Answer:
Check the explanation
Explanation:
223.1.17/24 indicates that out of 32-bits of IP address 24 bits have been assigned as subnet part and 8 bits for host id.
The binary representation of 223.1.17 is 11011111 00000001 00010001 00000000
Given that, subnet 1 has 63 interfaces. To represent 63 interfaces, we need 6 bits (64 = 26)
So its addresses can be from 223.1.17.0/26 to 223.1.17.62/26
Subnet 2 has 95 interfaces. 95 interfaces can be accommodated using 7 bits up to 127 host addresses can represented using 7 bits (127 = 27)
and hence, the addresses may be from 223.1.17.63/25 to 223.1.17.157/25
Subnet 3 has 16 interfaces. 4 bits are needed for 16 interfaces (16 = 24)
So the network addresses may range from 223.1.17.158/28 to 223.1.17.173/28
Answer:
k Nishant
Explanation:
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It's Insert. Insert > Page > Insert Page
Let P(n) be "a postage of n cents can be formed using 5-cent and 17-cent stamps if n is greater than 63".Basis step: P(64) is true since 64 cents postage can be formed with one 5-cent and one 17-cent stamp.Inductive step: Assume that P(n) is true, that is, postage of n cents can be formed using 5-cent and 17-cent stamps. We will show how to form postage of n + 1 cents. By the inductive hypothesis postage of n cents can be formed using 5-cent and 17-cent stamps. If this included a 17-cent stamp, replace this 17-cent stamp with two 5-cent stamps to obtain n + 1 cents postage. Otherwise, only 5-cent stamps were used and n 65. Hence there are at least three 5-cent stamps forming n cents. Remove three of these 5-cent stamps and replace them with two 17-cent stamps to obtain n + 1 cents postage.Hence P(n + 1) is true.
