# Prove n is divisible by 2^m

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• Mar 23rd 2008, 05:51 PM
mandy123
Prove n is divisible by 2^m
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
• Mar 23rd 2008, 11:13 PM
CaptainBlack
Quote:

Originally Posted by mandy123
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 $n$ into a number $k_1$ made from its last $m$ digits and the number
made by its remaining digits:

$n=k_2 10^{m} +k_1$

Now we are told that $2^m|k_1$, so to complete this problem therefore
it is sufficient to show that $2^m|k_210^m$

RonL
• Mar 24th 2008, 03:54 AM
topsquark
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