CS501 Advance Computer Architecture assignment no 5 fall January 2012

[Malaika khan]

Solution 1:

492/16 =30 ( rem=12), x0=12
30/16= 1 (rem=14), x1=14, x2=1
Thus 49210 = 1EC16


Solution 2:

X=0
X= x +D (=13) =13
X=16*13+4= 212

Hence D416=21210