# Thread: Prove n is divisible by 2^m

1. ## 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

2. 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
$n=k_2 10^{m} +k_1$
Now we are told that $2^m|k_1$, so to complete this problem therefore
it is sufficient to show that $2^m|k_210^m$