Prove that if n is a positive integer such that the integer which is made up from the last m digits of n in its decimal representation is divisible by 2^m then n is divisible by 2^m
Decompose $\displaystyle n$ into a number $\displaystyle k_1$ made from its last $\displaystyle m$ digits and the number
made by its remaining digits:
$\displaystyle n=k_2 10^{m} +k_1$
Now we are told that $\displaystyle 2^m|k_1$, so to complete this problem therefore
it is sufficient to show that $\displaystyle 2^m|k_210^m$
RonL
I know what the problem is asking and I don't dispute the result, but I don't like how its phrased. For example, 4|100, but "00" is divisible by anything. It probably should be mentioned as a special case for the problem to be stated correctly. (Yes, I'm in a picky mood this morning.)
-Dan