An IP datagram of size 1000 bytes arrives at a router. The router has to forward this packet on a link whose MTU (maximum transmission unit) is 100 bytes. Assume that the size of the IP header is 20 bytes. The number of fragments that the IP datagram will be divided into for transmission is ____________.

1. Total payload to be fragmented: The original IP datagram size is $1000\text{ bytes}$ with an IP header of $20\text{ bytes}$.

   $$\text{Data payload} = 1000 - 20 = 980\text{ bytes}$$

2. Maximum data payload per fragment:

   $$\text{Link MTU} = 100\text{ bytes}$$

   Each fragment requires its own $20\text{ byte}$ IP header, leaving:

   $$100 - 20 = 80\text{ bytes}$$

   Since the data payload in IP fragments (except possibly the last one) must be a multiple of $8\text{ bytes}$, and $80$ is divisible by $8$, the maximum payload per fragment is $80\text{ bytes}$.

3. Number of fragments:

   $$\text{Number of fragments} = \left\lceil \frac{980}{80} \right\rceil = \lceil 12.25 \rceil = 13$$

Thus, the IP datagram will be divided into 13 fragments (12 fragments carrying 80 bytes of data each, and 1 fragment carrying the remaining 20 bytes of data).