Data Representation and Boolean Logic — Question 22

(a) 1111101101100011

Grouping in bits of 4:

$\underlinesegment{1111} \quad \underlinesegment{1011} \quad \underlinesegment{0110} \quad \underlinesegment{0011}$

Therefore, (1111101101100011)₂ = (FB63)₁₆.

(b) 100000101011100

Grouping in bits of 4:

$\underlinesegment{0100} \quad \underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1100}$

Therefore, (100000101011100)₂ = (415C)₁₆.

(c) 1000111010100011

Grouping in bits of 4:

$\underlinesegment{1000} \quad \underlinesegment{1110} \quad \underlinesegment{1010} \quad \underlinesegment{0011}$

Therefore, (1000111010100011)₂ = (8EA3)₁₆.

(d) 111011111

Grouping in bits of 4:

$\underlinesegment{0001} \quad \underlinesegment{1101} \quad \underlinesegment{1111}$

Therefore, (111011111)₂ = (1DF)₁₆.