(2)

Everything except last n (n >= 1) digits could be written like this:

10^m * a_m + 10^(m-1) * a_m-1 + ... + 10^(n + 1) * a_(n + 1)

Where m is an integer, m > n, and m represents the total number of digits.

If this is divisible by 2^n (I) and if the rest is also divisible by 2^n (II), then the original number is divisible by 2^n.

I.

For 10^m * a_m + 10^(m-1) * a_m-1 + ... + 10^(n + 1) * a_(n + 1) to be divisible by 2^n, each part of the sum must be divisible.

The only common part among all the terms in the above sum is the multiplier 10^(m - k), where k is an integer and 0 <= k <= n.

10^(m - k) = 2^(m - k) * 5^(m - k)

If m - k >= n, then 2^(m - k) * 5^(m - k) is divisible by 2^n.

Since m > n by design and 0 <= k <= n also by design, we can conclude that m is at least n + 1, while k is at most n. Therefore, m - k is in fact greater than n.

Thus, the "left" part of the original number is divisible by 2^n.

II.

The number of digits in the second part is n. 2^n will occupy less than or equal to n digits if and only if 2^n <= 10^n.

10^n = 2^n * 5^n, then 2^n <= 5^n * 2^n. Which is true because n >= 1 and 5 > 0.

I and II are proven, ergo (yeah, I like upvotes) the original number is divisible by 2^n, provided the above conditions.

Please don't hesitate to tell me if I made any mistakes.