor\frac{720}{27}\right\rfloor+\left\lfloor\frac{720}{81}\right\rfloor+\left\lfloor\frac{720}{243}\right\rfloor=240+80+26+8+2=356\text{,}$$ it would have to be that $y=356$. 
So it remains to see if $3^x-3^{356}=720!$ has an integer solution in $x$. 

Side note: we can get a good approximation to $\log_3(720!)$ using Stirling's formula with one more term than is typically used: $$\log_3(720!)\approx\frac{1}{\ln3}\left(720\ln(720)-720 +\frac{1}{2}\ln(2\pi\cdot720)\right)\approx3660.3\ldots$$ Since the terms in Stirling's formula are alternating after this, we can deduce that this is correct to the tenths place. Values of $\ln(3)$, $\ln(720)$, and $\ln(2\pi)$ are easy to calculate by hand to decent precision if needed. 

$720!$ is a lot bigger than $3^{356}$. Since  $\log_3(720!)\approx3660.3$, in base $3$, $720!$ has $3661$ digits (trigits?), where as $3^{356}$ just has $357$. So $\log_3(720!+3^{356})$ and $\log_3(720!)$ must be very close together. Since the latter is $\approx3660.3$ though, it's not possible for the former to be an integer.
More formally, $$\log_3(720!)<\log_3(720!+3^{356})=x=\log_3(720!)+\log_3\mathopen{}\left(1+\frac{3^{356}}{720!}\right)\mathclose{}<\log_3(720!)+\frac{1}{\ln(3)}\frac{3^{356}}{720!}$$
$$3660.3\ldots<x<3660.3\ldots$$
and there is no integer $x$ between the values on the two ends.
