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Assume that there are about 60 students in your class. If every student is to be assigned a unique bit pattern, a) what is the minimum number of bits required to do this? b) How many more students can be admitted to the class without requiring additional bits for each student's unique bit pattern?

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Answer:

a) Therefore, the minimum number of bits required is 6.

b) Therefore, we can admit 4 more students to the class without requiring additional bit for each student's unique bit pattern

Step-by-step explanation:

a) The number of unique bit patterns using n bits is calculated using
2^(n).

In this case, there are 60 students, so, we need at least 60 unique bit pattern.


2^(n) = 60

Where n is the number of bit required; we are to find n

We take the logarithm of both side:


log_(2) 2^(n) = log_(2) 60\\n = log_(2) 60\\n = 6 (Approximately)

Therefore, the minimum number of bits required is 6

b) How many more students can be admitted to the class without requiring additional bits for each student's unique bit pattern?

With 6 bits, we can represent up to
2^(6) unique bit pattern which is 64 unique bit patterns.

To get the number of additional bit:

64 - 60 = 4

Therefore, we can admit 4 more students to the class without requiring additional bit for each student's unique bit pattern

User Karl Johan
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