(a) This can either be done using the long division method (easier) or convert everything to base ten and divide, then re-convert to base two. I'll be using the latter method. \(11111111_{two} = 1 \times 2^{7} + 1 \times 2^{6} + 1 \times 2^{5} + 1 \times 2^{4} + 1 \times 2^{3} + 1 \times 2^{2} + 1 \times 2^{1} + 1 \times 2^{0}\) = \(128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255_{ten}\) \(101_{two} = 1 \times 2^{2} + 0 \times 2^{1} + 1 \times 2^{0} \) = \(4 + 0 + 1 = 5_{ten}\) \(\frac{255}{5} = 51_{ten}\) 2 51 2 25 r 1 2 12 r 1 2 6 r 0 2 3 r 0 2 1 r 1 0 r 1 = \(110011_{two}\) (b) (i) Length of arc AXB = \(\frac{105}{360} \times 2 \times \frac{22}{7} \times 6cm\) = \(11cm\) \(\therefore Perimeter = 6cm + 11cm + 6cm = 23cm\) (ii) Area of sector = \(\frac{105}{360} \times \frac{22}{7} \times 6cm \times 6cm\) = \(33 cm^{2}\)