Finding a divisibility test for $2^j$ for all j greater than or equal to 3. I attempted doing this case by case for $2,3,4,..,9$ however if anyone has structured way of solving it, it would be very helpful. 
Thanks
 A: For very large numbers or small $j$ just go with the usual, a number is divisible by $2^j$ if the last $j$ digits are divisible by $2^j$
A: *

*The rule of congruence is blatantly needed here:


Checkability of $2^1$:
first digit:  $\ \ \ \ \ 2, 4, 6, 8 ,0$
second digit: $ 0, 0, 0, 0, 1$
the process is cyclic and first digit must be even

Checkability of $2^2$:
first digit:$\ \ \ \ \   4, 8, 2, 6 ,0 $
second digit: $ 0, 0, 1, 1, 2$
the process is cyclic and first digit in addition that it must be even, second digit is related to divisibility of first digit by 4.

Checkability of $2^3$:
first digit:$\ \ \ \ \   8, 6, 4, 2 ,0 $
second digit: $ 0, 1, 2, 3, 4$
the process is cyclic and first digit in addition that it must be even, second digit $Sec$ is related to first digit $Fir$, as $Fir=|(2*Sec)\mod 5-8|/2$.
...
I think it isnt transcendental as we move on ...
A: The standard method taught in school is 

A number is divisible by $2^n$ if the last $n$ digits are.

Here is another good one:

If  the first digit is even, examine the number formed by the last $n-1$ digits.
  If the first digit is odd, examine the number formed by the last $n-1$ digits plus $2^{n-1}$.

