I was reading Data Communication book by Forouzan in which there is a one restriction put on CIDR subnetting by Internet authorities:

First IP address in the block should be evenly divisible by the number of addresses. For example, below, the first address, when converted to a decimal number, is 3440387360, which when divided by 16 (block size) results in 215024210. enter image description here

Since this did not clear the practical significance behind this restriction I went on exploring and came across the video where it is explained as follows:

Whenever we divide any number by $k^{th}$ power of 2 (another restriction being that the block size should be $2^k$), the reminder is least significant k bits. So evenly divisible by $2^k$ means that least significant $k$ bits should all be 0. This way the first address of the allocated address block can be used as a block ID since it has all 0s in the host ID part.

Now that seems more logical despite the fact that I dont understand how

evenly divisible by the number of addresses $2^k$

translates to

least significant $k$ bits should all be 0

Can't it be just that the decimal equivalent of least significant $k$ bits should be divisible by 2. I mean if least significant 4 bits (for a block of $2^4$ addresses) is $(1010)_2$, then it translates to $(10)_{10}$ which is divisible by 2.

So in short why is evenly divisible translates to all 0 LSBs?


Since each place in a binary number is a power of two, the binary expansion of $2^k$ for any $k$ will be a 1 followed by $k$ zeroes. Examples:

$$1 = 2^0 = 1$$ $$16 = 2^4 = 1\underbrace{0000}_{4\ zeroes}$$ $$512 = 2^9 = 1\underbrace{000000000}_{9\ zeroes}$$

Multiplying a number that ends in zeroes by another number (integer) produces at least that many trailing zeroes. So any number divisible by a number ending in $k$ zeroes has to end in at least $k$ zeroes.


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