Math CentralQuandaries & Queries


Question from Ashish, a student:

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

Ashish, we have three responses for you.


I'll show you for n = 2 and you can do the general situation.

Suppose that your integer is k then you can write

k = p * 100 + q * 10 + r

where p is an integer and q and r are digits. For example 25654 = 25600 + 50 + 4 = 25600 + 54.

Since p * 100 is divisible by 22 = 4, k is divisible by 22 if and only if q * 10 + r is divisible by 22 .

I hope this helps,


In our March Problem of the Month this year, you find the following argument:

A number N  can be written K x (2n x 5n ) + (the number formed by the last n digits).



Another example: What you need to think about is the part of the number appearing before the last n digits. Look at 35640 for example.  This means 35 x 1000 + 640 = 35 x 103 + 640.  What do you know about 23 dividing 103 ?


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