Suppose that $m \ge 0$ show that $49 \mid 5\cdot3^{4m + 2} + 53\cdot2^{5m}$ I've re-written the equation in a few different ways hoping for a few different approaches:
$$49y = 5 \cdot 3^{4m + 2} + 53 \cdot 2^{5m} $$
I think the first equation has more potential, since it looks like I should be able to just solve for $m$, or get to a point where I can say that any positive $m$ will work, but there doesn't seem to be an easy way to proceed to solve for $m$.
or 
$5\cdot3^{4m + 2} + 53 \cdot 2^{5m} \equiv 0 \mod 49$
I'm not fabulous at congruence equations, but I was thinking I could reduce this down to something more manageable to solve the problem, but none of my congruence equation tricks really help.
Anyways I'm at a loss as to where to proceed, if you all have any tips on ways to proceed, that'd be great :D

Observe that 
\begin{align*}
2^{5} & \equiv 32 \pmod{49}\\
2^{5m} & \equiv 32^m \pmod{49}\\
53 \cdot 2^{5m} & \equiv 53 \cdot 32^m  \equiv 4 \cdot 32^m \pmod{49}
\end{align*}
Likewise 
\begin{align*}
3^{4} & \equiv 81 \equiv 32 \pmod{49}\\
3^{4m} & \equiv 32^m \pmod{49}\\
3^{4m+2} & \equiv 32^m \cdot 9 \pmod{49}\\
5 \cdot 3^{4m+2} & \equiv 32^m \cdot 45  \equiv 32^m \cdot (-4) \pmod{49}\\
\end{align*}
Now add the last two congruences of both systems to get
 $$53 \cdot 2^{5m}  +  5 \cdot 3^{4m+2} \equiv 4 \cdot 32^m + 32^m \cdot (-4) \equiv 0\pmod{49}$$
