Turn a power of $2$ into a multiple of $7$, maybe adding a $2$ but not a $7$ On St. Patrick's Day, the Lucky Charms leprechaun wagered me a bottle of Glenfiddich I couldn't solve this math problem before midnight:

There is a power of $2$ that can be turned into a multiple of $7$ with a simple rotation of a representation. Furthermore, that same power of $2$ can be turned into that same multiple of $7$ by adding a $+2$ to another representation. And if you really want to use a $7$ to get there, that can also be done, but you need a couple more things besides the $+2$ to accomplish the transformation.

I immediately said $1024$, since $2401 = 7^4$, but he said that's wrong, because that requires a rotation and a swap. "Think easier, lad," he said.
Midnight came and went and I didn't figure it out. But I still want to know the answer. I know it's something so easy I will feel stupid once the answer is revealed to me. The answer doesn't require heavy duty number crunching. I've been looking at the powers of $2$ up to $2^{64}$ but I can't figure it out. This riddle has me stumped.

The first thing that popped into my mind was bitwise rotation, like ROR, ROL, RCR, etc. But obviously a power of 2 in an integer data type rotated in whatever direction whatever number of bits simply gives another power of 2.
Your failed answer of 1024 and 2401 suggested to me that the answer involves a rotation of decimal digits, which after some head-scratching led me to this: $$2^{14} = 16384$$ and $$14^4 = 38416.$$
If we take $2^{14}$ and "flip" it we get $14^2 = 196$, which falls short of 38416. Maybe this is where the "$+2$" comes in: $14^{2 + 2} = 38416$. One way to use a digit 7 in the representation of $2^{14}$ is $(2^7)^2$. More changes are required to turn that into $14^2$, namely $(2 \times 7)^{2 + 2}$.
I don't read Chinese, but this webpage helped me figure things out http://wenwen.sogou.com/z/q469682270.htm
