Question

3) Based on the CIDR addressing scheme, What is the subnet mask for 125.124.0.0/21?

Answer 1

CIDR:   125.124.0.0/21
/21 indicates that number of 1's in subnet mask is 21
so, Subnet mask in binary is 11111111.11111111.11111000.00000000

11111111.11111111.11111000.00000000
Let's convert all octets to decimal separately
Converting 11111111 to decimal
=> 11111111
=> 1x2^7+1x2^6+1x2^5+1x2^4+1x2^3+1x2^2+1x2^1+1x2^0
=> 1x128+1x64+1x32+1x16+1x8+1x4+1x2+1x1
=> 128+64+32+16+8+4+2+1
=> 255
11111111 in decimal is 255

Converting 11111111 to decimal
11111111 in decimal is 255  (we've already calculated this above)

Converting 11111000 to decimal
=> 11111000
=> 1x2^7+1x2^6+1x2^5+1x2^4+1x2^3+0x2^2+0x2^1+0x2^0
=> 1x128+1x64+1x32+1x16+1x8+0x4+0x2+0x1
=> 128+64+32+16+8+0+0+0
=> 248
11111000 in decimal is 248

Converting 00000000 to decimal
=> 00000000
=> 0x2^7+0x2^6+0x2^5+0x2^4+0x2^3+0x2^2+0x2^1+0x2^0
=> 0x128+0x64+0x32+0x16+0x8+0x4+0x2+0x1
=> 0+0+0+0+0+0+0+0
=> 0
00000000 in decimal is 0

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||    11111111.11111111.11111000.00000000 in decimal notation is 255.255.248.0    ||
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Answer: 255.255.248.0