About this interactive
A device finds the network it is on by ANDing its IP address with its subnet mask, bit by bit: the result is 1 only where both bits are 1. Because a mask is ones and then zeros, every network bit of the address comes through unchanged and every host bit becomes 0. That result is the network address.
You only need binary in one octet. Where the mask octet is 255 the address octet copies across, and where it is 0 the result is 0. In the octet where the mask is neither, AND the two in binary. Take 10.5.83.20/21: the mask is 255.255.248.0, so the third octet does the work. 83 is 01010011 and 248 is 11111000, so the AND is 01010000, which is 80, and the network address is 10.5.80.0. The prefix and the dotted mask are the same thing written two ways: count the ones in the mask (8 + 8 + 5 = 21) and you have the prefix.
Checking someone else's work is the same AND, done for every row, plus one more check: that the prefix and the dotted mask agree. Flag every wrong row, then submit. Every wrong row is then explained, with the fix.
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