A Conversation for Divisibility

Divison by 4,8,16,32, etc.

Post 1

Apollyon - Grammar Fascist

Based on this entry, it would appear that the test for divisibility by any power of 2 can be generalised to:

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

Now, if only I knew how to prove that...

Divison by 4,8,16,32, etc.

Post 2

Icy North

Well, I think it's all in the entry. The number with the last n digits subtracted (i.e. replace them with zeroes) is divisible by 10^n, therefore it is divisible by both 5^n and 2^n. Therefore if the last n digits are also divisible by 2^n, then the whole number must be too.

Does this make sense?

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