Convert from base 5 to base 10 432\(_{five}\) = (4 x 5\(^2\)) + (3 x 5\(^1\)) + (2 x 5\(^0\)) = (4 x 25) + (3 x 5) + (2 x 1) = 100 + 15 + 2 = 117\(_{ten}\) Then convert from base 10 to base 3 3 117 3 39 r 0 3 13 r 0 3 4 r 1 3 1 r 1 0 r 1 Selecting the remainders from bottom to top: 117\(_{ten}\) = 11100\(_{three}\) Hence; 432\(_{five}\) = 11100\(_{three}\)