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NIMCET 2008

The addition of 4 bit, 2’s Given, √(224)ᵣ = (13)ᵣ, the value of radix r iscomplement binary numbers 1101 and 0100 results in

Solution

(13)ᵣ = r + 3

So,
√(224)ᵣ = r + 3

(224)ᵣ = 2r² + 2r + 4

(r + 3)² = 2r² + 2r + 4
r² + 6r + 9 = 2r² + 2r + 4
r² − 4r − 5 = 0

r = 5 (valid radix)
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