Suppose you have 3 Nbit unsigned integers a b and c What is

Suppose you have 3 N-bit unsigned integers: a, b, and c. What is the minimum number of bits required to present: a*b+c ?

Solution

Minimum number of bits required to present: a*b+c is (2*n + 1)

if two n bit numbers are multiplied then the result will 2*n bit.

and if two n bit numbers are added then the result will n+1 if there is a carry or else it will be n.

Therefore ,

a*b = 2*n bit

(a*b)+c = 2*n bit +n bit = (2*n + 1) (in case of most significant bit carry)

or (a*b)+c = 2*n bit +n bit = (2*n) (in case of no carry)