4. Convert each of the following numbers to their respective number systems equivale \( \begin{array}{ll}\text { a) } 12 \mathrm{DF}^{2} E_{10} \text { to binury, } & \text { (3 Marks } \\ \text { b) } 11110000_{2} \text { to octal, } & \text { ( } 2 \text { Mark } \\ \text { c) } 65,0 \text { to hexadecimal. } & \text { ( } 2 \text { Mark } \\ \text { d) } 673 \text { to decimal. } & \text { (2 Mark }\end{array} \)

Let's convert each of the given numbers to their respective number systems. ### a) Convert \( 12 \mathrm{DF}^{2} E_{10} \) to binary First, we need to convert \( 12 \mathrm{DF}^{2} E_{10} \) from hexadecimal (base 16) to decimal (base 10), and then from decimal to binary (base 2). 1. **Hexadecimal to Decimal:** - \( 12 \mathrm{DF}^{2} E_{10} \) can be broken down as follows: - \( 1 \times 16^5 + 2 \times 16^4 + D \times 16^3 + F \times 16^2 + 2 \times 16^1 + E \times 16^0 \) - Where \( D = 13 \), \( F = 15 \), and \( E = 14 \). - Calculating each term: - \( 1 \times 16^5 = 1 \times 1048576 = 1048576 \) - \( 2 \times 16^4 = 2 \times 65536 = 131072 \) - \( 13 \times 16^3 = 13 \times 4096 = 53248 \) - \( 15 \times 16^2 = 15 \times 256 = 3840 \) - \( 2 \times 16^1 = 2 \times 16 = 32 \) - \( 14 \times 16^0 = 14 \times 1 = 14 \) - Adding these values together: \[ 1048576 + 131072 + 53248 + 3840 + 32 + 14 = 1186202 \] 2. **Decimal to Binary:** - Now we convert \( 1186202 \) to binary. - Dividing by 2 and recording the remainders: - \( 1186202 \div 2 = 593101 \) remainder \( 0 \) - \( 593101 \div 2 = 296550 \) remainder \( 1 \) - \( 296550 \div 2 = 148275 \) remainder \( 0 \) - \( 148275 \div 2 = 74137 \) remainder \( 1 \) - \( 74137 \div 2 = 37068 \) remainder \( 1 \) - \( 37068 \div 2 = 18534 \) remainder \( 0 \) - \( 18534 \div 2 = 9267 \) remainder \( 0 \) - \( 9267 \div 2 = 4633 \) remainder \( 1 \) - \( 4633 \div 2 = 2316 \) remainder \( 1 \) - \( 2316 \div 2 = 1158 \) remainder \( 0 \) - \( 1158 \div 2 = 579 \) remainder \( 0 \) - \( 579 \div 2 = 289 \) remainder \( 1 \) - \( 289 \div 2 = 144 \) remainder \( 1 \) - \( 144 \div 2 = 72 \) remainder \( 0 \) - \( 72 \div 2 = 36 \) remainder \( 0 \) - \( 36 \div 2 = 18 \) remainder \( 0 \) - \( 18 \div 2 = 9 \) remainder \( 0 \) - \( 9 \div 2 = 4 \) remainder \( 1 \) - \( 4 \div 2 = 2 \) remainder \( 0 \) - \( 2 \div 2 = 1 \) remainder \( 0 \) - \( 1 \div 2 = 0 \) remainder \( 1 \) - Reading the remainders from bottom to top, we get: \[ 100100011010110011010 \] So, \( 12 \mathrm{DF}^{2} E_{10} \) in binary is \( 100100011010110011010_2 \). ### b) Convert \( 11110000_{2} \) to octal To convert from binary to octal, we group the binary digits into sets of three, starting from the right: - \( 11110000 \) can be grouped as \( 11\ 110\ 000 \