A subnet containing 100 total addresses cannot have exactly 100 addresses using the power-of-two subnet sizes described in the material. The smallest suitable block contains 128 addresses and is written in Classless Inter-Domain Routing (CIDR) notation as ____.

### Why this is right The listed subnet sizes double: 64, 128, 256, and so on. A 64-address block is too small for a requirement of 100 total addresses, so the next available size is 128. The material maps 128 addresses to `/25`. ### The concept A 32-bit IPv4 address with a `/25` prefix has 25 network bits and 7 variable bits. Seven variable bits allow $2^7=128$ total addresses. This question asks for a block that contains at least 100 total addresses, not 100 usable host addresses. In a conventional subnet, the network and broadcast addresses are reserved, so the number available to ordinary hosts is lower than the total block size. ### Worked numbers - **Given**: Required capacity, 100 total addresses. - **Find**: Smallest supported subnet size and its prefix. - **Formula**: Address count is a power of two; for prefix $p$, count is $2^{32-p}$. - **Solve**: 64 is less than 100; the next size is 128. Since $2^7=128$, there are 7 variable bits and $32-7=25$ network bits. - **Answer**: `/25`, containing 128 total addresses. - **Check**: 128 is at least 100, and the previous supported size, 64, is not. ### Terms - **CIDR**: Classless Inter-Domain Routing. Slash notation that expresses the network prefix length. - **Prefix length**: The count of leading bits that identify the network. - **Variable bits**: Bits available to represent addresses within a subnet. ### Remember Choose the next power-of-two block that is at least as large as the required total address count.