When is $\left\lfloor \frac {7^n}{2^n} \right\rfloor \bmod {2^n} \ne 0\;$? Is
$$\left\lfloor \frac {7^n}{2^n} \right\rfloor \bmod{2^n} \ne 0\;$$
always true when $n \ge 3$.
Baker's theorem on transcendental numbers that provide bounds for diophantine equations may be useful, but I will leave that to the experts.
 A: One approach would be $7^n = (8 - 1)^n = (2^3 - 1)^n$. Doing binomial expansion, we get $7^n/2^n = 2^n x + k$ (k is sum of all binomial terms $i < n /3$). Little improved because it leaves less terms than above, but essentially same methodology. 
A: The following may be a starting point:
Write $7^n \mod 4^n$: 
$$ 7^n =a\cdot 4^n + b \quad b<4^n $$
$$ \left\lfloor \frac{7^n}{2^n} \right\rfloor \equiv \left\lfloor \frac b{2^n} \right\rfloor \pmod{2^n}$$
and this is $0$ iff $b<2^n$.
Well, $7^n=(2^3-1)^n=\displaystyle\sum_{0\le i\le n} \binom{n}{i} 2^{3i} \cdot (-1)^{n-i} $, and those with indices $3i\ge 2n$ vanish, thus
$$7^n \equiv (-1)^n\sum_{ i < 2n/3 } \binom ni (-8)^{i} = ... \pmod{4^n}$$
A: A little re-arrangement of the floored division in the counter-example yielded:
$7^{n}\ mod\ 2^{n} = 7^{n} + k\cdot 4^{n}\ ;\ k < 0$ or:
$7^{n}\ mod\ 2^{n} \equiv 7^{n}\left (mod\ 4^{n}\right )$
The first form: $(7^{n}\ mod\ 2^{n}) - 7^{n} = k\cdot 4^{n}$, requires the lhs to be divisible by $4^{n}$, or have $(2n)$ trailing zeroes in binary. A search from $n = 3$ to $n = 2^{17}$ failed to find a candidate. I'm sure there's more to be done with the first form...

So: $7^{n}$ would require bits: [n, 2n-1] to be zero.
